A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C exists along the +x axis. A magnetic field also exists, and its x and y components are Bx = +1.9 T and By = +1.9 T. Calculate the force (magnitude and direction) exerted on the particle by each of the three fields when it is (a) stationary, (b) moving along the +x axis at a speed of 345 m/s, and (c) moving along the +z axis at a speed of 345 m/s.

Answer:

[tex]4.08cm/s^2[/tex]

Explanation:

The second equation of a uniformly accelerated motion could be used to solve this problem. This is given by equation (1);

[tex]s=ut+\frac{1}{2}at^2....................(1)[/tex]

where u is the particle's initial velocity, t is the time taken, a is the acceleration and s is the distance travelled.

Given;

u = 20cm/s

t = 7s

a = ?

s = ?

The particle moved from one point [tex]x_1[/tex] to another point [tex]x_2[/tex] along the x-axis, where [tex]x_1=10cm[/tex] and [tex]x_2=-30cm[/tex]. This information could be used to find the distance s covered by the object as follows;

[tex]s=x_1-x_2.................(2)\\s=10-(-30)\\s=10+30\\s=40cm[/tex]

We the make appropriate substitutions into equation (1) and then solve for the acceleration.

[tex]40=(20*7)+\frac{1}{2}*a*7^2\\40=140+\frac{1}{2}*a*49\\40=140+24.5a\\40-140=24.5a\\hence\\24.5a=-100\\a=\frac{-100}{24.5}\\a=-4.08cm/s^2[/tex]

The negative sign is an indication that the particle is decelerating.

Answer:

7.347 cm / s²

Explanation:

Using equation of linear motion

S = ut + 1/2 at²

where total displacement = final displacement - initial displacement

S = - 30 - 10 = - 40 cm

- 40 cm = (20 cm /s × 7 s) + 1/2 a (7²)

- 40 cm = 140 cm + 1/2 49 a

- 40 cm - 140 cm =  1/2 × 49 a

- 180 cm × 2 / 49 s² = a

a = -7.347 cm / s²

It is probably decelerating.

Answer:

Explanation:

a) To get the resistivity ρ at 50 Celsius, given the resitstivity at 20 Celsisus, use:

ρ = ρo(1 + α(T - To))

where To = 20 Celsius

b) Knowing the resistivity at 50 Celsius, and the (uniform) electric field E, you can determine the current density J using:

E = ρJ

(which is actually a density-averaged version of V = IR)

c) Assuming the current is uniform (which is should be in a uniform electric field and constant-diameter wire), the current i can be calculated using:

J = i/A --> i = JA

where A is the cross-sectional area of the wire (given by πr2); make sure to convert the given diameter to a radius, and the radius to base units

d) Since the electric field is given in volts per meter, and you have two meters of length in the wire, you can determine directly from that how many volts difference you need at the ends of the wire to get 0.2 volts per meter.

0.2 = V/d

with d = 2 m. This corresponds to a uniform electric field being related to voltage by V = Ed, where d is distance along the field line.

Explanation:

Below are attachments containing the solution.

Explanation:

The magnetic field about a straight length of current carrying wire is circular in shape. As we know that the electric current passing through the wire is producing magnetic field. One can find out the direction of the magnetic field lines by using right hand rule or a compass. If the conductor is a straight wire then the magnetic field lines will form a concentric circles around the wire. Put you thumb in the direction of the electric current in the wire then the fingers will curl to show the direction of magnetic field which thus form the circular loop.

Answer:

circular in shape.

Explanation:

This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. One way to explore the direction of a magnetic field is with a compass, as shown by a long straight current-carrying wire in.

Because the magnetic field created by the electric current in the wire is changing directions around the wire, it will repel both poles of the magnet by bending away from the wire.

Answer:

Time taken to reach your destination will be 10hours

Explanation:

Recall the formula for Speed;

speed=Total distance/Total time taken

Speed=2km/h

Total distance=20km

Time taken=x

let x be the unknown time taken

Input each values into the formula;

2=20/x

Making x subject of the equation

x=20/2

x=10

Total time taken =10hours.

Answer:

10 hours

Explanation:

The average speed of a body is given by;

[tex]v_{avg}=\frac{s}{t}..................(1)[/tex]

where s is the total distance travelled and t is the total time spent.

