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Kipish [7]
3 years ago
8

A man does 4,475 J of work in the process of pushing his 2.40 103 kg truck from rest to a speed of v, over a distance of 26.5 m.

Neglecting friction between truck and road, determine the following.
(a) the speed
(b) the horizontal force exerted on the truck
Physics
1 answer:
velikii [3]3 years ago
7 0

Answer:

1.9311 m/s

168.86792 N

Explanation:

v = Final velocity

m = Mass of truck = 2.4\times 10^3\ kg

s = Displacement = 26.5 m

Work done is given by

W=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow W=\dfrac{1}{2}m(v^2-0^2)\\\Rightarrow v=\sqrt{\dfrac{2W}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 4475}{2.4\times 10^3}}\\\Rightarrow v=1.9311\ m/s

The speed is 1.9311 m/s

Work done is given by

W=Fs\\\Rightarrow F=\dfrac{W}{s}\\\Rightarrow F=\dfrac{4475}{26.5}\\\Rightarrow F=168.86792\ N

The horizontal force is 168.86792 N

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Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

v_0-gt = 0.

where v_0 is the initial velocity.

Solving for t we get

t = \dfrac{v_0}{g}

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time t_0 since it had reached the highest point, the ball has traveled downwards and the velocity v_f it has gained is

v_f = gt_0,

and we are told that this is twice the initial velocity v_0; therefore,

v_f = 2v_0  = gt_0

which gives

t_0 = \dfrac{2v_0}{g}.

Thus, the total time taken to reach velocity 2v_0 is

t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}

t_{tot} = \dfrac{3v_0}{g}.

This t_{tot}, we are told, is 36 seconds; therefore,

36= \dfrac{3v_0}{g},

and solving for v_0 we get:

v_0 = \dfrac{36g}{3}

v_0 = \dfrac{36s(10m/s^2)}{3}

\boxed{v_0 = 120m/s}

which from the options given is choice e.

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3 years ago
A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te
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Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

\Delta L=\alpha L_{0}\Delta T

Where:

  • L(0) is the initial length
  • ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
  • ΔL is the final length minus the initial length (L(f)-L(0))
  • α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)  

So, we have:

L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})

L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})

L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)

L_{f}=0.117\: m

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

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3 years ago
A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t
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Answer:

v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t

Second stone:

y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

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3 years ago
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Answer:

The statements, observations, beliefs and suppositions all are the components of the pseudoscience.

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The claims included in the pseudoscience including the beliefs, statements and practices are claimed to be scientific but these are devoid of any scientific evidences provided by the experimental procedures.

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