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topjm [15]
3 years ago
15

A 950 N skydiver jumps from an altitude of 3000 m. What is the total work performed on the skydiver?

Physics
1 answer:
pentagon [3]3 years ago
6 0

Answer:

2850000J

Explanation:

Given parameters:

Weight of diver = 950N

Altitude  = 3000N

Unknown:

Total work done by the sky diver = ?

Solution:

To solve this problem, we use the expression:

        Work done  = Weight x height

Now insert the parameters:

        Work done  = 950 x 3000  = 2850000J

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Where is there potential energy throughout the loading, cocking, and releasing if the trebuchet?
pshichka [43]

Answer:

adapted from NOVA, a team of historians, engineers, and trade experts recreate a medieval throwing machine called a trebuchet. To launch a projectile, a trebuchet utilizes the transfer of gravitational potential energy into kinetic energy. A massive counterweight at one end of a lever falls because of gravity, causing the other end of the lever to rise and release a projectile from a sling. As part of their design process, the engineers use models to help evaluate how well their designs will work.

Explanation:

4 0
3 years ago
Larger animals use proportionately less energy than smaller animals; that is, it takes less energy per kg to power an elephant t
marin [14]

Answer:

Part a)

Q = 952 cal/day

Part b)

P = 46 Watt

Part c)

\Delta P = 54 W

Explanation:

As we know that 5000 kg African elephant requires 70,000 Cal for basic needs per day

so we will have

m = 5000 kg

Q = 70,000 Cal

so we have energy required per kg

E = \frac{70,000}{5000}

E = 14 cal/kg

Part a)

now we know that per kg the energy required will be same

so we have mass of the human is 68 kg

so energy required per day is given as

Q = 68 \times 14

Q = 952 cal/day

Part b)

Resting power is the rate of energy in Joule required per sec

so it is given as

P = \frac{952 \times 4186}{24\times 3600}

P = 46 Watt

Part c)

resting power given in the book is

P' = 100 W

so this is less than the power given

\Delta P = 100 - 46

\Delta P = 54 W

6 0
3 years ago
How are the direction of the net force on an object and the direction of the object’s acceleration related?
Oksanka [162]

They're the same direction.

8 0
3 years ago
Where is visible light located on the electromagnetic spectrum and why?
Marysya12 [62]
Radio waves, gamma-rays, visible light, and all the other parts of the electromagnetic spectrum are electromagnetic ... The different types of radiation are defined by the the amount of energy found in the photons.
5 0
3 years ago
In medieval warfare, one of the greatest technological advancement was the trebuchet. The trebuchet was used to sling rocks into
NNADVOKAT [17]

Answer:

2) a_y= -g  3) vₓ=constant v_y = v_{oy} - g t, 4)  vₓ = v₀ₓ - ax t

5)  changes the horizontal speed, should change range

7) changes the vertical speed change the maximum height

Explanation:

1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.

2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration

3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression

           v_y = v_{oy} - gt

at the point of maximum height, vy = 0 is equal to the maximum height

4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed

In the graph it would be directed to the left, therefore the velocity would be

           vₓ = v₀ₓ - ax t

5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.

the equations of motion are

           x = v₀ₓ t

           y = v_{oy} t - ½ g t²

7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,

the equations of motion are the same.

4 0
3 years ago
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