Given;

s = 20km

t = ?

[tex]v_{avg}=2km/h[/tex].

We substitute into equation (1) and then solve for t.

[tex]2=\frac{20}{t}\\2t=20\\t=\frac{20}{2}\\t=10hrs.[/tex]

Answer: v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s

Explanation:

Magnetic force(B) = 4.60×10^-3 T

Electric force(E) = 1.64×10^4 V/m

Both forces having equal magnitude ;

Magnetic force = electric force

qvB = qE

vB = E

v = (1.64×10^4) ÷ (4.60×10^-3)

v = 3.57×10^6 m/s

2.) Assume no electric field

qvB = ma

Where a = v^2 ÷ r

R = radius

a = acceleration

v = velocity

qvB = m(v^2 ÷ R)

R = (m×v) ÷ (|q|×B)

q=1.6×10^-19C

m = 9.11×10^-31kg

R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)

R = 32.5227×10^-25 ÷ 7.36×10^-22

R = 4.42×10^-3m

3.) period(T)

T = (2*pi*R) ÷ v

T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)

T = (27.775×10^-3) ÷(3.57×10^6)

T = 7.78×10^-9 s

Answer:

[tex]A = 1.056\,m^{2}[/tex]

Explanation:

The portion of solar energy converted into electric energy is given by the following equation:

[tex]\dot E = \eta \cdot I\cdot A[/tex]

The area needed to produce energy is derived by clearing the corresponding variable:

[tex]A = \frac{\dot E}{\eta \cdot I}[/tex]

[tex]A = \frac{75\,W }{(0.1)\cdot (710\,\frac{W}{m^{2}} )}[/tex]

[tex]A = 1.056\,m^{2}[/tex]

Answer:

Passed into the power grid for others to use the electricity

Explanation:

If a home uses a large supply of solar panels to generate electricity, but has no battery system, surplus electricity that is produced is usually passed into the power grid for others to use the electricity, generating a income to the homeowner

When a home using solar panels produces surplus electricity and has no battery system, the excess power is typically fed back into the electricity grid. This process, known as net metering, can often result in a credit from the electricity company.

If a home uses a large supply of solar panels to generate electricity, but has no battery system, surplus electricity that is produced is usually fed back into the grid. This feed-in process involves the excess electricity being transmitted to the local electricity network where it can be used by other homes or businesses. In many places, electricity companies provide a credit to the solar power producer for this extra electricity, which can lower the overall utility bill.

This concept is known as net metering which allows homeowners to offset the cost of power drawn from the utility grid by pushing their surplus electricity into the grid. However, it's crucial to note that this process might depend on the policies of the local utility or the regional grid operator.

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Given Information:  

Radius of circular loop = r = 9.1 cm = 0.091 m  

Change in time = Δt = 0.66 seconds

Change in magnetic field = ΔB = 0.90 T

Resistance of wire per unit length = R = 2.9x10⁻²  Ω/m

Number of turns = N = 1

Required Information:  

Electrical energy dissipated = E = ?  

Answer:  

Electrical energy dissipated = 50.09x10⁻³ Joules  

Step-by-step explanation:  

We know that energy is given by

E = Pt

Where power is given by

P = ξ²/R

Where ξ is the induced EMF in the wire and is given by

ξ = -NΔΦ/Δt

Where ΔΦ is the change in flux and is given by

ΔΦ = ΔBAcosφ

Where φ is the angle between magnetic field and circular loop

A = πr² and R = 2.9x10⁻²*2πr

Substituting the above relations into the energy equation and simplifying yields,

E = [-Nπr²cosφ(ΔB/Δt)²]*t/R

E = [-1*π(0.091)²*cos(0)(0.90/0.66)²*0.66]/2.9x10⁻²*2π*(0.091)

E = 0.050094 Joules

E = 50.09x10⁻³ Joules

Therefore, the average electrical energy dissipated in the circular loop of the wire is 50.09x10⁻³ Joules.

Final answer:

The problem about a parallel-plate capacitor requires the application of electromagnetism principles to determine changes in potential difference, initial and final stored energy, and the work required to separate the plates, all based on the capacitor's geometry and the effect of plate separation on capacitance.

