106 m/s in a uniform 1.9 x 105 N/C electric field. The field accelerates the Problem 6: An electron has an initial velocity of 5.25 electron in the direction opposite to its initial velocity. Part (a) What is the direction of the electric field? MultipleChoice 1) The field is in the direction of the electron's initial velocity 2) The field is in the direction to the right of the clectron's initial velocity 3) The ficld is in the opposite dircction of the elcctron's initial velocity 4) The field is in another direction not listed here Part (b) How far does the electron travel before coming to rest in m? Numeric : A numeric value is expected and not an expression Part (c) How long does it take the clectron to come to rest in s? Numeric A numeric value is expected and not an expression. Part (d) What is the magnitude of the electron's velocity (in m/s) when it returns to its starting point in the opposite direction of its initial velocity? Numeric : A numeric value is expected and not an expression

Answer:

The change in internal energy is zero.

(a) is correct option

Explanation:

Given that,

Work done W = -400 J

The First law of thermodynamics is law conservation of energy

Law of conservation energy is defined the total amount of energy is constant in isolated system.

The energy can not be created and destroyed but the energy can be changed from one form to other form.

The First law of thermodynamics is given by

[tex]\Delta U=Q+W[/tex]

Where,  

W = work done  by the system or on the system

[tex]\Delta U[/tex] = Total internal energy

Q = heat

In isothermal process, the temperature is constant

The energy can not be change and the internal energy is the function of temperature.

So, The internal energy will be zero .

Hence, The change in internal energy is zero.

Answer: a. 1.203 m/s   b.0.35m

Explanation:

The velocity [tex]V[/tex] of a wave is given by the following equation:

[tex]V=\frac{\lambda}{T}[/tex]  (1)

Where:

[tex]\lambda=6.50m[/tex] is the wavelength (the horizontal space between crest and crest)

[tex]T=2.70s.2=5.4s[/tex] is the period of the wave (If we were told it takes a time of 2.70 s for the boat to travel from its highest point to its lowest, the period is twice this amount, which is the time it takes to travel one wavelength)

Substituting the known values:

[tex]V=\frac{6.50m}{5.4s}=1.203m/s[/tex]  (2)

[tex]V=1.203m/s[/tex]  Velocity of the wave

The amplitude [tex]A[/tex] of a wave is defined is given by the following equation:

[tex]A=\frac{maximumpoint-minimumpoint}{2}[/tex]  (3)

If we know the total distance between the highest point to the lowest point is 0.7 m. This means:

[tex]maximumpoint-minimumpoint=0.7m[/tex]

Substituting this value in (3):

[tex]A=\frac{0.7m}{2}[/tex]  (4)

[tex]A=0.35m[/tex]  This is the amplitude of the wave

The waves for the boat of fisherman is travelling with the speed of 1.203 m/s and the amplitude of each wave is 0.35 m.

Speed of wave is the rate of speed by which the wave travel the distance in the time taken by it.

The distance traveled by the boat  from its highest point to the lowest point is 0.700 m. The time taken by the boat is 2.70.

The fisherman sees that the wave crests are spaced a horizontal distance of 6.50 m apart. As, the horizontal distance of crests is the wavelength of the wave. Thus, the wavelength of the wave is,

[tex]\lambda=6.50\rm s[/tex]

The period of the wave is twice the time taken from highest to lowest point. Thus, the time period of the wave is,

[tex]T=2\times2.7\\T=5.4\rm s[/tex]

The speed of the wave is the ratio of wavelength to the time period. Thus, the speed of the waves travelling is,

[tex]v=\dfrac{6.50}{5.4}\\v=1.203\rm m/s[/tex]

The amplitude of wave is the half of the difference of highest point to the lowest point of the wave.

Now, the distance traveled by the boat  from its highest point to the lowest point is 0.700 m. Thus, the amplitude of each wave is,

[tex]A=\dfrac{0.7}{2}A=0.35\rm m[/tex]

Thus, the waves for the boat of fisherman is travelling with the speed of 1.203 m/s and the amplitude of each wave is 0.35 m.

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Speed is defined as the ratio of the distance and time .The squirrel’s average speed will be 20.43 m/sec.

Power is defined as the rate of work done.Its unit is watt or joule per second.It is the ratio of work to the time.

The give data in the problem is;

P is the power of squirrel = 100 W

m is the mass of squirrel = 0.5 kg

h is the height climbed = 75 m

u is the potential energy = mgh = 0.5 x 9.81 x 75 = 367.875 J

Power is defined as the rate of work done.

