What is the total energy of a 175,000kg shuttle orbiting the Earth at 600km above Earth's surface?

Answers

Answer 1

Answer: m = 175,000 kg
g = 10 m/s^2
h = 600,000 m

Ep = mgh
Ep = 1050 * 10^9 [J]

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An incandescent light bulb produces light when electrons flow through thea. air.
b. glass.
c. filament.
d. vacuum.

Answers

Electriccurrent passes through a filament of an incandescent bulb, thereby increasingit temperature. When current flows, it contains electrons through the filamentto produce light. The answer is c. Typically, incandescent light bulb consistsof a glass enclosure that contains tungsten filament. The glass enclosure containseither a vacuum or an inert gas that serves as the filament protection fromevaporating. Incandescent light bulbs contain a stem attached at to its base toallow the electrical contacts to run through the envelope without gas or airleaks

Answer:filament

Explanation:

What is the smallest value of the force F such that the 8.0 kg block will not slide down the wall? The coefficient of static friction between the block and the wall is 0.4.Select one:

a. 10N

b. 100N

c. 200N

d. 98N

Answers

The smallest value of the force that will make the block not to slide down is 10 N.

We'll begin by calculating the normal reaction. This can be obtained as follow:

Mass (m) = 8 Kg

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (N) =?

N = mg

N = 8 × 10

N = 80 N

Finally, we shall determine the frictional force.

Coefficient of friction (μ) = 0.4

Normal reaction (N) = 80 N

Frictional Force (F) =?

F = μN

F = 0.4 × 80

F = 32 N

Since the frictional force is 32 N, therefore, a force lesser than the frictional force will make the blocknot to slide down.

From the options given above, only option A has a force that is lesserthan the frictional force.

Therefore, the correct answer to the question is Option A. 10 N

Learn more about frictional force:

brainly.com/question/20049999

Final answer:

The smallest value of the force that will not slide the 8.0 kg block down the wall is 31.36 N.

Explanation:

To determine the smallest value of the force such that the 8.0 kg block will not slide down the wall, we need to consider the static friction between the block and the wall. The formula for static friction is fs = μs * N, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force is equal to the weight of the block, which is mg = 8.0 kg * 9.8 m/s^2 = 78.4 N. Therefore, the smallest value of the force is equal to the maximum static friction force, which can be calculated as fs = 0.4 * 78.4 N = 31.36 N. So the correct answer is 31.36 N.

Learn more about Static Friction here:

brainly.com/question/13000653

#SPJ3

SPEAR is a storage ring at the Stanford Linear Accelerator which has a circulating beam of electrons that are moving at nearly the speed of light (2.998 108 m/s). If a similar ring is about 92.0 m in diameter and has a 0.40 A beam, how many electrons are in the beam

Answers

Answer:

$2.4* 10^(12)$

Explanation:

We are given that

Speed of light,v= $2.998* 10^8 m/s$

Diameter of ring,d=92 m

Radius,r= $(d)/(2)=(92)/(2)=46 m$

Current, I=0.40 A

We have to find the number of electrons in the beam.

We know that

Current,I= $(q)/(t)$

Where q= ne

$e=1.6* 10^(-19) C$

Using the formula

$0.40=(1.6* 10^(-19)n)/((2\pi r)/(v))$

$0.40=(1.6* 10^(-19)n* v)/(2\pi r)$

$0.40=(1.6* 10^(-19)n* 2.998* 10^8)/(2\pi* 46)$

$n=(0.40* 2\pi* 46)/(1.6* 10^(-19)* 2.998* 10^8)=2.4* 10^(12)$

Question no-2 (10 mark)
find the anguler momentum of a body if K.E = 4 jule and M.I = 2kg m2.

Answers

Answer:

KE = 1/2 I ω^2

Given I = 2 kg m^2 and KE = 4 J

ω = (2 KE / I)^1/2 = (2 * 4 / 2)^1/2 = 2^1/2 sec

P = I ω definition of angular momentum

P = 2 kg m^2 * (2 KE / I)^1/2 / sec = 2 * 2^1/2 kg m^2 / sec

The gases cast off by a dying star are called

Answers

Answer:

Planetary Nebula

Explanation:

A star has its own life cycle very similar to a human life cycle where it takes birth, grows up, becomes old and then dies. The life cycle of stars and human differ in the duration of cycle and end result.

When a star dies it can end up differently basis its mass. Here one of the ends of a star is mentioned where it casts off all the gases. Such type of situation will create a planetary nebula. One famous example of such nebula is Crab Nebula.

Planetary Nebulae hope that helps

8. A 740 kg car traveling 19 m/s comes to a complete stop in 2.0 s. What is the force exerted on the car during this stop?

Answers

Answer:

Force exerted on the car is 7030 N.

Explanation:

It is given that,

Mass of the car, m = 740 kg

Initial speed of the car, u = 19 m/s

Final speed of the car, v = 0

Time taken, t = 2 s

Let F is the force exerted on the car during this stop. We know that it is equal to the product of force and acceleration. Mathematically, it is given as :

$F=m* (v-u)/(t)$

$F=740* (0-19)/(2)$

F = -7030 N

So, the force exerted on the car during this stop is 7030 N. Hence, this is the required solution.

force = mass * acceleration

acceleration = change_in_velocity / time

so:

force = 740 kg * (19 m/s - 0 m/s) / 2.0 s
= 740 * 19 / 2 kg m per second^{2}
= 7030 kg m per second^{2}
= 7030 newtons of force

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