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Hitman42 [59]
3 years ago
13

A body of mass 80kg was lifted vertically through a distance of 5.0 metres. Calculate the work done on the body. ( Acceleration

due to gravity g=10ms²)​
Physics
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

80×5×10=4000J

so therefore, work done on the body is 4000J

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A 65-kg woman in an elevator is accelerating upward at a rate of 0.6 m/s2. The gravitational force is ___ N?
Nastasia [14]

Answer:

The gravitational force is 130.

Explanation:

During this problem you have to multiply the 65 and the 0.6.

5 0
3 years ago
A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is
valina [46]

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\  \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}

= 4.7476 m/sec

= 4.75 m/s

6 0
3 years ago
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Alenkasestr [34]

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<em><u>Wavel</u></em><em><u>ength</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>distan</u></em><em><u>ce</u></em><em><u> between</u></em><em><u> </u></em><em><u>one</u></em><em><u> </u></em><em><u>crest</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>one</u></em><em><u> </u></em><em><u>through</u></em><em><u> </u></em><em><u>,</u></em><em><u> </u></em><em><u>also</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>dist</u></em><em><u>ance</u></em><em><u> </u></em><em><u>after</u></em><em><u> </u></em><em><u>which</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>wave</u></em><em><u> </u></em><em><u>repe</u></em><em><u>at</u></em><em><u> </u></em><em><u>its</u></em><em><u>elf</u></em><em><u> </u></em><em><u>!</u></em>

<em><u>It's</u></em><em><u> </u></em><em><u>SI</u></em><em><u> </u></em><em><u>unit</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>meter</u></em><em><u> </u></em><em><u>!</u></em><em><u> </u></em>

<em><u>It</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>scalar</u></em><em><u> </u></em><em><u>quan</u></em><em><u>tity</u></em><em><u> </u></em><em><u>!</u></em><em><u>!</u></em><em><u> </u></em>

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5 0
3 years ago
What happens at night- describing air circulation
mash [69]

Answer:

The environment is warmed by the light throughout the day, such that the temperature increases. The weather is decreasing and the temperature decreases in the night as the sun falls. There was a misunderstanding. Thanks to the density, the atmosphere becomes densest on the earth. The air becomes colder and colder when you move up.

Explanation:

Answer is above

<em><u>Hope this helps.</u></em>

5 0
3 years ago
In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,
Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

Let mass of ball A and B\ is\ m  

Final velocity of ball A\ is\ v_1

Final velocity of ball B\ is\ v_2

initial velocity of ball A\ is\ u_1

Initial velocity of ball B\ is\ u_2

Momentum after collision =mv_1+mv_2

Momentum before collision = mu_1+mu_2

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

First Trial

mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0
3 years ago
Read 2 more answers
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