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3 years ago

Calculate the gravitational potential energy of a 2 kg banana hanging 5

Physics

1 answer:

Hunter-Best [27]3 years ago

7 0

Answer:

100 J

Explanation:

The potential energy is given by the formula ...

PE = mgh

= (2 kg)(10 m/s^2)(5 m) = 100 J

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You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

$I_i$ = Initial moment of inertia = 9.3 kgm²

$I_f$ = Final moment of inertia = 5.1 kgm²

$\omega_i$ = Initial angular speed = 1.8 rev/s

$\omega_f$ = Final angular speed

As the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s$

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J$

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

6 0

3 years ago

Magnetic Energy density affected by area?

Mademuasel [1]

Answer:

ya it is , affected!okay I think this

4 0

3 years ago

Three students have been studying relative motion and decide to do an experiment to demonstrate their knowledge. The experiment

miskamm [114]

Answer with Explanation:

We are given that

Constant speed of Jane=12.6 m/s

a.When Fred can throw the ball 30 m/s

We have to find the angle relative to the horizontal when he throw the ball in order for Sue to see the ball travel vertically upward.

Let $\theta$ be the angle .

Therefore,

$30 cos\theta=12.6$

$cos\theta=\frac{12.6}{30}=0.42$

$\theta=cos^{-1}(0.42)=65.165^{\circ}$

b.We have to find the height to which ball reach.

$v^2-v^2_0=2aS$

$S=\frac{v^2-v^2_0}{2a}=\frac{0-(30 sin65.165)^2}{2(-9.81)}$

$S=37.78 m$

7 0

3 years ago

A moving car skids to a stop with the wheels locked across a level roadway. Of the forces listed, identify which act on the car.

Vesnalui [34]

Answer:

Normal, Gravity, Friction, and Air Resistance.

Explanation:

When a moving car skid to stop and its wheels are locked across, then the following forces will be applied on the car:

Normal force: It will act counter to gravity that pushes an object against a surface and acts perpendicular to the contact surface.

Gravity: Gravity force acts in each and every object having mass and it can not be avoidable. So, the gravity force will also apply to the car and attract it to the earth's surface.

Friction: Friction is a force that acts opposite to the motion and stops or slows motion. Friction will be applied to the car that will oppose the motion of the car and stop it.

Air resistance: air resistance is defined as the forces exerted by air that acts opposite to the relative motion of an object. Air resistance will also be applied to the car when it will skid to stop as we are always surrounded by the air.

Hence, the correct answers are "Normal, Gravity, Friction, and Air Resistance."

4 0

3 years ago

An aluminum wire is 5.87 m long and has a diameter of 6.07 mm.

antoniya [11.8K]

Answer: 5.72 x 10-3Ω

Explanation:

Hi, to answer this question, first we have to calculate the cross sectional area of the cable:

Diameter (D)=6.07 mm

Since: 1000mm = 1m

6.07mm/ 1000mm/m = 0.00607 meters

Area of a circle : π (d/2)^2

A = π (0.00607/2)^2= 0.000028937 m2

Resistance formula:

Resistance (R) = P(resistivity) L (length)÷A (cross sectional area )

Replacing with the values given:

R = (2.82x10-8 x 5.87) / 0.000028937

R = 5.72 x 10-3Ω

Feel free to ask for more if needed or if you did not understand something.

7 0

3 years ago