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garik1379 [7]
3 years ago
12

A solid cylindrical object has a mass of 2.0 kg, a diameter

Physics
1 answer:
omeli [17]3 years ago
4 0

Answer:

I = 0.0025 kg.m²

Explanation:

Given that

m= 2 kg

Diameter ,d= 0.1 m

Radius ,R=\dfrac{d}{2}

R=\dfrac{0.1}{2}

R=0.05 m

The moment of inertia of the cylinder about it's axis same as the disc and it is given as

I=\dfrac{mR^2}{2}

Now by putting the all values

I=\dfrac{2\times 0.05^2}{2}

I = 0.0025 kg.m²

Therefore we can say that the moment of inertia of the cylinder will be  0.0025 kg.m².

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A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding
sveta [45]

Answer:

F1= 588 N

F2= 784 N

Explanation:

Please see the attached file.

5 0
3 years ago
Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire
Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

F = ma \rightarrow a = \frac{F}{m}

Replacing,

a =\frac{42N}{83000kg}

a =5.06*10^{-4}m/s^2

The total speed change

\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity } we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}

We have that,

t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }

t = 1403.16s

Therefore the Rocket should be fired around to 1403.16s

7 0
3 years ago
Two long, straight wires both carry current to the right, are parallel, and are 25 cm apart. Wire one carries a current of 2.0 A
Artemon [7]

Answer:

x = 7.14 meters

Explanation:

It is given that,

Current in wire 1, I_1=2\ A

Current in wire 2, I_2=5\ A  

Distance between parallel wires, r = 25 cm

Let at P point the net magnetic field equal to 0. The magnetic field at a point midway between the is given by :

B=\dfrac{\mu_oI}{2\pi r}

Let the distance is x from wire 1. So,

\dfrac{I_1}{r}=\dfrac{I_2}{(0.25-r)}

\dfrac{2}{r}=\dfrac{5}{(25-r)}

x=\dfrac{50}{7}\ m

x = 7.14 meters

So, the magnetic field will be 0 at a distance of 7.14 meters from wire 1. Hence, this is the required solution.

6 0
3 years ago
A rock is thrown downward from the top of a cliff with an initial speed of 20 m/s. If the rock hits the ground after 5.0 s, what
Vinil7 [7]

Answer

I hope it's helps you

7 0
2 years ago
What is the gravitational field theory and how does it work? four mark answer. Don't give me random answers just for points or i
MrMuchimi
It is a theory on a show that people try to solve.
6 0
2 years ago
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