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3 years ago

Amount of work done by a rotating object

2 answers:

5 0

The work done by a rotating object can be calculated by the formula Work = Torque * angle.

This is analog to the work done by the linear motion where torque is analog to force and angle is analog to distance. This is Work = Force * distance.

An example will help you. Say that you want to calculate the work made by an engine that rotates a propeller with a torque of 1000 Newton*meter over 50 revolution.

The formula is Work = torque * angle.

Torque = 1000 N*m

Angle = [50 revolutions] * [2π radians/revolution] = 100π radians

=> Work = [1000 N*m] * [100π radians] = 100000π Joules ≈ 314159 Joules of work.

riadik2000 [5.3K]3 years ago

5 0

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15 points! An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remai

alexandr1967 [171]

The alpha particle is emitted at 4235 m/s

Explanation:

We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:

$p_i = p_f\\ Mu=m_1 v_1 + m_2 v_2 =$

where:

$M =222u$ is the mass of the original nucleus

$v=420 m/s$ is the initial velocity of the nucleus

$m_1 = 4 u$ is the mass of the alpha particle

$v_1$ is the final velocity of the alpha particle

$m_2 = 222u-4u = 218 u$ is the mass of the daughter nucleus

$v_2 = 350 m/s$ is the final velocity of the nucleus

Solving for $v_1$ , we find the final velocity of the alpha particle:

$v_1 = \frac{Mu-m_2 v_2}{m_1}=\frac{(222)(420)-(218)(350)}{4}=4235 m/s$

Learn more about momentum:

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4 0

3 years ago

If the sun had twice the mace how would that affect the gravitational force of the sun

daser333 [38]

Answer: Gravity is the force that keeps planets in orbit around the Sun. Gravity alone holds us to Earth's surface.

Planets have measurable properties, such as size, mass, density, and composition. A planet's size and mass determines its gravitational pull.

A planet's mass and size determines how strong its gravitational pull is.

Models can help us experiment with the motions of objects in space, which are determined by the gravitational pull between them.

Explanation:

5 0

3 years ago

An astronomer is observing a star which puzzles her. The lines in the star's spectrum indicates that the star is very hot and sh

Firlakuza [10]

Answer:

the stars which are red in color are cool.

Explanation:

The stars which has reddish color are cool in nature while those stars which has white and blue in color are very hot in nature. The stars change its color when they becomes hotter , first the star color reddish when they are cool but with increasing temperature it changes the color from reddish to orange then yellow. After yellow it turns green and finally get blue color when the stars are very very hot.

5 0

3 years ago

A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t

mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

$K=U$

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}$

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}$

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

$a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}$

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

$T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s$

So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

7 0

3 years ago

A shopping cart is given an initial velocity of 4 m/s and experiences a constant acceleration of 4 m/s2. What is the magnitude o

NNADVOKAT [17]

Answer:

48m

Explanation:

Initial velocity = 4m/s
Acceleration = 4m/s²
Time = 4s
Displacement = ?

From Second equation of motion ,

$\longrightarrow$ s = ut + 1/2at²

$\longrightarrow$ s = 4*4 + 1/2*4*(4)² m

$\longrightarrow$ s = (16 + 2 * 16) m

$\longrightarrow$ s = ( 16 + 32 )m

$\longrightarrow$ s = 48 m

8 0

3 years ago