What is the acceleration of a 5kg mass pushed by a 10N force?

Answer:

Explanation:

A ) The surface is frictionless . If the box is at rest , there will be no tension in the rope.

B ) In this case , the box is moving with steady rate of 4.20 m /s . In this case also no force is acting on the box so tension in the rope will be nil.

C ) In this case the box is moving with acceleration of 5.8 m /s² so force on the box = mass x acceleration

= 57 x 5.8

= 330.6 N .

So tension in the rope will be equal to force acting on the rope . Hence tension = 330.6 N .

Final answer:

The tension in the rope tied to a 57.0 kg box on frictionless ice is 0 N when the box is at rest, 0 N when moving at a steady velocity of 4.20 m/s, and 331 N when the box is accelerating at 5.80 m/s².

Explanation:

When considering the tension in the rope connected to the 57.0 kg box on frictionless ice, different scenarios will result in different tensions:

Part A: Box at Rest

In the first scenario where the box is at rest, the tension in the rope will be zero Newtons (0 N). This is because, on frictionless ice, there is no other force opposing the box's motion that the rope would need to counteract.

Part B: Box Moving at Steady Velocity

For the second part, where the box is moving at a constant velocity (4.20 m/s), the tension is still zero Newtons. Even if the box is in motion, the lack of frictional forces on the ice means no net force is required to maintain the box's constant velocity.

Part C: Box Accelerating

In the third case, where the box is moving at 4.20 m/s and also accelerating at 5.80 m/s², we need to apply Newton's second law, F = ma. The tension (T) in the rope is calculated as the product of the mass (m) of the box and its acceleration (a). Thus:

T = m × a = 57.0 kg × 5.80 m/s² = 330.6 N

Therefore, the tension in the rope would be 331 N (rounded to the nearest integer).

Answer:

The change in temperature is [tex]\Delta T = 1795 K[/tex]

Explanation:

From the question we  are told that

   The temperature coefficient is  [tex]\alpha = 4 * 10^{-3 }\ k^{-1 }[/tex]

The resistance of the filament is mathematically represented as

           [tex]R = R_o [1 + \alpha \Delta T][/tex]

Where [tex]R_o[/tex] is the initial resistance

Making the change in temperature the subject of the formula

     [tex]\Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ][/tex]

Now from ohm law

           [tex]I = \frac{V}{R}[/tex]

This implies that current varies inversely with current so

           [tex]\frac{R}{R_o} = \frac{I_o}{I}[/tex]

Substituting this we have

       [tex]\Delta T = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ][/tex]

From the question we are told that

    [tex]I = \frac{I_o}{8}[/tex]

Substituting this we have

   [tex]\Delta T = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ][/tex]

=>     [tex]\Delta T = \frac{1}{3.9 * 10^{-3}} (8 -1 )[/tex]

        [tex]\Delta T = 1795 K[/tex]

The lamp filament's temperature rise can be calculated using the equations for resistance and the current. Given the temperature coefficient of resistivity and the ratio of initial and final currents, the change in temperature is calculated to be 875K.

The problem is solved by using the formula for the change in resistance due to temperature, R = R0(1 + αΔT), where R is the final resistance, R0 is the initial resistance, α is the temperature coefficient of resistivity, and ΔT is the change in temperature. We can relate the initial and final currents (I and If) using Ohm’s law: V = I*R = If Rf, where V is the constant voltage. Given that If = I/8, and substituting from R = R0(1 + αΔT), we get that ΔT = [(8 - 1)/(α*7)] = 875K.

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Answer:

t = 3.38 s

Explanation:

We have,

Initial speed of the ball that leaves the student's hand is 16 m/s

Initially, the hand is 1.90 m above the ground.

It is required to find the time for which the ball in the air before it hits the ground. We can use the equation of kinematics as :

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

Here, [tex]y_f=-1.9\ m, y_i=0[/tex] and a=-g

The equation become:

[tex]-1.9=16t-\dfrac{1}{2}\times 9.8t^2[/tex]

After rearranging we get the above equation as :

[tex]4.9t^2-16t-1.9=0[/tex]

It is a quadratic equation, we need to find the value of t. On solving the above equation, we get :

t = -0.115 s and t = 3.38 s (ignore t = -0.115 s )

So, the ball is in air for 3.38 seconds before it hits the ground.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:all of the above

Explanation:

because thoughts relate to your emotions and your behaviors reflect to your emotions

Mental health is a reflection of one's thoughts, emotions, and behaviors. Positive thinking, managed emotions and constructive behaviors contribute to a healthy mental state.