[tex]\rm p = \frac{W}{t} \\\\ \rm t= \frac{W}{P}[/tex]

[tex]\rm t=\frac{368.75}{100} \\\\ \rm t=3.67875 \ sec[/tex]

Speed is given as the distance and time;

[tex]\rm v = \frac{x}{v} \\\\ \rm v = \frac{75}{3.67} \\\\ \rm v = 20.43 m/sec[/tex]

Hence the squirrel’s average speed will be 20.43 m/sec.

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Final answer:

The squirrel must have an average speed of approximately 20.41 m/s to match the power output of a 100 W lightbulb while climbing a 75.0 m tall vertical maple tree.

Explanation:

To calculate the average speed a 0.5 kg squirrel must have to match the power output of a 100 W lightbulb while climbing a 75.0 m tall tree, we can use the formula for power (P), which is the rate at which work is done. The formula for power in terms of work (W) and time (t) is P = W/t. Work done by the squirrel is equal to the gravitational potential energy (mgh), where m is mass, g is acceleration due to gravity (9.8 m/s2), and h is the height the squirrel climbs.

Let's calculate the work done:

To match the power output of a 100 W lightbulb, the squirrel needs to do 367.5 J of work at a rate of 100 watts:

Now, it remains to find the average speed (v) the squirrel must have:

Therefore, the squirrel's average speed must be approximately 20.41 m/s to match the power output of a 100 W lightbulb while climbing the tree.

Answer:

Emf through a single coil of wire is 588 V.

Explanation:

We need to find the emf through a single coil of wire if the magnetic flux changes from -57 Wb to +43 Wb in 0.17 s

According to Faraday's law, emf of coil is directly proportional to the rate of change of magnetic flux. It can be written as :

[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]

Initial flux, [tex]\phi_1=-57\ Wb[/tex]

Final flux, [tex]\phi_2=+43\ Wb[/tex]

So, [tex]\epsilon=\dfrac{\phi_2-\phi_1}{dt}[/tex]

[tex]\epsilon=\dfrac{43\ Wb-(-57\ Wb)}{0.17\ s}[/tex]

[tex]\epsilon=588.23\ V[/tex]

or

[tex]\epsilon=588\ V[/tex]

So, the EMF through a single coil of wire is 588 V. Hence, this is the required solution.

Answer:

3 N

Explanation:

According to the Coulomb's law in electrostatics, the Coulomb force acting between two charges is directly proportional to the product of two charges and inversely proportional to the square of distance between them.

Case I :

F = 300 N, d = 33 cm

300 ∝ 1/ 3^2

300 ∝ 1/9    ..... (1)

Case II:

Let F be the force, d = 30 cm

F ∝ 1/30^2

F ∝ 1/900     ....(2)

Dividing equation (2) by equation (1), we get

F / 300 = 9 / 900

F = 300 / 100

F = 3 N

Answer:

C

Explanation:

Answer:

(a) 0.139 N/m

(b) 2.83 Hz

Explanation:

(a) m = 0.22 g = 0.22 x 10^-3 kg

f = 4 Hz

The formula for the frequency of spring is given by

f = [tex]f = \frac{1}{2\pi }\times \sqrt{\frac{k}{m}}[/tex]

Where, k be the spring constant

[tex]4 = \frac{1}{2\times 3.14 }\times \sqrt{\frac{k}{0.22\times 10^{-3}}}[/tex]

k = 0.139 N/m

(b)

m = 0.44 g = 0.44 x 10^-3 kg,

k = 0.139 N/m

Let the frequency be f.

f = [tex]f = \frac{1}{2\pi }\times \sqrt{\frac{k}{m}}[/tex]

[tex]f = \frac{1}{2\times 3.14 }\times \sqrt{\frac{0.139}{0.44\times 10^{-3}}}[/tex]

f = 2.83 Hz

In a simple harmonic system, the spring stiffness constant 'k' and frequency of oscillations are directly linked by the formula f = 1/(2π) * √(k/m). In the given scenario, the spring stiffness constant of the web is calculated as 34.5 N/m. When the mass of the insect doubles, the expected oscillation frequency drops to around 2.8Hz.

In the study of physics, particularly mechanics, harmonic motion is usually demonstrated using a simple spring-mass system. The spring stiffness constant 'k', also known as the force constant, is a key aspect to consider.

