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dezoksy [38]
3 years ago
5

A sensational 130 kg midfielder has 695 J of

Physics
1 answer:
antoniya [11.8K]3 years ago
4 0

Answer:

v = 3.27 m/s

Explanation:

KE = 1/2 mv^2

695 J = 1/2 (130kg)(v^2)

695 J / (1/2 x 130kg) = v^2

v^2 = square root of 10.69

v = 3.27 m/s

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Which one of the statements below is true about mechanical waves?
TEA [102]
They require a medium to travel through
5 0
2 years ago
A 61 kg skater is traveling at 2.5 m/s while carrying a 4.0 kg bowling ball. After he throws the bowling ball forward at twice t
gregori [183]

The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

p_i = (M+m)u

where

M = 61 kg is the mass of the skater

m = 4.0 kg is the mass of the ball

u = 2.5 m/s (forward) is the combined velocity of the skater and the ball

After, the ball is thrown at twice the velocity, so the final total momentum is

p_f = MV+mv

where

V is the final velocity of the skater

v = 2(2.5) = 5.0 is the final velocity of the ball

Since the total momentum must be conserved, we can write

p_i = p_f\\(M+m)u = MV+mv\\V=\frac{(M+m)u-mv}{M}=\frac{(61+4.0)(2.5)-(4.0)(5.0)}{61}=2.34 m/s

So, the skater is moving at 2.34 m/s (forward) after the shot.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

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#LearnwithBrainly

5 0
3 years ago
When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache
Masteriza [31]

Answer:

b) True.    the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

     fr - w = ma

Where the friction force has some form of type.

     fr = G v + H v²

Let's replace

     (G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

      fr - w = 0

      fr = mg

      (G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

     G v = mg

     v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

3 0
3 years ago
If you answer it I’ll love you forever!!!
Ilia_Sergeevich [38]

Answer:

Performance tests can be used to see if an implemented training program is working for the athlete or if the program needs alterations. They can also assess current abilities in specific athletic areas to help the athlete choose what to focus their energy on improving.

Explanation:

6 0
3 years ago
Read 2 more answers
Name the apparatus used to measure friction force? PLEASE HELP AS SOON AS POSSIBLE
melisa1 [442]
There is no apparatus for it. It is either use something like ruler and table and rub together or rub our hands, and friction force will be showed.
8 0
3 years ago
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