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3 years ago

12

If it actually hits the ground with a speed of 8.50 m/s , what is the magnitude of the average force of air resistance exerted o

n it?

1 answer:

kkurt [141]3 years ago

8 0

Well it will probably never hit the ground try to look over your answer and go from there I hope I helped

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James is planning a science fair project on sound waves. He places an alarm inside a jar which he can remove the

otez555 [7]

Answer: c

Explanation:

Sound waves cannot travel through a medium

3 0

3 years ago

Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0 incline. The coeffi cient of kinetic friction between the bl

UkoKoshka [18]

Answer:

a) $W_g =61.25J$

b) $W_k = -46.25J$

c) $W_N = 0$

d) $W_g$ would be the same.

$W_k$ would decrease.

$W_N$ would be the same.

Explanation:

a) On an inclined plane the force of gravity is the sine component of the weight of the block.

$F_g = mg\sin(\theta) = 5(9.8)\sin(30^\circ)\\W_g = F_g x = 5(9.8)\sin(30^\circ)2.5 = 61.25J$

b) The friction force is equal to the normal force times coefficient of friction.

$F_k = -mg\cos(\theta)\mu_k = -5(9.8)cos(30^\circ)0.436 = -18.5 N\\W_k = -F_kx = -46.25J$

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.

d) The relation between the vertical height and the distance on the ramp is

$h = x\sin(\theta)$

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of $x\sin(\theta)$ .

The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.

The work done by the normal force would still be zero.

5 0

3 years ago

Which of these is NOT an effect of humor?

IgorLugansk [536]

Feelings of jealousy and envy

7 0

3 years ago

A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational

a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: $F = 49.06N$

Pebble: $F = 29.44N$

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: $a =9.8m/s^{2}$

Pebble: $a =9.8m/s^{2}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{m1m2}{r^{2}}$ (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em> $m = 5.0 Kg$ <em>:</em>

m1 will be equal to the mass of the Earth $m1 = 5.972×10^{24} Kg$ and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth $r = 6371000m$ .

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}$

$F = 49.06N$

Newton's second law can be used to know the acceleration.

$F = ma$

$a =\frac{F}{m}$ (2)

$a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}$

$a =9.8m/s^{2}$

<em>Case for the pebble </em> $m = 3.0 Kg$ <em>:</em>

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}$

$F = 29.44N$

$a =\frac{F}{m}$

$a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}$

$a =9.8m/s^{2}$

3 0

3 years ago

Read 2 more answers

Which spacecraft carried the first Spacelab? -Columbia -Challenger -Discovery -Atlantis

Nat2105 [25]

The Atlantis spacecraft carried the first space lab.

Answer: Option 4

<u>Explanation: </u>

The first space lab is named as ATLAS 1 which is the abbreviation of Atmospheric Laboratory for Applications and Science. It is a short space lab set up in space to observe the atmospheric changes and other scientific experiments in the outer atmosphere of Earth from space.

It contains hi-tech instruments and facilities. It was a part of Phase I of NASA’s mission to planet Earth. This helped in better understanding of Earth’s outer and inner atmosphere. So, the spacecraft used to carry the ATLAS 1 is named as Atlantis.

4 0

3 years ago