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Galina-37 [17]
2 years ago
3

A spaceship in outer space has a doughnut shape with 440-m outer radius. The inhabitants stand with their heads toward the cente

r and their feet on an outside rim (see the figure).
Over what time interval would the spaceship have to complete one rotation on its axis to make a bathroom scale have the same reading for the person in space as when on Earth's surface?

Image for A spaceship in outer space has a doughnut shape with 440-m outer radius. The inhabitants stand with their head
Physics
1 answer:
mojhsa [17]2 years ago
8 0

Answer:

s

Explanation:

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A diver in the pike position (legs straight, hands on ankles) usually makes only one or one-and-a-half rotations. To make two or
Korolek [52]

Answer:

from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

Explanation:

This can be explained on the basis of conservation of angular momentum.

This means the initial and the final angular velocity is conserved. Consider initial position (1)in the pike and final position in the be truck position. So there inertia's will also be different.

⇒I_1\omega_1 = I_2\omega_2

\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}

also,

I_1= mr_1^2

I_2= mr_2^2

since, r_2^2

I_2^2

therefore,

\omega_1^2

So, from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

6 0
2 years ago
A baby elephant is stuck in a mud hole. to help pull it out, game keepers use a rope to apply a force f with arrowa, as part a o
Julli [10]

The two forces should be equal therefore:

2.10 * Fa = Fa + 2 * F * cos 18

simplifying the right side:

2.10 * Fa = Fa + 1.902 * F

1.10 Fa = 1.902 F

<span>F / Fa = 0.578</span>

3 0
2 years ago
Consider a 4-mg raindrop that falls from a cloud at a height of 2 km. When the raindrop reaches the ground, it won't kill you or
docker41 [41]

Answer:

The work done by the air resistance is -0.0782 J

Explanation:

Hi there!

The energy of the raindrop has to be conserved, according to the law of conservation of energy.

Initially, the raindrop has only gravitational potential energy:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the raindrop.

g = acceleration due to gravity (9.8 m/s²)

h = height.

Let´s calculate the initial potential energy of the drop:

(convert 4 mg into kg: 4 mg · 1 kg / 1 × 10⁶ mg = 4 × 10⁻⁶ kg)

PE = 4 × 10⁻⁶ kg · 9.8 m/s² · 2000  m

PE = 0.0784 J

When the drop starts falling, some of the potential energy is converted into kinetic energy and some energy is dissipated by the work done by the air resistance. On the ground all the initial potential energy has been either converted into kinetic energy or dissipated by the resistance of the air:

initial PE = final KE + W air

Where:

KE = kinetic energy.

W air = work done by the air resistance.

The kinetic energy when the raindrop reaches the ground is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass

v = velocity

Then:

KE = 1/2 ·  4 × 10⁻⁶ kg · (10 m/s)²

KE = 2 × 10⁻⁴ J

Now, we can calculate the work done by the air resistance:

initial PE = final KE + W air

0.0784 J = 2 × 10⁻⁴ J + W air

W air = 0.0784 J - 2 × 10⁻⁴ J

W air = 0.0782 J

Since the work is done in the opposite direction to the displacement, the work is negative, then, the work done by the air resistance is -0.0782 J.

5 0
2 years ago
A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction
yawa3891 [41]

Answer:

F'=\dfrac{F}{4}

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

For the slower car on the road,

F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

8 0
2 years ago
Which of the following can be reduced to a single number in standard form?
raketka [301]

Complete question is;

Which of the following can be reduced to a single number in standard form?

A) 3√3 + 5√8

B) 2√5 + 5√45

C) √7 + √9

D) 4√2 + 3√6

Answer:

Only option B) 2√5 + 5√45 can be reduced to a single number

Explanation:

A) For 3√3 + 5√8;

Let's simplify it to get;

3√3 + 5√(4 × 2)

From this, we get;

3√3 + (5 × 2)√2 = 3√3 + 10√2

This is 2 numbers and not a single number. Thus it can't be reduced to a single number in standard form.

B) 2√5 + 5√45

Simplifying to get;

2√5 + 5√(9 × 5)

This gives;

2√5 + (5 × 3)√5 = 2√5 + 15√5

Adding the surds gives;

17√5.

This is a single number and thus can be reduced to a single number

C) For √7 + √9

Simplifying, to get;

√7 + 3.

This is 2 numbers and not a single number. Thus it can't be reduced to a single number in standard form.

D) 4√2 + 3√6

Thus can't be simplified further because both numbers inside the square root don't have factors that are perfect squares.

Thus, it remains 2 numbers and not a single number and can't be reduced to a single number in standard form.

6 0
2 years ago
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