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3 years ago

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A child on a high dive has a mass of 40 kilograms. If the high dive is 10 meters in the air, what is the potential energy? GPE=m

gh

1 answer:

saw5 [17]3 years ago

7 0

Answer:

Ep = 3924 [J]

Explanation:

To calculate this value we must use the definition of potential energy which tells us that it is the product of mass by the acceleration of gravity by height.

$E_{p}=m*g*h\\$

where:

Ep = potential energy [J] (units of Joules)

m = mass = 40 [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 10 [m]

$E_{p} =40*9.81*10\\E_{p} = 3924 [J]$

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A rocket at rest with a mass of 942 kg is acted on by an average net force of 6,731 N upwards for 21 s. What is the final veloci

Galina-37 [17]

Answer:

Explanation:

Givens

m = 942

F = 6731

t = 21 seconds

vi = 0

vf = ?

Formula

F = m * (vf - vi ) / t

Solution

6731 = 942*(vf - 0)/21 Multiply both sides by 21

6731 * 21 = 942*vf

141351 = 942*vf Divide by 942

141351/942 = vf

vf = 151 m/s

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3 years ago

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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A blank is any disturbance that carries energy from one place to another through matter and space.

Eduardwww [97]

<h3>Answer;</h3>

<em>A wave </em>

<em><u>A wave</u></em> is any form of a disturbance that carries energy from one place to another through a matter and space

<h3>Explanation;</h3>

Waves carry energy from one point, the source to another point or place. The transmission of a wave may occur through the space or through a material medium.
Electromagnetic waves are those waves whose transmissions occurs through the space, they do not require material medium for transmission,for example, radio waves, while mechanical waves are those that require material medium for transmission, for example sound waves.
The energy of wave depends on the frequency of the wave and the wavelength of that particular wave.

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Bob is coming toward you at a speed of 75 km/hr. You throw a baseball in his direction at 75 km/hr. What does he see the ball do

KengaRu [80]

Answer:

He sees the ball coming at him at 150 km/hr.

Explanation:

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4 years ago

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