Explanation:

To answer the student's question about the parallel-plate capacitor, we need to apply concepts from electromagnetism, specifically the relationships between charge, voltage, capacitance, and energy in capacitors. When the parallel plates of a capacitor are separated, the capacitance changes, but the charge remains the same since it is isolated after being disconnected from the battery. The potential difference between the plates changes as a result of the changing capacitance. The initial energy stored in the capacitor can be calculated using the formula U = 1/2 CV^2, and the final energy stored after increasing the plate separation can be calculated with the same formula but with the new capacitance value. The work done to separate the plates is equivalent to the change in stored energy, which can either be the work done by an external force or the work lost to the system.

Answer:

By measuring the time taken for the stars line of sight velocity to cycle from peak to Peak, and by calculation using newtons version of Kepler's third law

Explanation:

The motion of orbiting planets using planet-hunting techniques can rely on doppler effect. The light from the stars they orbit, as seen from Earth. As the star moves back and forth, the Doppler shift causes a slight change in its apparent colour which can be detected using spectroscopy. The blue shift and the red shift.

The Doppler technique of blue shift and the red shift can be used to estimate the semi major axis of the planets orbit by

Measuring the time it takes for the stars line of sight velocity to cycle from peak to Peak, and using newtons version of Kepler's third law

Answer:[tex]1.53g[/tex]

Explanation:

average deceleration= ?

inial velocity: [tex]u=0[/tex]

final velocity: [tex]v=30m/s[/tex]

time: [tex]t=2seconds[/tex]

The first law of kinematics :

[tex]v=u+at[/tex]

find a the subject of the formula

[tex]a=v-u/ t[/tex]

[tex]a=\frac{0-30} 2[/tex]

[tex]a=-30/2[/tex]

[tex]a=-15m/s^{2}[/tex]

The deceleration about g(acceleration due to gravity) will be:

[tex]15/9.8[/tex]

[tex]1.53g[/tex]

The bungee jumper's average deceleration in g's is approximately -1.53 g's.

To find the average deceleration, we need to calculate the change in velocity and divide it by the time it took to come to a halt. The change in velocity is the final velocity minus the initial velocity, which is 0 m/s minus 30 m/s, giving us -30 m/s. The time is given as 2 s. Plugging these values into the formula for average deceleration, we get:

Average Deceleration = (Change in Velocity) / (Time)

Average Deceleration = (-30 m/s) / (2 s) = -15 m/s²

To convert this to g's, we need to divide by the acceleration due to gravity (9.8 m/s²).

Average Deceleration in g's = (-15 m/s²) / (9.8 m/s²) ≈ -1.53 g's

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3. Newton's third law

5. Conservation of momentum

Explanation:

Conservation of momentum is mostly used for describing collisions between objects. Here, the type of collision is inelastic collision in which the object when collides with the pendulum bob sticks to it and moves as a combined object. In this process the momentum is conserved.

Let the mass of the pendulum be m1 moving with a velocity v1.

Let the mass of the object be m2 moving with a velocity v2.

Since the momentum is conserved during collision, the equation will be

[tex]m1 v1 + m2 v2 = (m1 + m2) v[/tex]

Where, v is the velocity of the combined system.

Conservation of momentum is actually a direct consequence of Newton's third law.

Consider a collision between two objects, object A and object B. When the two objects collide, there is a force on A due to B. However, because of Newton's third law, there is an equal force in the opposite direction, on B due to A

FAB = -FBA

The mechanical energy is not conserved due to the fact that the kinetic energy is not the same before and after the collision.

The conservation of momentum is crucial for analyzing collisions, while the conservation of mechanical energy is ideal for studying swinging motion.

Conservation of momentum can best be used to analyze the collision between the object and the pendulum bob, while Conservation of mechanical energy is best suited to analyze the resulting swing.

Answer:

C

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

                                 t = v / a

                                 t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

                                 t = v / g

                                 t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

Answer:

The population mean are the same

Explanation:

Answer:

The population means are the same.

Explanation:

The hypotheses for a difference in two population means are similar to those for a difference difference two population proportions.

At null point, Ha=0

Let the mean population of the young one be u1

Let the mean population of the old one be u2.

Then, the difference between their mean population distance is given as

Ha=u2-u1

Since, Ha is null point, Ha=0

0=u2-u1

u2=u1

This shows that the mean population distance of the old is equal to the mean population distance of the young.

Therefore their mean population distance is the same

Since it is null alternative then, the population mean are the same.

We must sample the population using

1. Samples must be random to remove or minimize bias.

2. Sample must be representative of the populations in question.

Answer: The total energy created by the Alpha decay.

Explanation: The sum total of the kinetic energy of the alpha particle and the new nucleus is the total energy created by the alpha decay.

Consider the decay of a radioactive nuclide by the spontaneous emission from its nuclei of alpha particles. An alpha particle which is composed of two protons and two neutrons and has a charge of +2. With an appreciable mass and its ejection from the nuclide creates a certain amount of recoil energy in the nucleus. The total energy (Ex) created by alpha decay is therefore the sum of the kinetic energy of the particle, the recoil energy given to the new nucleus, and the total energy of any emitted gamma rays.

The sum of the kinetic energies of an alpha particle and the new nucleus after alpha decay equals the initial energy (Q-value) of the reaction, most of which is carried by the alpha particle due to mass differences.

The sum of the kinetic energies of the alpha particle and the new nucleus after an alpha decay can be determined by considering the conservation of energy and momentum. The energy released during the decay, known as the Q-value, is primarily carried away by the alpha particle due to its relatively smaller mass compared to the daughter nucleus.

The energy of the alpha particle can be measured experimentally, which then allows us to determine the energy of the new nucleus. According to conservation of energy, the sum of the kinetic energies of the alpha particle and the new nucleus is equal to the Q-value of the reaction. If the Q-value (Qa) is known to be 4.3 MeV, and assuming the potential energy is zero in the final state, this energy will be distributed between the alpha particle and the new nucleus.

Using the example values provided, if the kinetic energy of the new nucleus (KEnucleus) is calculated to be 23.3 eV, then the remainder of the Q-value minus the kinetic energy of the new nucleus will represent the kinetic energy of the alpha particle. Since the alpha particle carries away most of the kinetic energy, the sum of the kinetic energies will be very close to the initial Q-value.

Final answer:

To find the initial speed of the car, we use the work-energy principle and the equation μk × g × d = ½ × v². By substituting the given coefficient of kinetic friction (0.29) and skid mark distance (23.74 meters), we calculate the initial speed to be approximately 11.68 m/s.

Explanation:

The question involves calculating the initial speed of a car at the moment the driver applied the brakes, which resulted in skid marks on the road. The length of the skid marks and the coefficient of kinetic friction (μk) between the tires and the road are known.

To calculate the initial speed of the car, we use the work-energy principle, which states that the work done by the friction force is equal to the change in kinetic energy of the car:

Work done by friction = Change in kinetic energy

μk × m × g × d = ½ × m × v²

Since mass (m) cancels out and gravitational acceleration (g) is a constant (9.81 m/s²), we can simplify the equation to:

μk × g × d = ½ × v²

Now we can solve for the initial velocity (v):

v = √(2 × μk × g × d)

Plugging in the given values, μk = 0.29 and d = 23.74 meters, we have:

v = √(2 × 0.29 × 9.81 m/s² × 23.74 m)

v ≈ √(2 × 0.29 × 9.81 × 23.74)

v ≈ √(136.4)

v ≈ 11.68 m/s (approximately)

Thus, the initial speed of the car when the brakes were applied was approximately 11.68 meters per second.

Answer:

T- carrier

Explanation:

The T-carriers are frequently used for trunking between switching centers in a telephone network. It makes use if the same twisted pair copper wire that analog trunks employs. One pair for transmitting and the other pair for receiving.

Answer:

The correct answer to the question is

B. It always decreases

Explanation:

To solve the question, we note that the foce of gravity is given by

[tex]F_G=\frac{Gm_1m_2}{r^2}[/tex] where

G= Gravitational constant

m₁ = mass of first object

m₂ = mass of second object

r = the distance between both objects

If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have

[tex]F_{G2} =\frac{Gm_1(2m_2)}{(2r)^2}[/tex] = [tex]\frac{2}{4} \frac{Gm_1m_2}{r^2}[/tex]

Therefore the gravitational force is halved. That is it will always decrease

Answer:

Explanation:

The energy change during a chemical reation, i.e. the reaction energy, is equal to the chemical energy stored in the bonds of the products less the chemical energy stored in the bonds of the reactants.