In part (a) of your question, we can use the equation for the frequency of oscillations in a simple harmonic oscillator, which is formulated as f = 1/(2π) * √(k/m), where 'f' is the frequency, 'k' is the spring stiffness constant, and 'm' is the mass of the object causing the oscillations. Given the known values in your question, we would then rearrange this equation to solve for 'k': k = (f * 2π)² * m. Inserting the values given (f = 4.0Hz, m = 0.22g = 0.00022kg), we find that k = 34.5 N/m.

For part (b) of your question, if the mass of the insect trapped in the web is doubled (m = 0.44g = 0.00044kg), while the spring stiffness constant remains the same, we would expect the web to oscillate with a lower frequency. We can calculate this using the same formula as before, rearranged to solve for 'f': f = 1/(2π) * √(k/m), and find that the frequency would be approximately 2.8Hz.

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Answer:

A. because the Sun is aligned with the moon

Explanation:

There are two types of tides:

- Spring tides: spring tides occur when the Sun, the Earth and the Moon are aligned along the same line. When this occurs, the gravitational attractions exerted by the Sun and the Moon on the Earth are along the same line - as a result, the water of the oceans is pulled "stronger" along this line, causing a higher tide than normal.

- Neat tides: neat tides occur when the Sun and the Moon are placed at right angle with respect to the Earth - in this case, the tide level is lower, because the gravitational attractions exerted by the Sun and the Moon on the Earth are perpendicular to each other.

Answer:

Average velocity of water flow through the pipe = 3.98 m/s

Explanation:

Diameter of pipe = 20 mm

Discharge = 75 L/min

We know that

        Discharge = Area x Velocity

      [tex]\texttt{Area of pipe}=\frac{\pi d^2}{4}=\frac{\pi \times (20\times 10^{-3})^2}{4}=3.14\times 10^{-4}m^2[/tex]

       Discharge = 75 L/min

                         [tex]=\frac{75\times 10^{-3}}{60}=1.25\times 10^{-3}m^3/s[/tex]

Substituting

        [tex]1.25\times 10^{-3}=3.14\times 10^{-4}\times V\\\\V=3.98m/s[/tex]

Average velocity of water flow through the pipe = 3.98 m/s

Answer:

a lot my dude

Explanation:

cars are heavy

Explanation:

a) Calculate the z-score.

z = (x − μ) / σ

z = (1050 − 1100) / 93

z = -0.54

From a z-score table:

P(z<-0.54) = 0.2946

Therefore:

P(z>-0.54) = 1 - 0.2946

P(z>-0.54) = 0.7054

70.54% of steers are over 1050 pounds.

b) Calculate the z-score.

z = (x − μ) / σ

z = (1300 − 1100) / 93

z = 2.15

From a z-score table:

P(z<2.15) = 0.9842

98.42% of steers are under 1300 pounds.

c) Calculate the z-scores.

z = (x − μ) / σ

z = (1150 − 1100) / 93

z = 0.54

z = (1250 − 1100) / 93

z = 1.61

From a z-score table:

P(z<0.54) = 0.7054

P(z<1.61) = 0.9463

Therefore:

P(0.54<z<1.61) = P(z<1.61) - P(z<0.54)

P(0.54<z<1.61) = 0.9463 - 0.7054

P(0.54<z<1.61) = 0.2409

24.09% of steers weigh between 1150 pounds and 1250 pounds.

To calculate the proportion of steers in different weight categories, change the weights to z-scores, and use a standard normal distribution table to find the percentile ranks. For steers over 1050 pounds, about 70.43% meet this category. For steers under 1300 pounds, about 98.41% meet that mark. To find the proportion weighing between 1150 and 1250 pounds, find the percentile ranks for both weights and subtract.

This is a question related to the concept of Normal Distribution in statistics. First, you need to convert the weights to z-scores, which are measures of how many standard deviations an element is from the mean.

(a) To find the percent of steers weighing over 1050 pounds, first calculate the z-score: (1050-1100)/93= -0.5376. Using a standard normal distribution table, you find that the percentile rank for -0.5376 is approximately 0.2957 or 29.57%. However, this represents the proportion of steers that weigh less than 1050 pounds. To find how many weigh more, subtract this from 1. So about 70.43% of steers weigh more than 1050 pounds.

(b) Similarly, for under 1300 pounds, calculate the z-score: (1300-1100)/93= 2.1505. The percentile rank for 2.1505 is approximately 0.9841 or 98.41%, indicating 98.41% of steers will weigh less than 1300 pounds.

