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kicyunya [14]
3 years ago
13

You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 35 m/s. The ball has __________________

energy. Calculate it.
Physics
2 answers:
Anni [7]3 years ago
5 0

Answer:

<em>The ball has 1286.25 J of energy</em>

Explanation:

<u>Kinetic Energy</u>

Is the type of energy of an object due to its state of motion. It's proportional to the square of the speed.

The equation for the kinetic energy is:

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

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is often expressed in Joules (J).

The volleyball of a mass m=2.1 Kg is served at v=35 m/s, calculating its kinetic energy:

\displaystyle K=\frac{1}{2}\cdot 2.1\cdot 35^2

\displaystyle K=\frac{1}{2}\cdot2.1\cdot 1225

K = 1286.25 Joule

The ball has 1286.25 J of energy

Elden [556K]3 years ago
4 0

Answer:

The ball has kinetic energy

the kinetic energy is 945 J

Explanation:

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What would be the speed of an object just before hitting the ground if dropped 91.5 meters
Aleks04 [339]

Answer:

about 42.35 m/s

Explanation:

Use the equation for accelerated motion (g), and with zero initial velocity that doesn't include time:

v_f^2=v_i^2+2\,a\,\Delta x

which for our case would reduce to:

v_f^2=v_i^2+2\,a\,\Delta x\\v_f^2=0+2\,9.8\,(91.5)\\v_f^2= 1793.4\\v_f=\sqrt{1793.4} \\v_f \approx 42.35

then the velocity just before hitting would be about 42.35 m/s

5 0
3 years ago
if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav
allochka39001 [22]

Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, R_{max} = 3 m

The acceleration due to gravity on the Moon = 1 ÷ 16 of that on Earth

The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, R_{max}, of a projectile motion is found by using the following formula

R_{max} = \dfrac{u^2}{g}

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, g_{Moon} = 1/16 × g

∴ g_{Moon} = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

R_{max \ Moon} = \dfrac{u^2}{g_{Moon}}   = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is R_{max \ Moon} ≈ 48 m

8 0
3 years ago
Newtons third lawWhat action-reaction forces are involved when a rocket engine fires? Why doesnt a rocket need air to push on? A
dangina [55]

Answer: action forc roketorce

reaction force is engine fires

4 0
3 years ago
A 14.0 gauge copper wire of diameter 1.628 mm carries a current of 12.0 mA . A) What is the potential difference across a 1.80 m
Serga [27]

Answer: a) 139.4 μV; b) 129.6 μV

Explanation: In order to solve this problem we have to use the Ohm law given by:

V=R*I whre R= ρ *L/A  where ρ;L and A are the resistivity, length and cross section of teh wire.

Then we have:

for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω

and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω

Finalle we calculate the potential difference (V) for both wires:

Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V

V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V

8 0
3 years ago
Read 2 more answers
(c) A moving train has a kinetic energy of 8.1 x 10(power of 6)J.
tamaranim1 [39]

the friction force provided by the brakes is 30000 N.

<h3>What is friction force?</h3>

Friction force is the force that opposes the motion between two bodies in contact.

To calculate the average friction force provided by the brakes, we apply the formula below.

Formula:

  • K.E = F'd............. Equation 1

Where:

  • K.E = Kinetic energy of the train
  • F' = Friction force provided by the brakes
  • d = distance

Make F' the subject of the equation

  • F' = K.E/d............ Equation 2

From the question,

Given:

  • K.E = 8.1×10⁶
  • d = 270 m

Substitute these values into equation 2

  • F' = (8.1 ×10⁶)/270
  • F' = 30000 N

Hence, the friction force provided by the brakes is 30000 N

Learn more about friction force here: brainly.com/question/13680415

8 0
2 years ago
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