Your mental health is indeed a reflection of your thoughts, emotions and behaviors. Accumulative thoughts and feelings can shape our mental well-being to a great extent. Similarly, our behaviors, influenced by these thoughts and feelings, can either nurture or deteriorate our mental health. A balance in all three – positive thinking, managing emotions and constructive behaviors lead to a healthy mental state.

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Answer:2.4m

Explanation:

Velocity=360m/s

Frequency=150Hz

Wavelength=velocity ➗ frequency

Wavelength=360 ➗ 150

Wavelength=2.4m

Final answer:

The wavelength of a sound wave traveling at 360 m/s and a frequency of 150 Hz is 2.4 m.

Explanation:

The wavelength of a sound wave can be calculated using the equation:

λ = v / ƒ

where λ is the wavelength, v is the speed of sound, and ƒ is the frequency of the sound wave. Given that the speed of sound is 360 m/s and the frequency is 150 Hz, we can substitute these values into the equation:

λ = 360 m/s / 150 Hz = 2.4 m

Therefore, the wavelength of the sound wave is 2.4 m.

Answer:

A. The force exerted by the truth on the car is the same size as the force exerted by the car on the truck: Ftruck = Fcar

Explanation:

Both vehicles will experience equal both opposite forces one on the other.

Answer:

[tex]n_{T} = 31.68\,rev[/tex]

Explanation:

The angular acceleration is:

[tex]\ddot n_{1} = \frac{2.2\,\frac{rev}{s} -0\,\frac{rev}{s} }{8.8\,s}[/tex]

[tex]\ddot n_{1} = 0.25\,\frac{rev}{s^{2}}[/tex]

And the angular deceleration is:

[tex]\ddot n_{2} = \frac{0\,\frac{rev}{s}-2.2\,\frac{rev}{s} }{20\,s}[/tex]

[tex]\ddot n_{2} = -0.11\,\frac{rev}{s^{2}}[/tex]

The total number of revolutions is:

[tex]n_{T} = n_{1} + n_{2}[/tex]

[tex]n_{T} = \frac{\left(2.2\,\frac{rev}{s} \right)^{2}-\left(0\,\frac{rev}{s} \right)^{2}}{2\cdot \left(0.25\,\frac{rev}{s^{2}} \right)} + \frac{\left(0\,\frac{rev}{s} \right)^{2}-\left(2.2\,\frac{rev}{s} \right)^{2}}{2\cdot \left(-0.11\,\frac{rev}{s^{2}} \right)}[/tex]

[tex]n_{T} = 31.68\,rev[/tex]

Answer:

492.6 kN

Explanation:

Since the window as a radius of 2 m, its diameter is thus 4 m. Since the diving pool is 8 m deep, the height of water from the top of the window to the bottom of the pool is 8 - 4 = 4 m. The actual pressure acting on the window is thus

P = ρgh were ρ = density of water = 1000 kg/m³ g = 9.8 m/s² and h = 4 m

P = 1000 kg/m³ × 9.8 m/s² × 4 m = 39200 N/m².

Since P = F/A were F = force and A = area,

F = PA were A = area of window = πr² = π2² = 4π m² = 12.57 m²

F = 39200 N/m² × 12.57 m² = 492601.73 N = 492.6 kN

The force on the window of a diving pool due to the water pressure can be calculated by multiplying the pressure at the depth of the pool by the window's surface area. The pressure is found using density, gravity and depth, and the area is calculated from the radius of the circular window. The force comes out to be about 987 KN.

This problem deals with hydrostatic pressure in a diving pool. The force on the window can be calculated by multiplying the pressure at the depth of the window by the surface area of the window.

First, we calculate the pressure at the depth of the pool using the formula P = ρgh, where P is the pressure, ρ (rho) is the density of the water (typically 1000 kg/m³ for fresh water), g is acceleration due to gravity (9.81 m/s²), and h is the depth (8m). This yields P = 1000 kg/m³ * 9.81 m/s² * 8m = 78480 Pa (Pascal).

Second, we calculate the area of the circular window. The area A of a circle is given by πr², where r is the radius. In this case, the radius r is 2m, so A = π * (2m)² = 12.57 m².

