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3 years ago

PLEASE HELP ASAP I WILL GIVE BRAINLIEST!!!!!!!!

Physics

2 answers:

romanna [79]3 years ago

8 0

One fourth of the moon's lighted side is visible from Earth

when the Moon is at position-D in its orbit ... it looks like a

crescent in the sky.

There's a massive blunder in this drawing.

Look at the Moon at Third Quarter.

It's drawn with its dark side toward the sun, and its

lit-up side away from the sun ! That's just silly.

Teachers can make mistakes too.

Firlakuza [10]3 years ago

5 0

The answer would be D

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NEED HELP ASAP

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the Conservation of Momentum principle, 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

6 0

2 years ago

A virtual image formed by convex lens is always

natita [175]

Answer:

2. erect and magnified

4 0

3 years ago

Read 2 more answers

What should be done to lift the same load by applying less effort on an inclined plane

Scorpion4ik [409]

Answer:

Reduce the friction at the surface

Explanation:

If you can reduce the friction between the load and the plane less effort will be required as you are not having to apply effort to overcome friction.

5 0

1 year ago

How much force is needed to lift a 25-kg mass at a constant verlocity?

Troyanec [42]

-- In order to achieve constant verlocity, the net force on the mass must be zero. So if there ARE any forces acting on it, they must be balanced.

-- There is already a force on the mass that can't be eliminated . . . the force of gravity.

-- That force due to gravity is (mass x gravity) = (25 kg)(9.8 m/s²) = 245N in the downward direction.

-- In order to 'balance' the forces and make them add up to zero, we have to provide another force of 245N, all in the upward direction.

-- Then the forces on the object will be balanced, the NET force on it will be zero, and whichever way you start it moving, it will continue to move at a cornstant verlocity.

5 0

2 years ago

Read 2 more answers

a car moving at 5 m/s accelerates at a rate of 10 m/s2 for 25 seconds. How far does it move during this time?

Alexus [3.1K]

Answer:

t is time in s For example, a car accelerates in 5 s from 25 m/s to 3 5m/s. Its velocity changes by 35 - 25 = 10 m/s. Therefore its acceleration is 10 ÷ 5 = 2 m/s2

Explanation:

3 0

2 years ago