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Nimfa-mama [501]
1 year ago
13

How do I calculate the amount of work done using the information on the graphs given?​

Physics
1 answer:
Nataly [62]1 year ago
3 0

Answer:with the power of math and physic

Explanation:

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Analyze the explanation and identify any false statements. Each statement to be analyzed is identified by a capital letter. A ba
Sladkaya [172]

Answer:

Explanation:

The rotational kinetic energy remains constant as long as the length and angular speed are fixed.

Statement A is true.

When the ball is pulled inward and the length of the string is shortened, the rotational kinetic energy will remain constant due to conservation of energy,

Statement B is false .

Reason - Conservation of energy will not be there because external work is done on the system  by the force that pulls it inward.  

but the angular momentum L will not remain constant because there is an external force acting on the ball to pull it inward

Statement C is false .

Reason - the angular momentum L will remain constant because there is an external force acting on the ball which acts perpendicular to the velocity of the ball .

The moment of inertia I remains constant

Statement D is false

Reason - because distance from axis of rotation is changing.

the angular speed will remain the same throughout the process because the ball is rotating in the same plane throughout the motion.

Statement E is false

Reason - Since moment of inertia decreases , to conserve angular momentum , angular speed increases.

4 0
2 years ago
A stone with a weight of 5.30 N is launched vertically from ground level with an initial speed of 23.0 m/s, and the air drag on
frozen [14]

Answer:

a) h=25.7\ m

b) v'=21.8733\ m.s^{-1}

Explanation:

Given:

  • weight of the stone, w=5.3\ N
  • initial velocity of vertical projection, u=23\ m.s^{-1}
  • air drag acting opposite to the motion of the stone, D=0.266\ N

The mass of the stone:

m=\frac{w}{g}

m=\frac{5.3}{9.8}

m=0.5408\ kg

Now the acceleration of the stone opposite of the motion:

D=m.d

where:

d = deceleration

0.266=0.5408\times d

d=0.4918\ m.s^{-2}

<u>In course of going up the net acceleration on the stone will be:</u>

g'=g+d

g'=9.8+0.4918

g'=10.2918\ m.s^{-2}

a)

Now using the equation of motion:

v^2=u^2-2 g'.h

where:

v= final velocity when the stone reaches at the top of the projectile = 0

h = height attained by the stone before starting to fall down

0^2=23^2-2\times 10.2918\times h

h=25.7\ m

b)

during the course of descend from the top height of the projectile:

initial velocity, v=0\ m.s^{-1}

The acceleration will be:

g"=g-d

g"=9.8-0.4918

g"=9.3082\ m.s^{-2}

here the gravity still acts downwards but the drag acceleration acts in the direction opposite to the motion of the stone, now the stone is falling down hence the drag  acts upwards.

Using equation of  motion:

v'^2=v^2+2g".h (+ve acceleration because it acts in the direction of motion)

v'^2=0^2+2\times 9.3082\times 25.7

v'=21.8733\ m.s^{-1}

8 0
2 years ago
Ashley decides to enter her pet turtle in a race. She knows her turtle can travel at a rate of 2 meters per hour. The race track
Genrish500 [490]
It would probably be B
7 0
3 years ago
Newtons first law of motion states that the velocity of an object doesn’t change if the forces acting on the object are
Gnom [1K]
Balenced

Think of it this way if 2 same weight and same speed objects were racing toward each other perfectly straight then their forces will cancel out and the movement afterwords will be 0
4 0
2 years ago
A horizontal force of 300.0 N is used to push a 145-kg mass 30.0 m horizontally in 3.00 s. Calculate the power developed.
Svetradugi [14.3K]

Answer:

3 * 10³J/s

Explanation:

Given :

Force applied, F = 300 N

Distance, d = 30 m

Time, t = 3 seconds

Power, P = Workdone / time

Recall :

Workdone = Force * distance

Workdone = 300 N * 30 m = 9000 Nm

Workdone = 9 * 10³ J

Power = (9 * 10³ J) / 3s

Power = 3 * 10³J/s

4 0
3 years ago
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