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4 years ago

On a speed-time graph, a horizontal line shows the change in speed is ____.

1 answer:

4 0

On a speed/time graph, the height of the line at any point
shows the speed at that moment. If the line is horizontal,
then its height isn't changing, meaning that the speed isn't
changing. It's constant. The change is zero, until the line
starts rising or falling.

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Your car burns gasoline as you drive up a large mountain. What energy transformation is the car performing?

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Your car is performing a transformation of energy of:

Chemical energy to Mechanical energy

The chemical is the gasoline which is then converted to fire as the car runs thus creating the movement of the car which is mechanical energy.

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3 years ago

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Explain what white light is and what it’s composed of.

Nataly_w [17]

Answer and Explanation:

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3 years ago

What is the mass of an object that has a weight of 52N? (n=newtons)<br><br>Answer is in kilograms

polet [3.4K]

Hi there!

The answer you are looking for.....

<span>5.3025kg!</span>

Hope this helps you!

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4 years ago

What are the two major topics studied by a physicist

Alexxx [7]

Math and the study of motion

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3 years ago

You serve a tennis ball from a height of 1.80 m above the ground. The ball leaves your racket with a velocity of 18.0 m/s at an

Delicious77 [7]

Answer:

Yes, ball will clear the net

Explanation:

First we have to find the range of projectile motion.

Data given,

Ф = 7°

Initial velocity = 18 m/s

R = (V)^2.sin2Ф/g

Now by putting values

R = 7.99 m

Now for height

h = v^2.(sinФ)^2/2g

by putting values

h = 0.245 m

Since range is less than our distance (11.83 m) from net, so still it is not clear that ball will clear the net or not.

So, now from the maximum height, we have to calculate the horizontal distance of ball to net.

Now velocity in projectile motion is in two dimensions.

V(x) = 18 m/s

V(y) = 0 m/s (because at maximum height, ball will stop and then start again, so y-component of velocity will be 0 but since there will be no acceleration along x-axis, so V(x) will be 18 m/s)

Now, by formula S = V(y)t + (1/2)gt^2

we can calculate time which is required by the ball to reach net from the maximum height it has achieved.

Now, tricky part is to calculate S, because without it we can not calculate t.

So, by data given in question, we know that the ball is served at height of 1.8 m and it achieved the height of 0.245 m. But net is at height of 1.07 m.

So, the vertical distance downward, which ball will travel from maximum height to net will be

S = 1.8 + 0.245 - 1.07

S = 0.975 m

Since we know V(y) = 0 m/s

S = (1/2)gt^2

t = (2S/g)^(1/2)

t = 0.44 s

Now time for both vertical and horizontal distance are same,

So, for horizontal distance "D(x)"

D(x) = V(x) x t (Since, no acceleration along x axis, so we can use simple formula to calculate distance)

D(x) = 18 x 0.44

D(x) = 8.029 m

Now please notice that at maximum height, range was half, so at that point ball covered distance "a"

a = 3.99 m

From maximum height to net, as we calculated, ball covered

D(x) = 8.029 m

So, total distance covered by ball

a + D(x) = 3.99 + 8.029

a + D(x) = 12.024 m

which is more than your total distance from net which is 11.83 m. So, the ball will clear the net.

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3 years ago