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3 years ago

As the car falls off the cliff, what is happening to the kinetic energy of the falling car?

Physics

1 answer:

Airida [17]3 years ago

7 0

Answer:

The kinetic energy is increasing as potential energy is converted.

Explanation:

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What is the density of the block if it has a mass of 45.5 grams and a volume of 2.4 cm³?

rjkz [21]

Explanation:

density= mass/volume

therefore;45.5/2.4

=19.0 g/cm³

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3 years ago

Rank the following in terms of increasing inertia:

Naddik [55]

C. A 1200kg car is going 15m/s

7 0

3 years ago

Read 2 more answers

already did the work, i just need someone to see if i did the tangent line right? the line has to touch the point 0.6! thank you

Dmitrij [34]

The tangent looks good.

The curve is a bit crooked, at the 0.9 and 1.

But overall, cool graph.

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3 years ago

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

icang [17]

Answer:

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

Explanation:

There are two forces acting on the box along the horizontal direction:

- The force of push applied by the worker, in the forward direction, F

- The force of friction, $F_f$ , acting in the opposite direction (backward)

So the net force acting on the box is

$F_{net}=F-F_f$

According to Newton's second law of motion, the net force on an object is equal to the product between its mass (m) and its acceleration (a), so we can write:

$F_{net}=ma$

And so

$F-F_f = ma$

However, in this case the box is moving at constant speed; this means that its acceleration is zero:

$a=0$

Therefore we have:

$F-F_f=0$

Which means

$F_f=F$

And since we are told that

$F=25 N$

This means that the force of friction is also 25 N:

$F_f=F=25 N$

6 0

4 years ago

Read 2 more answers

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