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butalik [34]
3 years ago
7

An observer on the earth sees a spaceship approaching at 0.54c. The ship then launches an exploration vehicle that, according to

the earth observer, approaches at 0.82c. What is the velocity of the exploration vehicle relative to the spaceship? Give your answer as a decimal fraction of c, i.e. a β factor.
Physics
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

Explanation:

Expression for relative velocity

= \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2} }

= (.54 + .82 )c/ 1+ \frac{.54 \times.82}{1}

= 1.36 c / 1.4428

= .94 c

β = .94

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How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?
Anestetic [448]

Answer:

21 m

Explanation:

The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

d=vt

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d is the distance covered

v is the speed

t is the time

The frog in this problem has a speed of

v = 2.1 m/s

and therefore, after t = 10 s, the distance it covered is

d=(2.1)(10)=21 m

3 0
3 years ago
Four distinguishable particles move freely in a room divided into octants (there are no actual partitions). Let the basic states
mafiozo [28]

Answer:

Explanation:

Since the door that leads to the room is opened, this gives room for particles to move into the next identical room and divided into octants. Now the amount of space that can be occupied becomes double, the number of basic states has increased by 404916

8 0
3 years ago
when a circular plate of metal is heated in an oven, its radius increases at .03 cm/min, at what rate is the area increasing whe
Alinara [238K]

Answer:

Rate of change of area will be 9.796cm^2/min

Explanation:

We have given rate of change of radius \frac{dr}{dt}=0.03cm/min

Radius of the circular plate r = 52 cm

Area is given by A=\pi r^2

So \frac{dA}{dt}=2\pi r\frac{dr}{dt}

Puting the value of r and \frac{dr}{dt}

\frac{dA}{dt}=2\times 3.14\times 52\times 0.03=9.796cm^2/min

So rate of change of area will be 9.796cm^2/min

6 0
3 years ago
A sonar pulse sent out by a boat arrives back after 4 seconds. If the speed of sound in water is 1600m/s, how deep is the water?
Varvara68 [4.7K]

Answer:

the boat would be deeped by 3200 m

Explanation:

Given that

The boat arrives back after 4 seconds

And, the speed of the sound in water is 1,600 m/s

We need to find out how much deep is the water

So,

As we know that

Distance = ( speed × time) ÷ 2

Here we divided by 2 because the boat arrives back

= (1600 × 4) ÷ 2

= 3200 m

Therefore the boat would be deeped by 3200 m

7 0
3 years ago
21. A 60 kg student on a scooter is pushed with a constant force of 40 N across a horizontal concrete driveway at a constant spe
DedPeter [7]

The magnitude of the force of friction is 40 N

Explanation:

To solve the problem, we just have to analyze the forces acting on the student and the scooter along the horizontal direction. We have:

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- The frictional force, acting backward, F_f

Since the two forces are in opposite direction, the equation of motion is

F-F_f = ma

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m is the mass of the student+scooter

a is the acceleration

However, here the scooter is moving at constant speed: this means that its acceleration is zero, so

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which means that the magnitude of the force of friction is also equal to 40 N.

Learn more about force of friction here:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

3 0
3 years ago
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