Hence:

The change is positive, this is, the chemical energy of the products is greater than the chemical energy of the reactants.

That corresponds to the second graph, where the level of the energy of the products in the graph is higher than the level of the energy of the reactants. Therefore, the conclusion is that the reaction absorbed energy and it is endothermic.

Answer:

4,5

Explanation:

The resulting swing converts potential energy to kinetic energy and kinetic energy to potential energy when the swinging stops. This is line with the law of conservation of mechanical energy which deals with inter conversion of energy forms and energy not being able to get lost.

The conservation of momentum is most suited to the collision. This law states that when a collision occurs the initial momentum before and after the collision is the same(without any external force).

Answer:The amount of energy released by fusion in the Sun's core equals the amount of energy radiated from the Sun's surface into space.

Explanation: Energy balance is a used to describe the balance between the amount of energy input and the amount of energy output. When the amount of energy released is Equal to the amount of energy given back to the System we say there is an energy balance.

The Sun is the major source of energy in the Universe, and it produces energy through FUSION OF RADIOACTIVE MATERIALS IN ITS CORE, WHEN THE ENERGY RELEASED BY FUSION THE SUN IS EQUAL TO THE ENERGY RELEASED TO THE OUTER SPACE WE SAY THERE IS ENERGY BALANCE.

Answer:

The electric field between the plates is 2173 N/C

Explanation:

Given that,

Distance = 0.150 cm

Suppose,The initial speed of the electron is [tex]7.05\times10^6\ m/s[/tex]. The capacitor is 2.00 cm long,

We need to calculate the time

Using formula of time

[tex]t=\dfrac{d}{v}[/tex]

Put the value into the formula

[tex]t=\dfrac{2.00\times10^{-2}}{7.05\times10^{6}}[/tex]

[tex]t=2.8\times10^{-9}\ s[/tex]

We need to calculate the acceleration

Using equation of motion

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

[tex]a=\dfrac{2s}{t^2}[/tex]

Put the value into the formula

[tex]a=\dfrac{2\times0.150\times10^{-2}}{(2.8\times10^{-9})^2}[/tex]

[tex]a=3.82\times10^{14}\ m/s^2[/tex]

We need to calculate the electric field between the plates

Using formula of electric field

[tex]E=\dfrac{F}{q}[/tex]

[tex]E=\dfrac{ma}{q}[/tex]

Put the value into the formula

[tex]E=\dfrac{9.1\times10^{-31}\times3.82\times10^{14}}{1.6\times10^{-19}}[/tex]

[tex]E=2173\ N/C[/tex]

Hence, The electric field between the plates is 2173 N/C

Answer:

The answer to the question is

The luminosity of stars A is four times that of star B

Explanation:

Flux (F) produced by a source of light is directly proportional to the brightness or Luminosity (L), and varies inversely to its distance d, that is [tex]F \alpha \frac{L}{d^2}[/tex]

Therefore if the two stars present the same flux then we have

[tex]\frac{L_1}{d_1^2} = \frac{L_2}{(2d_1)^2}[/tex] then crossing out like terms gives [tex]\frac{L_1}{1} = \frac{L_2}{2^2}[/tex] or 4·L₁ = L₂

The luminosity of  star A is 4 times the that of star B

Final answer:

Star A, being twice as far away from Earth as Star B but appearing equally bright, must have a luminosity that is four times greater than Star B's luminosity, due to the inverse square law of light.

Explanation:

When comparing the luminosities of two stars that appear equally bright from Earth, but one is twice as far away as the other, we must account for the inverse square law of light. This law dictates that the intensity (brightness) of light from a source (in this case, a star) decreases proportionally to the square of the distance from the source.

Therefore, if Star A is twice as far away from Earth as Star B, yet they appear to have the same brightness, Star A must have a luminosity four times greater than that of Star B. This is because, to compensate for the increased distance, Star A must emit more light to be perceived as equally bright as Star B from Earth.

Final answer:

To calculate the flashlight's power usage, we use power and current formulas based on given charge and voltage values. Similarly, to estimate battery life for a flashlight bulb, we divide the battery's Amp-Hour capacity by the bulb's current draw.

