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3 years ago

What is the work done by 20 Newton force applied at an angle of 45.0° to move a box a horizontal distance of 40 meters

Physics

2 answers:

ludmilkaskok [199]3 years ago

4 0

Hope this helps! let me know if you need clarification

zzz [600]3 years ago

3 0

Answer:

5.6 × 10^2 joules

Explanation:

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Light traveling though air strikes the surface of the four different materials shown. Which material reflects light but does not

dangina [55]

The spoons .................

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3 years ago

Which of the following situations has more kinetic energy than potential energy?

san4es73 [151]

Answer: I think its C

Explanation:

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2 years ago

A boxed 12.0 computer monitor is dragged by friction 7.50 up along the moving surface of a conveyor belt inclined at an angle of

sasho [114]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below is the solution:

W done by Normal = 0. (make the incline flat, Normal force goes directly up: no work done)
<span>W done by gravity = w*displacement = (11kg*9.8) * 7.5sin(35) = -463J </span>
<span>W done by friction is the opposite of the work done by weight because the object is not moving. Therefore W done by friction = 463J</span>

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3 years ago

A cheetah accelerates from rest to 30m/s in 3 seconds. Calculate the acceleration of the cheetah.

fiasKO [112]

30m/s times 3s= 90 m/s

4 0

3 years ago

After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48

masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

$15min*\frac{1hr}{60min}=0.25hr$

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

$V=\frac{distance}{time}$

the plane traveled a distance to the north of 48km so the velocity is:

$V=\frac{48km}{0.25hr}$

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

$V_{x}=42km/hr*cos(-19^{o} )=39.71 i$

and

$V_{y}=42km/hr*sin(-19^{o} )=-13.67 j$

So the velocity of the wind can be expressed as a vector as:

$V_{wind}=(39.71i - 13.67j)km/hr$

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

$V_{plane x}=V_{plane/wind x}+V_{wind x}$

$V_{plane/wind x}=V_{plane x}-V_{wind x}$

$V_{plane/wind x}=(0-39.71 i)km/hr$

$V_{plane/wind x}= -39.71 i km/hr$

and

$V_{plane y}=V_{plane/wind y}+V_{wind y}$

$V_{plane/wind y}=V_{plane y}-V_{wind y}$

$V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr$

$V_{plane/wind x}= 205.67 j km/hr$

So the velocity of the plane with respect to the wind can be rewritten as:

$V_{plane/wind x}= (-39.71i + 205.67 j) km/hr$

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

$speed=\sqrt{(-39.71)^{2}+(205.67)^{2} }$

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0

3 years ago