Finally, we calculate the force F by multiplying the pressure P by the area A. This gives F = P * A = 78480 Pa * 12.57 m² = 986857.6 N or approximately 987 KN.

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Answer:

8.33*10^-16 Watt

Explanation:

Given that

Length of the rod, l = 2 m,

Area of the rod, A = 2 x 2 mm² = 4*10^-6 m²

resistivity of the rod, p = 6*10^-8 ohm metre,

Potential difference of the rod, V = 0.5 V

Let R be the resistance of the rod, then

R = p * l / A

R = (6*10^-8 * 2) / (4*10^-6)

R = 3*10^14 ohm

Heat generated per second = V² / R Heat = (0.5)² / (3*10^14)

Heat = 0.25 / 3*10^14

Heat = 8.33*10^-16 Watt

Therefore, the rate at which heat is generated is 8.33*10^-16 Watt

The current density of the wire is approximately[tex]2.04 x 10^6 A/m²[/tex]elocity of the electrons in the wire is approximately [tex]1.51 x 10^-3 m/s.[/tex]

To find the magnitude of the current density, we need to calculate the cross-sectional area of the wire and divide the current by that area. The current density (J) is given by J = I/A, where I is the current and A is the cross-sectional area. Since the wire is cylindrical, we can use the formula for the area of a circle, A = πr², where r is the radius of the wire.

Given that the diameter of the wire is [tex]1.02mm[/tex]radius can be calculated as [tex]r = (1.02mm)/2 = 0.51mm = 0.00051m[/tex]

Using this value, the cross-sectional area is [tex]A = π(0.00051m)² = 8.18 x 10^-7 m².[/tex]

Dividing the current of[tex]1.67A[/tex]e cross-sectional area, we get the current density [tex]J = (1.67A)/(8.18 x 10^-7 m²) ≈ 2.04 x 10^6 A/m².[/tex]

To find the drift velocity, we can use the formula v_d = J/(nq), where n is the density of free electrons and q is the charge of an electron. Given that the density of free electrons is [tex]8.5 x 10^28 electrons/m³[/tex]

the charge of an electron is [tex]q = 1.60 x 10^-19 C[/tex]

we can substitute these values into the formula: [tex]v_d = (2.04 x 10^6 A/m²) / (8.5 x 10^28 electrons/m³ * 1.60 x 10^-19 C/electron) ≈ 1.51 x 10^-3 m/s.[/tex]

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The current density in the wire is approximately 2.04 × 10^6 A/m^2, and the drift velocity of the electrons is approximately 1.51 × 10^-4 m/s.

Sure, let's break down the calculation step by step:

Given data:

Diameter of the wire, d = 1.02 mm = 1.02 × 10^-3 m

Current, I = 1.67 A

Power of the lamp, P = 200 W

Density of free electrons, n = 8.5 × 10^28 electrons/m^3

Calculate the cross-sectional area of the wire:

Radius, r = d/2 = 1.02 × 10^-3 m / 2 = 0.51 × 10^-3 m

Cross-sectional area, A = π * r^2

A = π * (0.51 × 10^-3 m)^2 ≈ 8.19 × 10^-7 m^2

Calculate the current density (J):

Current density is the current per unit area.

J = I / A

J = 1.67 A / 8.19 × 10^-7 m^2 ≈ 2.04 × 10^6 A/m^2

Calculate the charge carrier density (n):

Given density of free electrons, n = 8.5 × 10^28 electrons/m^3

Calculate the drift velocity (v_d):

Drift velocity is given by the formula: J = n * e * v_d

Where e is the charge of an electron (approximately 1.6 × 10^-19 C)

Rearranging for drift velocity: v_d = J / (n * e)

Substituting values: v_d = (2.04 × 10^6 A/m^2) / (8.5 × 10^28 electrons/m^3 * 1.6 × 10^-19 C)

v_d ≈ 1.51 × 10^-4 m/s

Answer:

gasoline zone    P_net = 6860 Pa

Water zone        P_net = 26460 Pa

Force direction is out of tank  

Explanation:

The pressure is defined

         P = F / A

         F = P A

let's write Newton's equation of force

            F_net = F_int - F_ext

            P_net A = (P_int - P_ext) A

The P_ext is the atmospheric pressure

         P_ext = P₀

the pressure inside is

gasoline zone

         P_int = P₀ + ρ' g h'

water zone

         P_int = P₀ + ρ' g h' + ρ_water h_water

we substitute

Zone with gasoline

            P_net = ρ' g h'

            P_net = 700 9.8  1

            P_net = 6860 Pa

Water zone

             P_net = rho’ g h’ + rho_water g h_water

             P_net = 6860 + 1000 9.8  2

             P_net = 26460 Pa

To find the explicit value of the force, divide by a specific area.

Force direction is out of tank

Answer:

4.36 rad/s

Explanation:

Radius of platform r = 2.97 m

rotational inertia I = 358 kg·m^2

Initial angular speed w = 1.96 rad/s

Mass of student m = 69.5 kg

Rotational inertia of student at the rim = mr^2 = 69.5 x 2.97^2 = 613.05 kg.m^2

Therefore initial rotational momentum of system = w( Ip + Is)

= 1.96 x (358 + 613.05)

= 1903.258 kg.rad.m^2/s

When she walks to a radius of 1.06 m

I = mr^2 = 69.5 x 1.06^2 = 78.09 kg·m^2

Rotational momentuem of system = w(358 + 78.09) = 436.09w

Due to conservation of momentum, we equate both momenta

436.09w = 1903.258

w = 4.36 rad/s

Answer:

D

Explanation:

The most likely reason that a material that is a poor thermal conductor makes a good oven mitt is because its electrons do not move freely.

The most likely reason that a material that is a poor thermal conductor makes a good oven mitt is because its electrons do not move freely.

In a good thermal conductor, such as metal, the electrons are able to move freely and transfer heat quickly. However, in a poor thermal conductor, such as the material of an oven mitt, the electrons are not able to move freely, which means they cannot transfer heat easily and thus provide insulation.

Therefore, option D: its electrons do not move freely is the most likely reason that a material makes a good oven mitt.

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Answer:

(a) 64.6 kg/ms

(b) 4088.61 N

Explanation:

(a)

I = mΔv.................. Equation 1

Where I = impulse acting on the ball, Δv = change in velocity.

But,

Δv = v-u.............. Equation 2

Where v = final velocity, u = initial velocity.

Substitute equation 2 into equation 1

I = m(v-u).................. Equation 3

Assuming: upwards to be positive

Given: m = 1.9 kg, v = 9 m/s, u = -25 m/s (downward)

Substitute into equation 3

I = 1.9[9-(25)]

I = 1.9(9+25)

I = 1.9(34)

I = 64.6 kgm/s.

(b)

F = I/t................... Equation 4

Where F = Average force on the floor from the ball, t =  time of contact of the floor with the ball.

Given: I = 64.6 kgm/s, t = 0.0158 s

Substitute into equation 4

F = 64.6/0.0158

F = 4088.61 N

Answer:

A) Impulse = 64.6 N.s

B) Force = 4088.6 N

Explanation:

We are given;

Mass of ball;m = 1.9 kg

Initial velocity of ball;u = -25 m/s (negative value because ball was drop downward)

Final velocity of ball;v = 9 m/s

From Newton's second law of motion, we know that;

Impulse = Change in momentum

Thus;

Impulse = final momentum - initial momentum = mv - mu

Thus;

Impulse = m(v - u)

Plugging in the relevant values, we have;

Impulse = 1.9(9 - (-25)

Impulse = 1.9(9 + 25)

Impulse = 1.9(34)

Impulse = 64.6 N.s

B) As said earlier,

impulse = 64.6 N.s

Now, we are looking for the magnitude of the average force.

Now, Impulse is also expressed as Force x time.

Thus,

Force x time = 64.6 N.s

We are given time as 0.0158 s

Thus, making Force the subject of the formula, we now have;

Force = 64.6/0.0158

Force = 4088.6 N

Answer:

B = 0.204T

Explanation:

To find the value of the magnetic force you use the following formula:

[tex]F_B=ILBsin\theta[/tex]

I: current of the copper rod = 45A

B: magnitude of the magnetic field

L: 0.78m

you assume that magnetic field B and current I are perpendicular between them.

The magnetic force must be, at least, equal to the friction force, that is:

[tex]F_{f}=F_{B}\\\\\mu N=\mu Mg=ILB\\\\B=\frac{\mu Mg}{IL}[/tex]

M: mass of the rod = 1.20kg

μ: coefficient of static friction = 0.61

g: gravitational acceleration constant = 9.8m/s^2

By replacing the values of the parameters you obtain:

[tex]B=\frac{(0.61)(1.20kg)(9.8m/s^2)}{(45A)(0.78m)}=0.204T[/tex]

Answer:

The magnitude of the smallest magnetic field is [tex]B = 0.1744 \ T[/tex]  

Explanation:

  From the question we are told that

      The mass of the copper is  [tex]m = 1.20 kg[/tex]

       The distance of separation for the rails is  [tex]d = 0.78 \ m[/tex]

        The current is  [tex]I = 45 A[/tex]

        The coefficient of static friction is [tex]\mu = 0.61[/tex]

The force acting along the vertical axis is mathematically represented as

        [tex]F = mg - F_y[/tex]

Where  [tex]F_y[/tex] is the force acting on copper rod due to the magnetic field generated this is mathematically represented as

         [tex]F_y = I * d * B_1[/tex]

The magnetic field here is  acting towards the west because according to right hand rule magnetic field acting toward the west generate a force acting in the vertical axis

So the equation becomes

         [tex]F = mg - I * d * B_1[/tex]

Here [tex]B_1 = B sin \theta[/tex]

          [tex]F = mg - I * d * Bsin(\theta )[/tex]

 The in the horizontal axis is mathematically represented as

         [tex]F_H = ma + F_x[/tex]

 Since the rod is about to move it acceleration is  zero

     Now [tex]F_x[/tex] is the force acting in the horizontal direction due to the magnetic field acting downward this is because a  according to right hand rule magnetic field acting downward generate a force acting in the horizontal positive horizontal direction.  this mathematically represented as

         [tex]F_H = 0 + I * d * B_2[/tex]

So the equation becomes

            [tex]F_H = I * d * B_2[/tex]

Here   [tex]B_2 = B cos \theta[/tex]

          [tex]F_H = I * d * Bcos (\theta)[/tex]

    Now the frictional force acting on this rod is mathematically represented as

          [tex]F_F = \mu * F[/tex]

         [tex]F_F = \mu * (mg -( I * d * Bsin(\theta )))[/tex]

Now when the rod is at the verge of movement

          [tex]F_H = F_F[/tex]

So     [tex]I * d * Bcos (\theta) = \mu * (mg -( I * d * Bsin(\theta )))[/tex]

=>    [tex]B = \frac{\mu mg }{I * d (cos \theta + \mu (sin \theta ))}[/tex]

   Now [tex]\theta[/tex] is the is the angle of the magnetic field  makes with the vertical and the horizontal and this can be mathematically evaluated as

                [tex]\theta = tan^{-1} (\mu )[/tex]

Substituting value

                [tex]\theta = tan^{-1} ( 0.61 )[/tex]

                [tex]\theta = 31.38^o[/tex]

Substituting values into the equation for B

          [tex]B = \frac{0.61 (1.20) (9.8)}{(45) (0.78) (cos (31.38) + 0.61 (sin (31.38)) )}[/tex]

               [tex]B = 0.1744 \ T[/tex]

Answer:

[tex]D_{1}=11.32 m[/tex]    

Explanation:

We will need to use the ideal gas equation. The equation is given by:

[tex]PV=nRT[/tex]

As we have the same amount of molecules in the initial and final steps, therefore we can do this:

[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex] (1)

- P(1) is the atmospheric pressure (P(1) = 1 atm) and P(2) is 0.028 atm

- T(1) is 300 K and T(2) is 190 K

- V(1) is the volume of the balloon in the first step, we can consider a spherical geometry so:

[tex]V_{1}=\frac{4}{3}\pi (\frac{D_{1}}{2})^{3}[/tex] (2)

[tex]V_{2}=\frac{4}{3}\pi (\frac{D_{2}}{2})^{3}[/tex] (3)

- D(2) = 32 m

So [tex]V_{2}=17157.3 m^{3}[/tex]

Let's solve the equation (1) for V(1)

[tex]V_{1}=\frac{T_{1}P_{2}V_{2}}{P_{1}T_{2}}[/tex]

[tex]V_{1}=\frac{300*0.028*17157.3}{1*190}[/tex]

[tex]V_{1}=758.53 m^{3}[/tex]

And using the equation (2) we can find D.    

