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photoshop1234 [79]
3 years ago
14

A rock is dropped from a window 5 m above the ground. the rock hits the ground 1 s later with a speed of 10 m/s. what is the ave

rage speed of the rock during this time
Physics
1 answer:
Leviafan [203]3 years ago
6 0
Average speed=total distance/total
time
=5m/s
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What are the factors that effect the strength of a frictional force???
seropon [69]

If you write down the formula for friction, you will get an answer.

Ff = u * N               Where N is a push down force that an object experiences.

                              u (mu) is a constant and has no units

It may not be accelerating and still experience friction. A is not correct.

Color and Density will not affect the frictional force. B is not so.

Buoyant forces are a different thing altogether. Generally friction has nothing to do with them. C is incorrect.

The last one is your answer. Technically mg should be the answer and not mass, but the second part is correct.

5 0
3 years ago
Tarzan has foolishly gotten himself into another scrape with the animals and must be rescued once again by Jane. The 60.0 kg Jan
Brut [27]

Complete part of Question: What is Jane's (and the vine's) angular speed just before she grabs Tarzan

Answer:

Jane's (and the vine's) angular speed just before she grabs Tarzan, w = 1.267 rad/s

Explanation:

According to the law of energy conservation:

Total change in kinetic energy = Total change in potential energy

Mass of Jane = 60 kg

Mass of the vine = 32 kg

Mass of Tarzan = 72 kg

Height of Tarzan = 5.50 m

Length of the vine = 8.50 m

Jane's change in gravitational potential energy,

U_J = 60 * 9.8 * 5.5\\U_J = 3234 J

Vine's gravitational potential energy,

U_v = Mgh/2\\U_v = 32*9.8*5.5/2\\U_v = 862.4J

Vine's Kinetic energy :

KE_V = 0.5 I w^{2} \\I_V = \frac{ML^2}{3} = \frac{32 * 8.5^2}{3} = 770.67 kg m^2\\ KE_V = 0.5 *770.67 * w^{2}\\KE_V = 385.33 w^{2}

Jane's Kinetic energy:

KE_J = 0.5m(wL)^2\\KE_J = 0.5*60(w * 8.5)^2\\KE_J = 2167.5 w^2

U_J + U_V = KE_J + KE_V

3234 + 862.4 = 2167.5w² + 385.33w²

4096.4 = 2552.83w²

w² = 4096.4/2552.83

w² = 1.605

w = √1.605

w = 1.267 rad/s

5 0
3 years ago
When a pendulum is at the midpoint of its oscillation, hanging straight down, which statement is true?
Svetach [21]
When a pendulum is at the midpoint of its oscillation, hanging straight down ...

-- that's the fastest it's going to swing, so its kinetic energy is maximum;
and
-- that's the lowest it's going to get, so its potential energy is minimum.

'c' is your choice.
5 0
3 years ago
Read 2 more answers
A 15.0cm string is how many times shorter than the 30.0cm string?
Alex Ar [27]

Easy. answers B. Because 15.00x2=30

3 0
3 years ago
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You set out to design a car that uses the energy stored in a flywheel consisting of a uniform 80-kg cylinder of radius r that ha
Oliga [24]
E = 300km * 2000J/km = 600000J

The rotational energy is given by:
E = 0.5 * I * ω² 

I for a uniform cylinder is given by:
I  = 0.5 * m * r²

Resulting equation:
E = 0.25 * m * r² * ω²

Given values:
ω = 430 rev/s = 430 * 2π / s
E = 600000

Solve for r.
8 0
3 years ago
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