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romanna [79]
3 years ago
7

A 2-kg ball rolls at a speed of 10 m/s on the ground. Is this an example of Kinetic Energy, Potential Energy, or both?

Physics
1 answer:
nadya68 [22]3 years ago
3 0

If a 2kg ball rolls at a speed of 10 m/s on the ground, this is an example of kinetic energy. When a mass of the body moves at a specific rate, there is movement. The equation for kinetic energy is 1/2mv^2. Therefore the kinetic energy of the ball is (1/2)(2kg)(10m/s)^2 is 100 Joules.

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The energy achieved I think
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If an object is being pulled by two forces, one 4 N to the left and the other 2 N to the right, what is the net
nika2105 [10]

Answer:

2N

Explanation:

subtract  rthe two forces to see which is greater

4-2=2

6 0
2 years ago
4. The Mariana trench is in the Pacific Ocean and has a depth of approximately 11,000 m. The density of seawater is approximatel
Alex73 [517]

Explanation:

It is known that relation between pressure and density is as follows.

            P = \rho gh

where,    P = pressure

     \rho = density

            g = acceleration due to gravity

            h = height

Putting the given values into the above formula as follows.

              P = \rho gh

                 = 1025 \times 9.8 \times 11000

                 = 110495000 Pa

Now, relation between pressure and force is as follows.

                P = \frac{F}{A}

or,            F = PA

                F = 110495000 \times \pi \times (0.1)^{2}

                   = 3.47 \times 10^{6} N

Thus, we can conclude that a force of 3.47 \times 10^{6} N can be  experienced at such depth.

3 0
2 years ago
Bill is learning to play tennis. He does pretty well hitting the ball back to his opponent but, many times he misses the ball wh
KiRa [710]

Answer: Place his feet parallel to the baseline prior to tossing the ball

5 0
2 years ago
A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The
Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2

Kx^2=mv^2

v=\sqrt{\dfrac{kx^2}{m}}

By putting the values

v=\sqrt{\dfrac{kx^2}{m}}

v=\sqrt{\dfrac{1300\times 0.264^2}{25}}

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

V_f^2=V_i^2-2gh

By putting the values

0^2=1.9^2-2\times 10\times h

h=0.1805 m

h=18.05 cm

4 0
3 years ago
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