[tex]D_{1}=2(\frac{3}{4}V_{1})^{1/3}[/tex]

[tex]D_{1}=2(\frac{3}{4\pi}*758.53)^{1/3}[/tex]        

[tex]D_{1}=11.32 m[/tex]        

I hope it helps you!

Answer:

1.56 × 10^-3 cm.

Explanation:

So, we are given the following parameters from the question above;

Length = 3.67 cm, breadth = 2.93 cm, and the number of embedded transistors = 3.5 million.

Step one: find the area of the computer chip.

Therefore, Area = Length × breadth.

Area = 3.67 cm × 2.93 cm.

Area of the computer chip = 10.7531 cm^2. = 10.75 cm^2.

Step two: find the area of one transistor

The area of one transistor is; (area of the computer chip) ÷ (number of embedded transistors).

Hence;

The area of one transistor= 10.7531/4.4 × 10^6.

The area of one transistor= 2.44 × 10^-6 cm^2.

=> Note that We have our transistors as square, therefore;

The maximum dimension = √ (2.44 × 10^-6) cm^2.

The maximum dimension= 1.56 × 10^-3 cm.

Answer:40000kg•m/s

Explanation:

momentum=20000kg•m/s

Since they velocity is doubled

Momentum would also be doubled, therefore momentum would be 40000kg•m/s

Answer:

C: 40,000 kg · m/s

Explanation:

I got it right on edg :)

Answer:

Force of friction is (-30 N).

Explanation:

The force applied on the  box across the floor is 30 N.

The force of gravity is (-8 N) and the the normal force is 8 N.

It is based on Newton's third law of motion. Newton's third law of motion states that the force acting on object 1 to object 2 is equal in magnitude of the force from object 2 to 1 but in opposite direction.

Here there is force of 30 N is applied in horizontal direction. The frictional force act in opposite direction. So, the force of friction is -30 N so that box across the floor.

Answer:

The third force  = [tex]-(26.0 \, N \, \hat{i} + 12.00 \, N \, \hat{j})[/tex]

Explanation:

Here we note that, the formula for the resultant force is as follows;

[tex]\Sigma F = \Sigma F_x + \Sigma F_y[/tex]

Also

∑F = m×a

Where:

[tex]F_x[/tex] = Force components in the x direction

[tex]F_y[/tex] = Force components in the y direction

∑F = Resultant force vector

m = Mass of the object

a = Acceleration vector ob the object =

[tex]a = 8.00 \, m/s^2 \, \hat{i} + 6.00 \, m/s^2 \, \hat{j}[/tex]

[tex]F_1 = 30.0 \, N \, \hat{i} + 16.0 \, N \, \hat{j}[/tex]

[tex]F_2 = 12.0 \, N \, \hat{i} + 8.00 \, N \, \hat{j}[/tex]

Therefore, since ∑F = m×a, we have;

[tex]\Sigma F = 2 kg \times (8.00 \, m/s^2 \, \hat{i} + 6.00 \, m/s^2 \, \hat{j})[/tex]

[tex]\Sigma F = (16.00 \, m/s^2 \, \hat{i} + 12.00 \, m/s^2 \, \hat{j})[/tex]

Hence from [tex]\Sigma F = \Sigma F_x + \Sigma F_y[/tex], we have;

[tex]F_{x1} + F_{x2} + F_{x3} = 16[/tex]

That is 30 + 12 + [tex]F_{x3}[/tex] = 16

∴  [tex]F_{x3}[/tex] = 16 - (30 + 12) = -26

Similarly,

[tex]F_{y1} + F_{y2} + F_{y3} = 12[/tex]

Therefore,  [tex]F_{y3}[/tex] = 12 - (16 + 8) = -12

Hence, [tex]F_3 = -26.0 \, N \, \hat{i} - 12.00 \, N \, \hat{j} = -(26.0 \, N \, \hat{i} + 12.00 \, N \, \hat{j})[/tex]

The third force, [tex]F_3, = -(26.0 \, N \, \hat{i} + 12.00 \, N \, \hat{j})[/tex]

The third force applied on the object calculated using Newton's second law provides a magnitude of (-26.00 N î - 12.00 N ĵ).

The subject of this problem is Newton's second law (F=ma), where F is the total force, m is the mass of the object, and a is acceleration. Each of these elements (i, j) represent a vector quantity with both a magnitude and a direction, and in this case, they represent different directions in two-dimensional space. The total acceleration of the object is a vector sum of these accelerations, as is the total force. Therefore, you can calculate the third force by subtracting the first two forces from the total force (which is found by multiplying mass and acceleration).  