Explanation:

The question is about determining the average current used by a flashlight bulb over a time period and how long the battery would last. To calculate the power usage, we apply the formula P = IV, where P is the power in watts, I is the current in amperes, and V is the voltage in volts. Given that 600 C (coulombs) of charge passes through the flashlight in 0.500 hours (which is 1800 seconds) and the voltage is 3.00 V, we can find the average current using the formula I = Q/t, where Q is the charge in coulombs and t is the time in seconds.

To find out how long a battery will keep a flashlight bulb burning, we divide the battery capacity in ampere-hours (Ah) by the current drawn by the bulb. If a 1.00-W bulb is used and the battery is rated at 1.00 Ah and 1.58 V, we first need to use the power formula P = IV to find the current drawn by the bulb.

The flashlight bulb that draws 0.60 A can be lit for approximately 1.67 hours using a 1.00 Ah alkaline battery. This is calculated by dividing the battery capacity by the current drawn by the bulb.

To determine how long a flashlight bulb that draws 0.60 A can be lit, we must consider the battery's capacity and voltage. Here is a step-by-step explanation:

First, identify the battery capacity. Assume we have an alkaline battery rated at 1.00 Ah.

Next, understand that 1.00 Ah means the battery can supply 1.00 ampere for 1 hour.

Since the flashlight draws 0.60 A, we calculate the time the battery can light the bulb using the formula:

Time (hours) = Battery Capacity (Ah) / Current (A)

Substitute the values:

Time = 1.00 Ah / 0.60 A = 1.67 hours

Thus, the flashlight bulb can be lit for approximately 1.67 hours or 1 hour and 40 minutes before the battery is depleted.

Yes, a vibrating proton would produce an electromagnetic wave, as accelerating charges emit radiation. This principle is central to many technological and scientific applications, including radio transmissions and the study of galactic structures in astronomy.

Accelerating charges such as protons, when they vibrate, indeed produce electromagnetic waves. This effect is due to the fact that a changing electric field generates a magnetic field, and a changing magnetic field, in turn, generates an electric field. As a proton oscillates, it experiences acceleration and therefore can emit radiation. This principle is extensively used in various technologies, like radio transmission, where an alternating current in an antenna accelerates charges and creates electromagnetic waves.

The production and detection of electromagnetic waves are crucial in many fields, including communications and astronomy. Just like an electron, a proton is also a spin 1/2 particle with a magnetic moment and can emit radiation that can be detected, such as the 21-cm line in the hydrogen spectrum, which allows astronomers to map the spiral arms of galaxies.

Considering the mass and charge of particles, a vibrating proton can generate electromagnetic radiation, albeit at different frequencies compared to electrons due to their larger mass. This forms the basis of nuclear magnetic resonance (NMR) utilized in various scientific and medical applications.

Answer:

Explanation:

Applying the exponential function of decay

M=Cexp(-kt)

At t =0 the mass is 13mmg

Therefore

13=Cexp(0)

C=13

M=13exp(-kt)

After 12mins, M=8mmg

8=13exp(-K×12)

8/13=exp(-12k)

0.615=exp(-12k)

Take In of both sides

In(0.615)=-12k

-0.4855=-12k

Then, k=0.0405

Then the equations become

M=13exp(-0.0405t)

We need to find t at M=2mmg

M=13exp(-0.0405t)

2=13exp(-0.0405t)

2/13=exp(-0.0405t)

0.1538=exp(-0.0405t)

Take In of both sides

In(0.1538)=-0.0405t

-1.872=-0.0405t

Then t=-1.82/-0.0405

t=46.22mintes

Answer:

4. Both points have the same instantaneous angular velocity

Explanation:

Angular velocity is a measure of the the number of rotations per unit time. This does not depend on the radius of the wheel. Hence, all points on the wheel have the same angular velocity. This invalidates option 1.

The centripetal acceleration is given by the product to the square of the angular velocity and the radius or distance from the centre. A and B are located at different distances from the centre. Hence, they have different centripetal acceleration. This invalidates option 2.

The tangential acceleration depends on the linear velocity which itself is a product of the angular velocity and the distance from the centre. Hence, it is different for both points because they are at different distances from the centre.

