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Nezavi [6.7K]
2 years ago
7

A uniform solid sphere has mass m= 7 kg and radius r= 0. 4 m. What is its moment of inertia about an axis tangent to its surface

?
Physics
1 answer:
lilavasa [31]2 years ago
4 0

The moment of inertia of a uniform solid sphere is equal to 0.448 kgm^2.

<u>Given the following data:</u>

Mass of sphere = 7 kg.

Radius of sphere = 0.4 meter.

<h3>How to calculate moment of inertia.</h3>

Mathematically, the moment of inertia of a solid sphere is given by this formula:

I=\frac{2}{5} mr^2

<u>Where:</u>

  • I is the moment of inertia.
  • m is the mass.
  • r is the radius.

Substituting the given parameters into the formula, we have;

I=\frac{2}{5} \times 7 \times 0.4^2\\\\I=2.8 \times 0.16

I = 0.448 kgm^2.

Read more on inertia here: brainly.com/question/3406242

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Nutka1998 [239]
The movement of air flows from high pressure to low pressure
3 0
3 years ago
The capitary action of water is caused the cohesion of water molecules to different types of molecules and the adhesion of water
s344n2d4d5 [400]

Answer:

This is True

Explanation:

I just did this exact unit in bio last week I hope this helps ;)

5 0
3 years ago
A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

4 0
3 years ago
a 75-kg refrigerator is located on the 70th floor of a skyscraper (300 meters above ground). what is the potential energy of the
mart [117]
Here, you can calculate it's potential energy with respect to ground.
We know, U = mgh
Here, m = 75 Kg
g = 9.8 m/s²       [ constant value for earth system ]
h = 300 m

Substitute their values into the expression:
U = 75 × 9.8  × 300
U = 220500 J

In short, Your Final Answer would be 220,500 J 

Hope this helps!
8 0
3 years ago
What happened to the weight of an object when it is taken from Earth to the Moon? why?<br>​
Sholpan [36]

Answer:

the weight of the object decreases when it is taken from the Earth to the Moon

Explanation:

The weight of an object is defined as the product of the mass of the object with the acceleration due to gravity of the Planet.

W =mg

where,

W = weight of the object

m = mass of the object

g = acceleration due to gravity on the planet

The mass of an object remains constant everywhere in the universe. Therefore, the weight is directly proportional to the value of acceleration due to gravity.

The value of acceleration due to gravity on the Moon is lesser than its value on the Earth.

<u>Hence, the weight of the object decreases when it is taken from the Earth to the Moon </u>

6 0
3 years ago
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