(c) To find the proportion weighing between 1150 and 1250 pounds, find the z-scores for both weights and their respective percentile ranks. Then subtract the smaller percentile from the larger one.

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Answer:

[tex]5.5\cdot 10^{-11} C[/tex]

Explanation:

The capacitance of the parallel-plate capacitor is given by:

[tex]C=\epsilon_0 \frac{A}{d}[/tex]

where

[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity

[tex]A=190 mm^2 = 190 \cdot 10^{-6} m^2[/tex] is the area of the plates

[tex]d=1.2 mm = 0.0012 m[/tex] is the separation between the plates

Substituting,

[tex]C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F[/tex]

The energy stored in the capacitor is given by

[tex]U=\frac{Q^2}{2C}[/tex]

Since we know the energy

[tex]U=1.1 nJ = 1.1 \cdot 10^{-9} J[/tex]

we can re-arrange the formula to find the charge, Q:

[tex]Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C[/tex]

The charge needed to be transferred between the plates to store 1.1 nJ of energy is approximately 55.5 pC.

To determine the charge that must be transferred between two initially uncharged parallel plates to store 1.1 nJ of energy, we use the concept of a parallel plate capacitor. The energy stored in a capacitor is given by the formula:

U = 0.5 * C * V²

Where U is the energy (1.1 nJ = 1.1 × 10⁻⁹J), C is the capacitance, and V is the potential difference between the plates.

Step 1: Calculate the Capacitance

Capacitance (C) for a parallel-plate capacitor is given by:

C = (0 * A) / d

Where 0 is the permittivity of free space (8.85 × 10⁻¹² C²/N·m²), A is the area of the plates (190 mm² = 190 × 10⁻⁶ m²), and d is the separation between the plates (1.2 mm = 1.2 × 10⁻³ m).

Substitute the values:

C = (8.85 × 10⁻¹²* 190 × 10⁻⁶) / 1.2 × 10⁻³

C ≈ 1.4 × 10⁻¹² F

Step 2: Calculate the Voltage

Using the energy formula, solve for V:

1.1 × 10⁻⁹ J = 0.5 * 1.4 × 10⁻¹² F * V²

V² = (1.1 × 10⁻⁹ J) / (0.5 * 1.4 × 10⁻¹²F)

V² ≈ 1571 J/F

V ≈ 39.65 V

Step 3: Calculate the Charge

Use the relationship Q = C * V:

Q = 1.4 × 10⁻¹² F * 39.65 V

Q ≈ 5.55 × 10⁻¹¹ C

Therefore, approximately 55.5 pC (picoCoulombs) of charge must be transferred from one plate to the other.

The resulting current density in the  section of silver of uniform cross section be 1.54 × 10^11  A/m^ 2.

Electrical resistivity is a measurement of a material's degree of resistance to current flow. The SI unit for electrical resistivity is the ohm metre (m). It is frequently represented by the Greek letter rho.

Materials that easily transmit current and have a low resistance are called conductors. Insulators have a high resistance and do not conduct electricity. Resistivity is defined as the size of the electric field across it that results in a particular current density.

Resistivity at any temperature T is given by:  ρ = ρ₀[1 + α(T-T₀)]  

Given:

ρ₀ = 1.59x10^-8 ohms*m

reference temperature (T₀) = 20°C.

T =  3°C and α = 0.0038/°C.

So, resistivity  ρ =   1.59x10^-8[ 1 +  0.0038×( 3 -20)] ohm.m

= 7.39 × 10^-9 ohm.m.

Again: electric field : E = 1139 V/m

So,  resulting current density  = E/ρ = 1139/ 7.39 × 10^-9  A/m^2

=1.54 × 10^11  A/m^ 2.

Hence, The resulting current density in the  section of silver of uniform cross section be 1.54 × 10^11  A/m^ 2.

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To determine the current density in the silver specimen at 3 C, you must calculate its resistivity at that temperature using the temperature coefficient. Then, find the conductivity and use Ohm's law with the given electric field to calculate the current density.

To find the resulting current density (J), we need to first adjust the resistivity ( ) of silver for the given temperature of 3 C. The resistivity at this temperature can be found using the formula (T) = 0[1 + ( T - 20 C)], where (T) is the resistivity at temperature T, 0 is the resistivity at 20 C, and is the temperature coefficient of the material.

Substituting the given values into the formula, we get: (3 C) = 1.59 10^-8 [ 1 + 0.0038( 3 C - 20 C) ] = 1.59 ₓ 10⁻⁸ [ 1-0.0646] = 1.59 ₓ 10⁻⁸ (0.9354 m) = 1.49 ₓ 10⁻⁸ .