Since both A and B are fixed points on the wheel, they move through equal angles in the same time. In fact, for any other fixed point, they all move through the same angle in the same time. This invalidates option 5.

Final answer:

All points on a rotating wheel share the same instantaneous angular velocity; however, points farther from the axis will experience greater centripetal acceleration. The correct statement is that both points have the same instantaneous angular velocity.

Explanation:

The question concerns the properties of points located at different radii of a rotating wheel, specifically relating to angular velocity, centripetal acceleration, and tangential acceleration. We can address the situation by considering the principles of circular motion. When a rigid wheel is rotating about a fixed axis, all points on the wheel have the same instantaneous angular velocity since every point on the wheel rotates through the same angle in the same amount of time.

Centripetal acceleration is proportional to the radius and the square of the angular velocity. Since point A, being at the rim, is farther from the axis than point B, point A experiences a greater centripetal acceleration. On the other hand, tangential acceleration is related to the angular acceleration and the radius. If the wheel is rotating with decreasing angular velocity, both point A and B experience the same tangential acceleration because it is a property of the wheel's rotation, not the points' individual locations.

The correct statement in this scenario is that both points have the same instantaneous angular velocity, which makes option 4 the true statement.

Answer:

[tex]\omega_f = 1 rad/s[/tex]

Explanation:

Given,

After two revolutions, angular speed = 0.5 rev/s

Angular speed after 8 revolution = ?

Using equation of circular motion

[tex]\omega_f^2 = \omega_0^2 + 2\alpha \theta[/tex]

[tex]0.5^2 =0^2 + 2\times \alpha\times 2[/tex]

[tex]\alpha = 0.0625 rad/s^2[/tex]

Now, Angular speed after 8 revolution

[tex]\omega_f^2 = \omega_0^2 + 2\alpha \theta[/tex]

[tex]\omega_f^2 =0^2 + 2\times 0.0625 \times 8[/tex]

[tex]\omega_f = 1 rad/s[/tex]

Hence, the angular speed after 8 revolution is equal to 1 rad/s.

The angular acceleration of the fan blades is 0.25 rev/s². Using this constant angular acceleration, the angular speed of the blades after 8 revolutions is calculated to be 2 rev/s.

To solve this problem, we use the equation of motion for rotational systems.

Δω = α * Δθ Where Δθ is the change in angle (in rev), Δω is the change in angular speed (in rev/s), and α is the angular acceleration (in rev/s²). Given that the blades have an angular speed of 0.5 rev/s after 2 revolutions, we can solve for α: α = Δω / Δθ = (0.5 rev/s) / (2 rev) = 0.25 rev/s²

Then, knowing this constant angular acceleration, we can find the angular speed after 8 revolutions: Δω = α * Δθ

Δω = (0.25 rev/s²) * (8 rev) = 2 rev/s

So after 8 revolutions, the angular speed of the ceiling fan is 2 rev/s.

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Answer:

It will require 14.715 N of force to hold the cartoon beneath the water.

Explanation:

Given the the volume of cartoon is 1.5 liters.

We need to find the force required to hold this cartoon beneath the water.

As we know from the Archimedes principle that the net force is equal to the volume of liquid displaced.

Given volume of the cartoon is 1.5 liters. So, 1.5 liters of water will be displaced.

And we know the density of the water is [tex]1000\ kg/m^3[/tex]. That is [tex]1\ kg/L[/tex]

And [tex]g=9.81\ m/s^2[/tex]

[tex]F_N=\rho Vg\\F_N=1\times 1.5\times 9.81\\F_N=14.715\ N[/tex]

So, it will require 14.715 N of force to hold 1.5 liter volume of cartoon beneath the water.

Answer:

The total velocity of the ball will be 14.14 m/s.

Explanation:

Horizontal Velocity component = 10 m/s

Vertical Velocity component = -10 m/s

Total velocity of the ball will be found from the following equation:

(Total velocity) ^2 = (Horizontal Velocity) ^2 + (Vertical Velocity) ^2

Total Velocity ^2 = 10^2 + (-10)^2

Total Velocity^2 = 100 + 100

Total Velocity = [tex]\sqrt{200}[/tex]

Total Velocity = 14.14 m/s