Total Force (F) = m*a = (2 kg)*(8.00m/s² î + 6.00m/s² ĵ) = 16.00 N î + 12.00 N ĵ.

We know two forces, F1 = 30.00 N î + 16.00 N ĵ and F2 = 12.00 N î + 8.00 N ĵ. Adding them, we get F1 + F2 =  42.00 N î + 24.00 N ĵ.

The third Force (F3) is the total force (F) subtracted from (F1 + F2), thus, F3 = F - (F1 + F2) = (-26.00 N î - 12.00 N ĵ).

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Answer:

The maximum amount of work is  [tex]W = 1563.289 \ J[/tex]

Explanation:

From  the question we are told that

   The temperature of the environment is  [tex]T = 280\ K[/tex]

    The volume of container A is  [tex]V_A = 2 m^3[/tex]

    Initially the number of moles  is  [tex]n = 1.2 \ moles[/tex]

     The volume of container B is [tex]V_B = 3.5 \ m^3[/tex]

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

             [tex]W = P_A V_A ln[ \frac{V_B}{V_A} ][/tex]

Now from the Ideal gas law

          [tex]P_A V_A = nRT[/tex]

So substituting for [tex]P_A V_A[/tex] in the equation above

          [tex]W = nRT ln [\frac{V_B}{V_A} ][/tex]

Where R is the gas constant with a values of  [tex]R = 8.314 \ J/mol[/tex]

Substituting values we have that

            [tex]W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ][/tex]

          [tex]W = 1563.289 \ J[/tex]

The maximum amount of work done on the turbine can be found by calculating the change in internal energy of the gas as it reaches equilibrium. The work done on the turbine is given by the difference in pressure between the initial and final states of the gas, multiplied by the change in volume. In this case, the maximum work done on the turbine is (nRT/VA)(VB - VA).

When the valve is opened and the gas flows from container A to container B, the gas will expand and reach equilibrium. Since the process is isothermal, the temperature remains constant at T = 280K. To find the maximum work done on the turbine, we need to calculate the change in internal energy of the gas using the equation ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the gas, and W is the work done on the gas. For an ideal monatomic gas, the internal energy is given by U=1.5nRT, where n is the number of moles, R is the gas constant, and T is the temperature. In this case, we have n = 1.2 moles, V = 2m3 for container A, and V = 3.5m3 for container B. Therefore, the total volume is V = VA + VB = 5.5m3. Using the ideal gas law PV = nRT, we can find the initial pressure of the gas in container A, which is P = nRT/VA. Since the gas expands to fill both containers at equilibrium, the final pressure is P = nRT/V. The work done on the turbine is given by W = P(VB - VA). Plugging in the values, we find that the maximum work done on the turbine is W = (nRT/VA)(VB - VA).

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Answer:(c)

Explanation:

The electric field is a distribution is a distribution of vectors at point due to the presence of one or more charged objects.

If two or more charged particles are present in a system then the net electric field at a point is the vector addition of all the charges.

For example if a negative and a positive charge is present then electric field of negative charge is towards  the negative charge while it is away for positive charge.  

option d and e are wrong as Electric field is a vector quantity

In an electric circuit, resistance and current are ____

A. directly proportional

B. inversely proportional

C. have no effect on each other

Explanation:

A

Answer:

Options A, D and E....make up cell theory

Rudolf Virchow was the first to propose the concept of Omnis cellula-e cellula in relation to cell division.

It is a scientific theory that was proposed in the mid-nineteenth century.

It was proposed by Theodore Schwann a British Zoologist and Matthias Schleiden a German botanist.

The three principles are:

Hence options A, D, E are correct.

