Ask question

Ask question

All categories

3 years ago

5

A 1.0 kilogram ball is thrown into the air with an initial velocity of 30m/s. How much kinetic energy does the ball have? (B) ho

w much potential energy does the ball have when it reaches the top of its accent? (C) how high into the air did the ball travel?

1 answer:

Gala2k [10]3 years ago

7 0

Answer:

450 joules ; 450 joules ; 45.9 m

Explanation:

Given that :

Initial Velocity, u = 30m/s

Mass, m = 1 kg

Kinetic Energy of ball (KE) = 0.5mu²

K. E = 0.5 * 1 * 30^2

K.E = 0.5 * 900

K.E = 450 Joules

B.) Potential Energy (P. E)

P. E = mgh

At the highest point, all kinetic energy has would have become potential energy, hence

K. E = P. E = 450 Joules

C) Height of the ball :

From ; P. E = mgh

Where ; g = acceleration due to gravity = 9.8m/s² ; h = height

450 = 1 * 9.8 * h

450 = 9.8h

h = 450 / 9.8

h = 45.918

h = 45.9 m

You might be interested in

What is the term for the overall change in position?

mart [117]

That's "displacement". It only depends on the beginning and ending locations, and doesn't care about the route between them.

6 0

3 years ago

Read 2 more answers

X-Rays at different wavelengths provide scientists studying objects in space with information about which of the following?

olganol [36]

Answer:

object composition

Explanation:

6 0

2 years ago

Read 2 more answers

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

elena55 [62]

<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird decided to take a pit stop and slows down for 250 m. She spent 5 seconds in the pit stop.

Here final velocity $v=0 \ m/s$

initial velocity $u= 71 m/s$ distance

Distance covered in the slowing down phase = $250 m$

$v^2-u^2=2as$

$a= \frac {(v^2-u^2)}{2s}$

$a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2$

$v=u+at$

$t= \frac {(v-u)}{a}$

$= \frac {(0-71)}{(-10.082)}=7.042 s$

$t_1=7.042 s$

The car is in the pit stop for 5s $t_2=5 s$

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

$a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2$

$v=u+at$

$t= \frac{(v-u)}{a}$

$t= \frac{(71-0)}{6.81}= 10.425 s$

$t_3=10.425 s$

total time= $t_1 +t_2+t_3=7.042+5+10.425=22.467 s$

Distance covered by the Mercedes Benz during this time is given by $s=vt=71 \times 22.467= 1595.157 m$

Distance covered by the Thunderbird during this time= $250+350=600 m$

Difference between distance covered by the Mercedes and Thunderbird

= $1595.157-600=995.157 m$

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

6 0

3 years ago

Density: Two blocks, A and B, are put in a tank of water. Block A has a density of 1.21 g/cm³. Block B has a density of 1.37 g/c

stira [4]

Answer:

Block A

Explanation:

Block A will float higher in the water compared to the second Block.

The density of water is 1g/cm³.

According to the principle of floatation "an object that floats in a liquid will displace equal amount of fluid to the weight of the object".

A body will become more submerged in water if it has more density because density is the mass per volume of body.

An object with a higher density than another will sink in the liquid of the one with lesser density.

Object A has lesser density and will float higher up and displace very little water.
Object B has higher density and will be more submerged.

4 0

3 years ago

A 77.0−kg short-track ice skater is racing at a speed of 12.6 m/s when he falls down and slides across the ice into a padded wal

sukhopar [10]

Answer:

-6112.26 J

Explanation:

The initial kinetic energy, $KE_i$ is given by

$KE_i=0.5mv_1^{2$ } where m is the mass of a body and $v_i$ is the initial velocity

The final kinetic energy, $KE_f$ is given by

$KE_f=0.5mv_f^{2}$ where $v_f$ is the final velocity

Change in kinetic energy, $\triangle KE$ is given by

$\triangle KE=KE_f-KE_i=0.5mv_f^{2}-0.5mv_1^{2}=0.5m(v_f^{2}-v_i^{2})$

Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for $v_f$ and 12.6 m/s for $v_i$ and 77 Kg for m we obtain

$\triangle KE=0.5*77*0^{2}-0.5*77*(0^{2}-12.6^{2})=-6112.26 J$

From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26 J</u>

3 0

2 years ago