Now that we have the resistivity at 3 C, we can find the conductivity ( s) using s = 1/ which equals 1/(1.49 ₓ 10⁻⁸ ) = 6.71 ₓ 10⁷ S/m.

The current density J can then be calculated using Ohm's law, J = sE, where E is the electric field intensity. Using the provided electric field of 1139 V/m, J = 6.71 ₓ10⁷ S/m 1139 V/m = 7.64ₓ 10¹⁰ A/m.

Which body is in equilibrium?

(1) a satellite orbiting Earth in a circular orbit .  No.  The forces on it are unbalanced.  There's only one force acting on it ... the force of gravity, pulling it toward the center of the Earth.  That's a centripetal force, and the satellite is experiencing centripetal acceleration.

(2) a ball falling freely toward the surface of Earth.  No.  The forces on it are unbalanced.  There's only one force acting on it ... the force of gravity, pulling it toward the center of the Earth.  The ball is accelerating toward the ground.  

(3) a car moving with a constant speed along a  straight, level road.  YES. We don't even need to analyze the forces, just look at the car.  It's moving in a straight line, and its speed is not changing.  The car's acceleration is zero !  That  right there tells us that the NET force ... the sum of all forces acting on the car ... is zero.  THAT's called 'equilibrium'.

(4) a projectile at the highest point in its trajectory.  No.  The forces on it are unbalanced.  There's only one force acting on it ... the force of gravity, pulling it toward the center of the Earth.  The projectile is accelerating toward the ground.

A car moving with a constant speed along a straight line road is the body in equilibrium.  

• When the forces are balanced, then the object is considered to be in equilibrium.  

• In equilibrium, the net force is zero and the acceleration of the body is also zero.  

• The acceleration of 0 m/s does not signify that the body is at rest. An object is considered to be in equilibrium when it is either at rest and staying or rest, or is in motion, however, having same speed and direction.  

• Therefore, when the car is moving with a similar speed along a straight line, it is considered to be having zero acceleration due to constant velocity and is moving in a straight line showing thus the car is in equilibrium.  

Thus, a car moving with a constant speed along a straight level road is in equilibrium.  

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Answer:

145.8 m

Explanation:

M = mass of the runaway train car = 15,000 kg

V₀ = initial speed of the train = 5.4 m/s

V = final speed of the train = 0 m/s

F = force applied to bring the car to rest = 1500 N

d = distance traveled before stopping

Using work-change in kinetic energy theorem

Work done by applied force = Change in kinetic energy

- F d = (0.5) M (V² - V₀²)

- (1500) d = (0.5) (15000) (0² - 5.4²)

d = 145.8 m

We apply Newton's second law to find the deceleration of the train car. Then, we use the work-energy principle to compute the distance traveled during the stopping process.

The problem at hand involves using concepts of physics, specifically Newton's second law and work-energy principle. The mass of the train car is 15000 kg and its initial speed is 5.4 m/s. The external force acting on it is 1500 N which brings the car to rest, implying that its final speed is 0m/s.

To solve the problem, we first need to find the deceleration of the train car using the formula F=ma (where F=force, m=mass, a=acceleration). The negative sign indicates deceleration. After calculating the deceleration, we will use the equation of motion to find the distance traveled during the stopping period. The equation is v2 = u2 + 2as where v is final speed, u is initial speed, a is acceleration and s is distance.

After substituting the known values into the equation, we will be able to solve for s (the distance covered while stopping).

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Explanation:

Three balances can be written.

Mass of A into the column = mass of A out of the column

Mass of B into the column = mass of B out of the column

Mass of C into the column = mass of C out of the column

There are four unknowns: mass flow of the first feed stream, mass flow of the overhead stream, mass flow of the bottom stream, and percent B or C of the middle stream.

Since there are more unknowns than equations, there is not enough information to solve the equations.

Final answer:

The system described with three species allows for three material balances to be written, one for each chemical species A, B, and C.

Explanation:

The number of material balances that can be written on the overall system depends on the number of chemical species in the system. In the scenario provided, there are three species: A, B, and C. Therefore, three material balances can be written, one for each species.

The general form for a material balance is the sum of inputs minus the sum of outputs plus generation minus consumption equals accumulation. Since there is no indication of chemical reactions or accumulation (steady-state operation is assumed), the material balances for each species become:

For real-world applications, these material balances could be more complex, accounting for multiple units, reactions, and phase changes. However, with the information given and the assumption of steady-state, the total number of material balances that can be written for this system is three.

