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yulyashka [42]
2 years ago
14

What torque will increase angular velocity of a solid cylinder of mass 16 kg and diameter 1 m from zero to 120 rpm in 8 s?

Physics
1 answer:
const2013 [10]2 years ago
7 0
This is what I found online! Hope it helps:)

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PLEASE HELP ME ASAP PLEASEEEEE
AleksAgata [21]
I believe its the law of inertia
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A 0.48 kg circus monkey is about to be shot from a cannon as part of his thrilling circus act. Draw a free body diagram labeling
Alenkinab [10]
What is it that you need help on?
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2 years ago
What is transferred by a radio wave?
Alex

Answer: B. energy

Explanation:

4 0
3 years ago
A small rocket to gather weather data is launched straight up. Several seconds into the flight, its velocity is 120 mis and it i
padilas [110]

The net force on the rocket is 864 N

Explanation:

The net force acting on the rocket can be calculated by using Newton's second law of motion, which states that:

F = ma

where

F is the net force on an object

m is the mass of the object

a is its acceleration

For the rocket in this problem, at a certain instant we have:

m = 48 kg

a=18 m/s^2

Therefore, the force on the rocket is

F=(48)(18)=864 N

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

8 0
2 years ago
The route followed by a hiker consists of three displacement vectors, X, Y and Z. Vector X is along a measured trail and is 1430
poizon [28]

Answer:

  • magnitude : 1635.43 m
  • Angle: 130°28'20'' north of east

Explanation:

First, we will find the Cartesian Representation of the \vec{X} and \vec{Y} vectors. We can do this, using the formula

\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | its the magnitude of the vector and θ the angle. For  \vec{X} we have:

\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )

\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )

where the unit vector \hat{i} points east, and \hat{j} points north. Now, the \vec{Y} will be:

\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )

Now, taking the sum:

\vec{X} + \vec{Y} + \vec{Z} = 0

This is

\vec{Z} = - \vec{X} - \vec{Y}

(Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )

(Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 2200 m \ - \ 956.86 m \ )

(Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 1243.14 m\ )

Now, for the magnitude, we just have to take its length:

|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}

|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}

|\vec{Z}| = 1635.43 m

For its angle, as the vector lays in the second quadrant, we can use:

\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m})

\theta = 180\° - arctan( -1.1720)

\theta = 180\° - 45\°31'40''

\theta = 130\°28'20''

5 0
3 years ago
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