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kipiarov [429]
3 years ago
6

A solid aluminum sphere of radius R has moment of inertia I about an axis through its center. What is the moment of inertia abou

t a central axis of a solid aluminum sphere of radius 2R
Physics
1 answer:
frutty [35]3 years ago
6 0

Answer:

I1 = 2/5 M1 R^2   for a sphere about its center

I2 = 2/5 M2 (2 R)^2 = 2/5 M2 R^ * 4 = 8/5 M2 R^2

Remember that M2 is greater than M1 by a factor 0f 2^3 = 8

Then I2 exceeds I1 by a factor of 32

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Compare the freezing point of water in the aquanaut’s apartment to its value at the surface. Is it higher, lower, or the same?
Lelu [443]

Answer:

Freezing Point - Lower

Boiling Point - Higher

Solid- liquid transition line in the phase diagram has a negative slope, but the liquid-gas transition line has a positive slope. Since there is more air pressure at 100m it will take less to freeze the water but more to boil it since it requires a larger temperature under larger pressures

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3 years ago
What is the length of AR?​
AlladinOne [14]

Answer:Assault rifle?

Explanation:

20 m16 size and a 14.5 barrel and get a pump shot and get the blue version and go to tilted and one pump even tho forkknife is garbage.:)

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3 years ago
A climber pulls herself 8 meters upwards with a force of 150 Newtons. If it takes her 16 seconds to cover the 8 meters, how much
mr Goodwill [35]

Answer:

 P = 75 W

Explanation:

given,

Distance, L = 8 m

Force,F = 150 N

Time, t = 16 s

Work by the climber

Work done = Force x displacement

W = F. L

W = 150 x 8

W = 1200 J

We know,

Power =\dfrac{Work\ done}{time}

P =\dfrac{1200}{16}

 P = 75 W

Hence, Power climber is using to climb is equal to 75 W.

3 0
3 years ago
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in
wel

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

What is the change in the potential energy (in Joules) of the mass as it goes up the incline?  

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?

Given Information:  

Mass = m = 0.5 kg

Horizontal distance = d = 40 cm = 0.4 m

Vertical distance = h = 7 cm = 0.07 m

Normal force = Fn = 1 N

Required Information:  

Potential energy = PE = ?

Work done = W = ?

Answer:

Potential energy = 0.343 Joules

Work done = 0.39 N.m

Explanation:

The potential energy is given by

PE = mgh

where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.

PE = 0.5*9.8*0.07

PE = 0.343 Joules

As you can see in the attached image

sinθ = opposite/hypotenuse

sinθ = 0.07/0.4

θ = sin⁻¹(0.07/0.4)

θ = 10.078°

The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

Fx = 0.984 N

Work done is given by

W = Fxd

where d is the horizontal distance

W = 0.984*0.4

W = 0.39 N.m

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3 years ago
Which is an example of qualitative data?
LiRa [457]
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