Answer:

28

Explanation:

X-13=15

X=15+13

X=28

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

where we have

[tex]\alpha = -80.0 rad/s^2[/tex] is the angular acceleration (negative since the rotor is slowing down)

[tex]\omega_f [/tex] is the final angular speed

[tex]\omega_i = 2000 rad/s[/tex] is the initial angular speed

t = 10.0 s is the time interval

Solving for [tex]\omega_f[/tex], we find the final angular speed after 10.0 s:

[tex]\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s[/tex]

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

If we re-arrange it for t, we get:

[tex]t = \frac{\omega_f - \omega_i}{\alpha}[/tex]

where here we have

[tex]\omega_i = 2000 rad/s[/tex] is the initial angular speed

[tex]\omega_f=0[/tex] is the final angular speed

[tex]\alpha = -80.0 rad/s^2[/tex] is the angular acceleration

Solving the equation,

[tex]t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s[/tex]

Answer:

271 m

Explanation:

We can divide the motion into two stages.  In the first stage, the rocket is accelerating upward.  In the second stage, the fuel has exhausted, and the rocket is in free fall.

For the first stage, we are given:

y₀ = 0 m

v₀ = 0 m/s

a = 45.3 m/s²

t = 1.46 s

Find: y, v

y = y₀ + v₀ t + ½ at²

y = (0) + (0) (1.46) + ½ (45.3) (1.46)²

y = 48.3 m

v = at + v₀

v = (45.3) (1.46) + (0)

v = 66.1 m/s

In the second stage, we are given:

y₀ = 48.3 m

v₀ = 66.1 m/s

v = 0 m/s

a = -9.8 m/s²

Find: y

v² = v₀² + 2a(y - y₀)

(0)² = (66.1)² + 2(-9.8) (y - 48.3)

y = 271 m

The rocket reaches a maximum height of 271 m.

Answer:

Part a)

a = 1.37 m/s/s

Part b)

since the force of gravity is more than the thrust force of rocket so it will not lift off the surface of Earth

Explanation:

Part a)

Net force on the Module due to thrust of engine is given as

[tex]F = 30,000 N[/tex]

now net force while is ejected out of the surface of moon is given as

[tex]F_{net} = F - F_g[/tex]

here we know that

[tex]F_g = mg_{moon}[/tex]

where we have

[tex]g_{moon} = \frac{g}{6}[/tex]

[tex]F_{net} = 30,000 - (10000)(\frac{9.8}{6})[/tex]

[tex]F_{net} = 13666.67 N[/tex]

now the acceleration of the module on the moon is

[tex]a = \frac{F_{net}}{m}[/tex]

[tex]a = \frac{13666.67}{10,000} = 1.37 m/s^2[/tex]

Part b)

Now on the surface of earth the force of gravity on the module is given as

[tex]F_g = mg [/tex]

[tex]F_g = 10,000 \times 9.8[/tex]

[tex]F_g = 98000 N[/tex]

since the force of gravity is more than the thrust force of rocket so it will not lift off the surface of Earth

Acceleration is the rate of change of velocity. The acceleration of the module is 1.365 m/s².

Given to us

Mass of the Module = 10,000 kg

The thrust of the engine, Fₓ = 30,000 N

A.)

We know in order to lift the module the thrust produced by the engine must be greater than the gravitational pull by the module. therefore,

[tex]F_{N} = F_x - W[/tex]

where W is the weight of the moon,

[tex]F_{N} = F_x - (m\times g_{moon})[/tex]

Also, the acceleration on the moon is 1/6 of the acceleration on the earth,

[tex]F_{N} = 30,000- (10,000\times \dfrac{9.81}{6})\\F_N = 13,650\ N[/tex]

We know that force is the product of mass and acceleration, therefore,

[tex]F_N = m \times a\\\\13,650 = 10,000 \times a\\\\a = 1.365 m/s^2[/tex]

Hence, the acceleration of the module is 1.365 m/s².

B.)

If we need to lift the module on earth, we need a thrust that is greater than the weight of the module,

[tex]Weight = mass \times acceleration\\W = 10,000 \times 9.81\\W = 98,100\ N[/tex]

As the weight of the module is greater than the thrust produced by engines. therefore, the module can not take off from the earth.

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Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

Answer:

The answer is A.

Explanation:

Just trust me ok.

The acceleration that car B appears to have to an observer in car A is equal to [tex]-16\;km/h/s[/tex].

Given the following data:

Relative acceleration can be defined as the acceleration of an observer B with rest to the rest frame of another observer A.

Mathematically, relative acceleration is given by this formula:

[tex]A_{BA}=A_B-A_A[/tex]

Substituting the given parameters into the formula, we have;

[tex]A_{BA}=-8-8\\\\A_{BA}=-16\;km/h/s[/tex]

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The acceleration that car A appears to have to an observer in car A is -8 km/h per second.

To determine the acceleration that car A appears to have to an observer in car A, we need to consider the relative motion between the two cars.

Since car A has a constant velocity of 100 km/h, the observer in car A would perceive car B's motion as a decrease in velocity with a rate of 8 km/h each second. This means that car B's acceleration as perceived by the observer in car A would be -8 km/h per second.

Therefore, the acceleration that car A appears to have to an observer in car A is -8 km/h per second.

To calculate the time required to deposit 0.500 g of metallic nickel on a custom car part using a current of 3.00 A, you can use the equation Time (s) = Mass (g) / (Current (A) × Charge (C/g)). Calculations involving the charge of nickel and the given values can be used to determine the time in seconds. Converting the time to minutes or hours is also possible.

To calculate the time required to deposit 0.500 g of metallic nickel using a current of 3.00 A, we need to use the equation:

Time (s) = Mass (g) / (Current (A) × Charge (C/g))

For nickel, the charge is 2+ and the atomic mass is approximately 58.69 g/mol. From these values, we can calculate the charge in C/g:

Charge (C/g) = (2 × 1.602 × 10-19 C) / (58.69 g/mol)

Using this value and the given mass and current, we can calculate the time required in seconds:

Time (s) = 0.500 g / (3.00 A × Charge (C/g))

Converting the time to minutes or hours can be done by dividing by 60 or 3600, respectively.

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Answer:

The volume charge density of the sphere is [tex]1.58\times 10^{-13}\ C/m^3[/tex]

Explanation:

It is given that,

Charge on sphere, [tex]q = -353\ e =-353\times 1.6\times 10^{-19}\ C=-5.64\times 10^{-17}\ C[/tex]

Radius of sphere, r = 4.4 cm = 0.044 m

We need to find the volume charge density in that sphere. It is given by total charge divided by total volume of the sphere.

[tex]\rho=\dfrac{q}{V}[/tex]

[tex]\rho=\dfrac{5.64\times 10^{-17}\ C}{\dfrac{4}{3}\pi (0.044\ m)^3}[/tex]

[tex]\rho=1.58\times 10^{-13}\ C/m^3[/tex]

So, the volume charge density of the sphere is [tex]1.58\times 10^{-13}\ C/m^3[/tex]. Hence, this is the required solution.

Answer:

200 KJ

Explanation:

h = 20 m, m = 1000 kg, g = 10 m/s^2

As the rock is sliding and it starts from rest. it falls from a f=height of 20 m

So, by using the energy conservation law

Potential energy at the top of hill = Kinetic energy at the bottom of hill

So, Kinetic energy at the bottom of hill = Potential energy at the top of hill

K.E = m g h

K. E = 1000 x 10 x 20 = 200,000 J

K.E = 200 KJ

Answer:

power, P = 90 hp

Explanation:

It is given that,

Mass of the car, m = 1500 kg

Initial velocity of car, u = 0

Final velocity of car, v = 25 m/s

Time taken, t = 7 s

We need to find the average power delivered by the engine. Work done divided by total time taken is called power delivered by the engine. It is given by :

[tex]P=\dfrac{W}{t}[/tex]

According to work- energy theorem, the change in kinetic energy of the energy is equal to work done i.e.

[tex]W=\Delta E=\dfrac{1}{2}m(v^2-u^2)[/tex]

[tex]P=\dfrac{\dfrac{1}{2}m(v^2-u^2)}{t}[/tex]

[tex]P=\dfrac{\dfrac{1}{2}\times 1500\ kg\times (25\ m/s)^2}{7\ s}[/tex]

P = 66964.28 watts

Since, 1 hp = 746 W

So, P = 89.76 hp

or

P = 90 hp

So, the average power delivered by the engine is 90 hp. Hence, the correct option is (E) " 90 hp".                                                                                                  

The average power delivered by the engine of a 1500-kg car accelerating from 0 to 25 m/s in 7.0 s is about 90 horsepower (hp). This is calculated by determining the work done as the change in kinetic energy divided by time, and then converting watts to horsepower.

