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3 years ago

A 62 kg skydiver moving at terminal speed falls 50 m in 1 s. What power is the skydiver expending on the air?

Physics

1 answer:

IgorLugansk [536]3 years ago

3 0

Answer:

P = 30380 W

Explanation:

given,

mass of the skydiver, m = 62 Kg

distance, s = 50 m

time, t = 1 s

we know,

$Power= \dfrac{work\ done}{time}$

$Power= \dfrac{F.s}{t}$

F = m g

$Power= \dfrac{m g.s}{t}$

inserting all the values

$P = \dfrac{62\times 9.8\times 50}{1}$

P = 30380 W

hence, Power the skydiver expending on the air is 30380 W

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Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Ganezh [65]

Answer:

242.85 Hz

Explanation:

For maximum intensity of sound, the path difference,ΔL = (n + 1/2)λ/2 where n = 0,1,2...

Since Abby is standing perpendicular to one speaker, the path length for the sound from the other speaker to him is L₁ = √(2.00² + 5.50²) = √(4.00 + 30.25) = √34.25 = 5.85 m.

The path difference to him is thus ΔL = 5.85 m - 5.50 m = 0.35 m.

Since ΔL = (n + 1/2)λ/2 and for lowest frequency n = 0,

ΔL = (n + 1/2)λ/2 = (0 + 1/2)λ/2 = λ/4

ΔL = λ = v/f and f = v/4ΔL where f = frequency of wave and v = velocity of sound wave = 340 m/s.

f = 340/(4 × 0.35) = 242.85 Hz

5 0

3 years ago

Here's a question from ~ [ AIEEE 2002 ]

lyudmila [28]

r=150m
coefficient of friction= $\mu$ =0.6

As car is avoid skidding

$\\ \sf\hookrightarrow \dfrac{mv^2}{r}=\mu mg$

Cancel m

$\\ \sf\hookrightarrow \dfrac{v^2}{r}=\mu g$

$\\ \sf\hookrightarrow v^2=\mu rg$

$\\ \sf\hookrightarrow v^2=0.6(10)(150)$

$\\ \sf\hookrightarrow v^2=60(150)$

$\\ \sf\hookrightarrow v^2=900$

$\\ \sf\hookrightarrow v=30ms^{-1}$

Done

7 0

2 years ago

Read 2 more answers

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

sveta [45]

Answer:

the two ice skater have the same momentum but the are in different directions.

Paula will have a greater speed than Ricardo after the push-off.

Explanation:

Given that:

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

A. Which skater, if either, has the greater momentum after the push-off? Explain.

The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off

The law of conservation of momentum states that the total momentum of two or more objects acting upon one another will not change, provided there are no external forces acting on them.

So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.

Momentum is the product of mass and velocity.

SO, from the information given:

Let represent the mass of Paula with $m_{Pa}$ and its initial velocity with $u_{Pa}$