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Crazy boy [7]
3 years ago
13

A box of mass 64 kg is at rest on a horizontal frictionless surface. A constant horizontal force F~ then acts on the box and acc

elerates it to the right. It is observed that it takes the box 5.4 seconds to travel 32 meters. What is the magnitude of the force?
Physics
1 answer:
iren2701 [21]3 years ago
3 0

Answer:

The correct answer is;

The magnitude of the force is 35.12 N

Explanation:

To solve the question, we note that the friction is zero and the force causes motion of a stationary mass

One of the equations of motion is required such as

v² = u² + 2× a× s

Where

v = Final velocity = 5.93 m/s

u = Initial velocity = 0 m/s , object at rest

a = acceleration

s = distance moved = 32 meters

But v = Distance/Time = 32 m /5.4 s = 5.93 m/s

Therefore

5.93² = 2×a×32

or a = 35.12/ 64 = 0.55 m/s²

Therefore Force F = Mass m × Acceleration a

Where mass m = 64 kg

Therefore F = 64 kg×0.55 m/s² = 35.12 N

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Answer:

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Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

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In a motor, the combined effect of electric currents and magnetic forces turn electrical energy into which of the following?
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A motor is built to use all those things and produce mechanical energy.
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Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

m_1v_1+m_2v_2 = (m_1+m_2)V_f

Where,

m_{1,2}= Mass of each object

v_{1,2}= Initial Velocity of Each object

V_f= Final Velocity

Our values are given as

m_1 = 2190Kg

v_1 =10.5m/s

m_2 = 3220kg

V_f = 4.74m/s

Replacing we have that

m_1v_1+m_2v_2 = (m_1+m_2)V_f

(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)

v_2 = 0.8224m/s

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0
4 years ago
What is the equation for an inelastic collision
abruzzese [7]
M1 v1 = (m1 + m2)v2.

All of the exponents should be lowered to the bottom right of the letters.
7 0
3 years ago
Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the a
Naily [24]

Answer:

The angular acceleration of the pencil<em> α  = 17 rad·s⁻²</em>

Explanation:

Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:

    τ = I α                              (1)

    W r = I α                          (2)

The weight is that the pencil has is,

   sin 10 = r / (L/2)

   r = L/2(sin(10))

 

The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:

    I = 1/3 M L²

Thus,

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   α  = 3/2(9.8) / 0.150sin(10)

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Therefore, the angular acceleration of the pencil<em> </em>is<em> 17 rad·s⁻²</em>

3 0
3 years ago
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