The kinetic energy (KE) of the car at 25 m/s, given that it started from rest, is:

KE = ½ m v^2

KE = ½ × 1500 kg × (25 m/s)^2

KE = 468750 J

The average power, P, delivered by the engine is the work done divided by the time:

P = Work / time = KE / time

P = 468750 J / 7 s

P = 66964.29 W

To convert the power to horsepower:

Power (hp) = Power (W) / 746

Power (hp) = 66964.29 W / 746

Power (hp) = 89.74 hp

Therefore, rounding to the nearest whole number, the average power delivered by the car’s engine is about 90 hp.

C) Staring At Someone's Body, Giving Them The “Once Over”

All of the following are examples of verbal sexual harassment EXCEPT:

c) Staring At Someone's Body, Giving Them The “Once Over”

Verbal harassment is considered any conscious and repeated attempt to humiliate, demean, insult, or criticize someone with words. Verbal sexual  harassment  is harassing someone by saying something or verbally disrespecting someone in a way that you are discomforting them . It is a type of non - physical harassment that  makes someone humiliated or threatened

hence ,a) Telling “Dirty" Jokes Or Stories b) Making Kissing Sounds, Howling, Or Smacking Lips d)Making Repeated Requests For A Date all are example of verbal sexual harassment all three are non-physical and only words are being used in harassing someone

c) Staring At Someone's Body, Giving Them The “Once Over” is an example of visual sexual harassment as staring will not be considered as verbal , it means that with your vision you are discomforting someone

correct option c)Staring At Someone's Body, Giving Them The “Once Over”

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Answer:

The total resistance is 5000 N. So the net force left for acceleration is 7 000 N (12 000 - 5 000).

7000 N applied to 2x700 kg gives a maximum acceleration of

a = F/m = 7000/1400kg = 5 m/s^2

At 5m/s^2, it will take 8 s to reach 40 m/s. In 8 second the distance covered will be

d = 1/2 at^2 = 1/2 x 5 x 8^2 = 160 m

Because each glider has the same drag and inertia, the tension in the rope between the gliders will be exactly half of the tension between plane and the two gliders:

12 000/2 = 6 000N.

The second law of Newton and kinematics allows to find the answers for the distance and tension are:

A) The distance traveled is 127.2 m

B) The tension in the second glider is 6003 N

Kinematics studies the movement of bodies establishing relationships between the position, velocity and acceleration of bodies

               v² = [tex]v_o^2 + 2 a x[/tex]

Where v is the velocity, v₀ the initial velocity, a the acceleration and x the distance traveled

Newton's second law states that the net force is proportional to the mass and the acceleration of the body

         F = m a

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body

A) Ask the length needed for takes off

We look for the acceleration that the system has using Newton's second law,

The free body diagram is a representation of the system without the details of the bodies, in the adjoint we can see a free body diagram of the forces, let's write Newton's second law for each axis

Glider 1

x-axis

          T₁ - T₂ -fr = m a

y-axis  

         N₁ - W = 0

         N₁ = W

Glider 2

x-axis

         T₂ -fr = m a

y-axis  

         N₂ - W =

         N₂ = W

We write the system of equations

        T₁ - T₂ -fr  = m a

              T₂ - fr = m a

We solve the system

         T₁ - 2 fr = 2 m a

         a = [tex]\frac{T_1 - 2 fr}{2m}[/tex]

Indicates that the friction force for each glider is 1600 N

Calculate

         a = [tex]\frac{12000 - 2 \ 1600}{2 \ 700}[/tex]

         a = 6.29 m / s²

Taking the acceleration we can use the kinematics relationship to find the distance traveled

          v² = [tex]v_o^2+2ax[/tex]

As part of rest the initial velocity is zero

           v² = 0 + 2 ax

           x = [tex]\frac{v^2}{2a}[/tex]v² / 2 a

indicate that the speed of 40 { m/s} is required for takeoff

           x = [tex]\frac{40^2}{2 \ 6.29}[/tex]

           x = 127.2 m

B) The tension (T2) between the two gliders is requested

We write the second Newton law for the second glider, see attached

                  T₂ - fr = m a

                  T₂ = m a + fr

                  T₂ = 700 6.29 + 1600

                  T₂ = 6003 N

In conclusion using Newton's second law and kinematics we can find the answers for the distance and tension are;

A) The distance traveled is 127.2 m

B) The tension in the second glider is 6003 N

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