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3 years ago

A 100 kg cyclist travels at 4.5 m/s. A 50 kg gorilla is dropped straight down onto her. What is the new speed of the system?

Physics

1 answer:

krek1111 [17]3 years ago

6 0

Mass of cyclist = 100 kg

Mass of gorilla = 50 kg

Initial speed = 4.5 m/s

Let Final speed = s

100 x 4.5 = (100 + 50) x s

450 = 150s

S = 450/150

S = 3

The new speed is 3m/s

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4 years ago

19. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to

Aleks [24]

Recall this gas law:

$\frac{P₁V₁}{T₁}$ = $\frac{P₂V}{T₂}$

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T₁ and T₂ are the initial and final temperatures.

Given values:

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8 0

3 years ago

Which of the following statements about diodes and triodes is correct

nata0808 [166]

There are no correct statements on your list.

4 0

3 years ago

The dimensions of the disk of our milky way galaxy are:.

Katen [24]

Answer:

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With that being said, it is safe to say that the dimensions are somewhere around 100,000 by 1,000

6 0

3 years ago

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was

tigry1 [53]

Answer:

$\Delta V=316167V$

Explanation:

The difference of electric potential between two points is given by the formula $\Delta V=Ed$ , where d is the distance between them and E the electric field in that region, assuming it's constant.

The electric field formula is $E=\frac{F}{q}$ , where F is the force experimented by a charge q placed in it.

Putting this together we have $\Delta V=\frac{Fd}{q}$ , so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight W, giving a net force $N=W-F$ . On the first drop only W acts, while on the second drop is N that acts.

Using the equation for accelerated motion (departing from rest) $d=\frac{at^2}{2}$ , so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting: