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USPshnik [31]
1 year ago
12

A 1,350 kg package is dropped from a plane from a height of 760meters. What is its potential energy right before it starts to fa

ll?
Physics
1 answer:
larisa [96]1 year ago
7 0

Given:

Mass (m) of package = 1350 kg

Height, h = 760 meters

Let's find the potential energy right before the object starts to fall

To find the potential energy, apply the formula below:

P.E=m\ast g\ast h

Where:

m = 1350 kg

h = 760 meters

g is the acceleration due to gravity = 9.8 m/s^2

Substitute values into the equation:

\begin{gathered} PE=1350\ast9.8\ast760 \\  \\ PE=10054800\text{ J }\approx10.0548\text{ MJ} \end{gathered}

Therefore, the potential energy right before the object starts of fall is 10.0548 megajoules.

ANSWER:

10.0548 MJ

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Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
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The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
3 years ago
PLEASE HELP ME!!!! i will give brainliest to whoever gets it right
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Answer:

I believe the answer is B.)

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2 years ago
The generator at a power plant produces AC at 20,000 V. A transformer steps this up to 355,000 V for transmission over power lin
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Answer:

Number of coil in the output is 39938

Explanation:

We have given a step up transformer

Input voltage of transformer, that is primary voltage v_p=20000volt

Output voltage, that is secondary voltage v_s=355000volt

Number of turns in primary N_p=2250

For transformer we know that \frac{V_p}{V_s}=\frac{N_p}{N_s}

\frac{20000}{355000}=\frac{2250}{N_s}

N_s=39937.5

As the number of turns can not be in fraction so number of turns in the output coil is 39938

7 0
2 years ago
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