To learn more about cell theory and cells here,

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Answer:

The acceleration of its center of mass is  

The frictional force is  [tex]f = 21.65 \ N[/tex]

Explanation:

From the question we are told that

      The mass of the ball is [tex]m = 17 kg[/tex]

     The radius of the ball is [tex]r = 50cm = \frac{50}{100} = 0.5m[/tex]

     the angle with the horizontal is [tex]\theta = 19 ^ o[/tex]

    The of the ball at the base is  [tex]v = 5.0 \ m/s[/tex]

This setup is shown on the first uploaded image

looking at the diagram we see that the force acting on the ball can be mathematically evaluated as

       [tex]mg sin \theta -f = ma[/tex]

Where f is the frictional force

The torque on the ball is mathematically represented as

      [tex]\tau = f * r[/tex]

This torque can also be mathematically represented as

       [tex]\tau = I \alpha[/tex]

where I is the moment of inertia of the ball which is mathematically represented as

              [tex]I = \frac{2}{3} m r^2[/tex]

While [tex]\alpha[/tex]  is the angular acceleration which is mathematically represented as

          [tex]\alpha = \frac{a}{r}[/tex]

So   [tex]\tau = \frac{2}{3} m r^2 * \frac{a}{r}[/tex]

Equating the both formula for torque

        [tex]f * r = \frac{2}{3} m r^2 * \frac{a}{r }[/tex]

  =>  [tex]f = \frac{2}{3} ma[/tex]

Substituting this for f in the above equation

      [tex]mg sin \theta = ma + \frac{2}{3} ma[/tex]

      [tex]g sin \theta = \frac{5}{3} a[/tex]

     [tex]a = \frac{3}{5} * g * sin \theta \alpha[/tex]

Substituting values

     [tex]a = 1.91 m/s^2[/tex]          

    Now substituting into the equation frictional force equation

             [tex]f = \frac{2}{3} * 17 * 1.91[/tex]

             [tex]f = 21.65 \ N[/tex]

Answer:

[tex]a=-1.92 m/s^{2}[/tex]

[tex]F_{f}=-21.76 N[/tex]

Explanation:

We can use the definition of the torque:

[tex]\tau=I\alpha[/tex]

When:

Now, torque is the product of the friction force times the radius.

[tex]F_{f}*R=\frac{2}{3}mR^{2}*\frac{a}{R}[/tex]

[tex]F_{f}=\frac{2}{3}ma[/tex] (1)

Now, let's analyze the force acting over the sphere using the Newton's second law.

[tex]F=ma[/tex]

[tex]-mgsin(\theta)-F_{f}=ma[/tex] (2)  

Let's put F(f) of the equation (1) into the equation (2):

[tex]-mgsin(\theta)-\frac{2}{3}ma=ma[/tex]

[tex]a=-\frac{3}{5}gsin(\theta)[/tex]

[tex]a=-\frac{3}{5}*9.81*sin(19)[/tex]

[tex]a=-1.92 m/s^{2}[/tex]

Hence: [tex]F_{f}=\frac{2}{3}ma=\frac{2}{3}*17*(-1.92)[/tex]

[tex]F_{f}=-21.76 N[/tex]

I hope it helps you!  

Answer:

L = 0 m

Therefore, the cricket was 0m off the ground when it became Moe’s lunch.

Explanation:

Let L represent Moe's height during the leap.

Moe's velocity v at any point in time during the leap is;

v = dL/dt = u - gt .......1

Where;

u = it's initial speed

g = acceleration due to gravity on Mars

t = time

The determine how far the cricket was off the ground when it became Moe’s lunch.

We need to integrate equation 1 with respect to t

L = ∫dL/dt = ∫( u - gt)

L = ut - 0.5gt^2 + L₀

Where;

L₀ = Moe's initial height = 0

u = 105m/s

t = 56 s

g = 3.75 m/s^2

Substituting the values, we have;

L = (105×56) -(0.5×3.75×56^2) + 0

L = 0 m

Therefore, the cricket was 0m off the ground when it became Moe’s lunch.

Answer:

Find the attachments for complete solution

Answer:

The mass of the blood is 5.6286 kg

Explanation:

Given:

V = volume of blood = 5.31 L = 0.00531 m³

ρ = density = 1060 kg/m³

Question: What is the mass of the blood mblood in Jeff's body, m = ?

To calculate the mass of the blood you just have to multiply the density of the blood by the volume occupied by it:

[tex]m=\rho *V=1060*0.00531=5.6286kg[/tex]

Answer:

c. an aperture

Explanation:

Aperture: It relates to the size of the opening, like a doorway, through which light moves into the eye, camera lens or a telescope. In human eye aperture is known as pupil, the black part in the center of the eye. The size of the pupil can increase or decrease depending upon the amount of light available. The same thing happens with a camera as well. The amount of light passing through the lens can be varied by varying the size of the aperture.