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xeze [42]
3 years ago
11

An engine performs 6400 j of work on a motorbike the motorbike and the rider have a combined mass of 200 kg if the bike started

at rest what is the speed of the bike after the work is performed
Physics
1 answer:
spin [16.1K]3 years ago
4 0

The work done by the engine is converted into kinetic energy of the motorbike+rider system:

W=K=\frac{1}{2}mv^2

where

W=6400 J is the work

m=200 kg is the mass of the system

v is the speed acquired by the motorbike


Rearranging the equation and substituting the numbers in, we find

v=\sqrt{\frac{2W}{m}} =\sqrt{\frac{2(6400 J)}{200 kg}} =8 m/s

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olganol [36]

Answer:

B Relesing the string

Explanation:

got it right on edge

5 0
3 years ago
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You apply a net force on a soccer ball of 15 N. If the acceleration it has is 5 m/s2 what is the mass of the ball?​
Ipatiy [6.2K]

Answer:

<h2>3 kg </h2>

Explanation:

The mass of the ball can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

We have

m =  \frac{15}{5}  = 3 \\

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

4 0
3 years ago
g A change in the initial _____ of a projectile changes the range and maximum height of the projectile.​
docker41 [41]

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= (u^{2}*sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= (u^{2} * sin^{2}θ )/2g

the maximum distance also depends upon square of the initial velocity.

​

​

​

7 0
3 years ago
A ball is thrown up into the air with an initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed
kaheart [24]

Answer:

B) t = 1.83 [s]

A) y = 16.51 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f} =v_{o} -g*t

where:

Vf = final velocity = 0

Vo = initial velocity = 18 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.

A) The maximum height is reached when the final velocity of the ball is zero.

0 = 18 - (9.81*t)

9.81*t = 18

t = 18/9.81

t = 1.83 [s], we found the answer for B.

Now using the following equation.

y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\

where:

y = elevation [m]

Yo = initial elevation = 0

y = 18*(1.83) - 0.5*9.81*(1.83)²

y = 16.51 [m]

7 0
3 years ago
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amm1812

Answer: D. They are the coldest stars.

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