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vredina [299]
3 years ago
7

A child in a tree house uses a rope attached to a basket to lift a 24 N dog upward through a distance of 4.9 m into the house.

Physics
1 answer:
jenyasd209 [6]3 years ago
3 0
<h3>Answer:</h3>

117.6 Joules

<h3>Explanation:</h3>

<u>We are given;</u>

  • Force of the dog is 24 N
  • Distance upward is 4.9 m

We are required to calculate the work done

  • Work done is the product of force and distance
  • That is; Work done = Force × distance
  • It is measured in Joules.

In this case;

Force applied is equivalent to the weight of the dog.

Work done = 24 N × 4.9 m

                  = 117.6 Joules

Hence, the work done in lifting the dog is 117.6 Joules

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A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = \frac{1}{2}mv^{2} + 0

2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

7 0
3 years ago
Read 2 more answers
The amount of work done by two boys who apply 200 N of force in an
Aleksandr [31]

The amount of work done by two boys who apply 200 N of force in an unsuccessful attempt to move a stalled car is 0.

Answer: Option B

<u>Explanation: </u>

Work done is the measure of work done by someone to push an object from its present position. We can also define work done as the amount of forces needed to move an object from its present position to another position. So the amount of work done is directly proportionate to the product of forces acting on the object and the displacement of the object.  

           \text { Work done }=\text { Force } \times \text { displacement }

So in this present case, as the two boys have done an unsuccessful attempts to push a stalled car so that means the displacement of the car is zero as there is no change in the position of the car. But they have applied a force of 200 N each. So the amount of work done will be

           \text { Work done }=200 \mathrm{N} \times 0=0

Thus, the amount of work done by two boys will be zero due to their unsuccessful attempt to move a stalled car.

8 0
2 years ago
1.- Se desea elevar un cuerpo de 1500kg utilizando una elevadora hidráulica de plato grande
kvv77 [185]

Answer:

181.48 N

Explanation:

Calculate the area :

Area = pi * r² ;

pi = 3.14 ; r1 = 90cm /100 = 0.9m ; r2 = 10/100 = 0.1m

Area 1, A1 = 3.14 * 0.1² = 0.0314 m²

Area 2, A2 = 3.14 * 0.9² = 2.5434 m²

Force, F = mass * acceleration due to gravity

F2 = 1500 * 9.8 = 14700 N

Force 1 / Area 1 = Force 2 / Area 2

Force 1 = (Force 2 / Area 2), * Area 1

Force 1 = (14700 / 2.5434) * 0.0314

Force = 5779.6650 * 0.0314

= 181.48 N

5 0
3 years ago
U235 + n → Xe134 + Sr100 + 2n
xenn [34]
Nuclear fission formula by the looks of it. Possibly how Professor Lisa Meitner realised that she had split the atomic nucleus. The Xenon and the Strontium (Xe and Sr) would presumably show up in a radio chemical assaying test at her university. 
A few years later, Professor J Robert Oppenheimer watched a nuclear test somewhere near Los Alamos, US and lamented "I am become death, the destroyer of worlds". Shortly thereafter, Hiroshima and Nagasaki were razed to the ground and annihilated by nuclear bombs. Professor Meitner, probably inadvertently, had got the keys to the doors to "nuclear hell", and JRO ended up turning them. Something like that maybe, and a very harrowing and tumultuous period in human history.
Note in the fission equation, that out come two neutrons. They go off and produce a similar fission in another U235 nucleus into a chain reaction which, i not moderated by, say, Boron, can end up as a "mushroom cloud".
8 0
2 years ago
Which formula correctly expresses the property density?
Olenka [21]

Answer:

Option A

D = m/v

Explanation:

Density is defined as mass per unit volume of an object. Therefore, D=m/v where m is the mass of the object and v is the volume

Therefore, option A is the right option

6 0
2 years ago
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