PHYSICS COLLEGE

A 7600 kg satellite is in a circular orbit around Earth at a height of 2300 km above Earth's surface. What is this satellite's speed

Answers

Answer 1
Answer:

Answer:

6779.7m/s

Explanation:

Using

GMm/(Re +h)² = mv²/ (Re+h)

So making v subject we have

V= √GM/Re+h

So

V = √ 6.67*10^-11 x 5.97*10^24/(6371+2300)*10^3

V= 6779.7m/s

Note h = height of satellite

Re= radius of the earth

M = mass of the earth

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A force of 240.0 N causes an object to accelerate at 3.2 m/s2. What is the mass of the object?

Answers

the mass would be 75kg

1. The workpart in a turning operation is 88 min in diameter and 400 mm long. A feed of 0.25 mm/rev is used in this operation. If cutting speed is 3.5 m/s, the too should be changed in every 3 workparts, but if the cutting speed is 2.5 m/sec, the tool can be used to produce 20 pieces between the tool changes. Determine the cutting speed that will allow the tool to be used for 50 parts between tool changes.

Answers

Find the given attachments

What is the maximum distance allowed between the center of hole #2 and datum B as seen in the front view?

Answers

Answer:

4.003" (inches )

Explanation:

The maximum distance allowed between the center of hole #2 and datum B can be calculated by adding 4.000" + 0.003" ( perpendicularity of the of hole #2) as seen from the front view of the diagram .

Note :The hole 2 is sited below the workpiece when viewed from the front view while the Datum B is positioned on the left end of the workpiece also note that the diameter is

An 800-kHz sinusoidal radio signal is detected at a point 6.6 km from the transmitter tower. The electric field amplitude of the signal at that point is 0.780 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the amplitude of the magnetic field of the signal at that point

Answers

Answer:

0.0000000026 T

Explanation:

E_0 = Maximum electric field strength = 0.78 V/m

B_0 = Maximum magnetic field strength

c = Speed of light = 3* 10^8\ m/s

Relation between amplitudes of electric and magnetic fields is given by

E_0=B_0c\n\Rightarrow B_0=(E_0)/(c)\n\Rightarrow B_0=(0.78)/(3* 10^8)\n\Rightarrow B_0=0.0000000026\ T

The amplitude of the magnetic field is 0.0000000026 T

You're driving along at 25 m/s with your aunt's valuable antiques in the back of your pickup truck when suddenly you see a giant hole in the road 55 m ahead of you. Fortunately, your foot is right beside the brake and your reaction time is zero! Can you stop without the antiques sliding and being damaged? Their coefficients of friction are μs=0.6 and μk=0.3. Hint: You're not trying to stop in the shortest possible distance. What's your best strategy for avoiding damage to the antiques?

Answers

Answer:

Explanation:

initial velocity, u = 25 m/s

distance, s = 55 m

coefficient of static friction = 0.6

coefficient of kinetic friction = 0.3

Let the acceleration is a.

Use third equation of motion

v² = u² + 2as

0 = 25 x 25 - 2 x a x 55

a = 5.68

a = μg

μ = 5.68 / 9.8 = 0.58

so, the coefficient of friction is less then the coefficient of static friction so the antiques are safe.

Part A A microphone is located on the line connecting two speakers hat are 0 932 m apart and oscillating in phase. The microphone is 2 83 m from the midpoint of the two speakers What are the lowest two trequencies that produce an interflerence maximum at the microphone's location? Enter your answers numerically separated by a comma

Answers

Answer:

a) f=368.025\ \textup{Hz}

b) f_2=736.051\ \textup{Hz}

Explanation:

Given:

The distance between two speakers (d) = 0.932 m

The distance of the microphone from the midpoint = 2.83 m

Thus, distance of microphone from the nearest speaker (L) = 2.83 - (0.932/2) = 2.364 m

also, the distance of the microphone from the farther speaker (L') = 2.83 + (0.932/2) = 3.296 m

Now,

The path difference is calculated as

L' - L = d = 0.932 m

Now,for a maxima to be produced at the microphone, the waves must constructively interfere.

for this to happen the path difference should be integral multiple of the wavelength.

thus,

\textup d = n\lambda

hence, the largest wavelength will be for n = 1,

therefore,

0.932 = 1 × λ

or

λ = 0.932 m

now, the velocity of sound is given as c = 343 m/s

thus, the frequency will be

f=(c)/(\lambda)

on substituting the values, we get

f=(343)/(0.932)=368.025\ \textup{Hz}

now, the 2nd largest wavelength will be for n = 2

0.932 = 2 × λ

or

λ = 0.466

thus, the frequency will be

f_2=(343)/(0.466)=736.051\ \textup{Hz}

hence, these are the lowest first two frequencies.
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