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3 years ago

3. A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What

Physics

1 answer:

WITCHER [35]3 years ago

6 0

Answer:

Friction between the box and the floor is 25N to the left.

Explanation:

According to Newton's second law of motion, the net force acting on an object is equal to the produce between the object's mass and its acceleration:

$F_{net}=ma$

where

m is the mass of the object

a is its acceleration

In this problem, we have two forces acting on the object:

- The applied force, F = 25 N, to the right

- The force of friction $F_f$ , opposing the motion of the box, so to the left

So we can write the net force as

$F_{net}=F-F_f$

Also, we know that the box is moving at constant speed: this means its acceleration is zero, so

$a=0$

Therefore

$F_{net}=0$

WHich means:

$F-F_f=0$

And therefore,

$F_f=F=25 N$

which means that the force of friction is also 25 N.

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3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

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Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

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3 years ago

You yell into a canyon. You hear the echo in 3 seconds. How long did the sound of your voice travel before bouncing off a cliff?

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Answer:

The sound travelled 516 meters before bouncing off a cliff.

Explanation:

The sound is an example of mechanical wave, which means that it needs a medium to propagate itself at constant speed. The time needed to hear the echo is equal to twice the height of the canyon divided by the velocity of sound. In addition, the speed of sound through the air at a temperature of 20 ºC is approximately 344 meters per second. Then, the height of the canyon can be derived from the following kinematic formula:

$2\cdot h = v\cdot t$ (1)

Where:

$h$ - Height, measured in meters.

$v$ - Velocity of sound, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $v = 344\,\frac{m}{s}$ and $t = 3\,s$ , then the height of the canyon is:

$h = \frac{v\cdot t}{2}$

$h = \frac{\left(344\,\frac{m}{s} \right)\cdot (3\,s)}{2}$

$h = 516\,m$

The sound travelled 516 meters before bouncing off a cliff.

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Answer:

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1 year ago

To complete a project, 200,000 Joules of work is needed. The time taken to complete the project is 20 seconds. How much power is

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Answer:

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Explanation:

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Hope this Helps!

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3 years ago

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