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Mumz [18]
3 years ago
5

An astronaut in a space craft looks out the window and sees an asteroid move pas a backward direction at 68 mph relative to the

space craft. If the velocity of the space craft is 126 mph relative to the position of the sun, what is the velocity of the asteroid relative to the sun?
Physics
1 answer:
Annette [7]3 years ago
6 0

Answer: -194 mph

Explanation:

Taking into account the <u>Sun as the center </u>(origin, point zero) <u>of the reference system</u>, the velocity of the spacecraft relative to the Sun V_{R-S} is:

V_{R-S}=126mph  Note it is <u>positive</u> because the spacecraft is moving <u>away</u> from the Sun

Taking into account the <u>spacecraft as the center of another reference system</u>, the velocity of the asteroid relative to the spacecraft V_{A-R} is:

V_{A-R}=-68mph Note it is <u>negative</u> because the asteroid is moving<u> towards</u> the spacracft.

Now, the velocity of the asteroid relative to the Sun V_{A-S} is:

V_{A-S}=V_{A-R}-V_{R-S}

V_{A-S}=-68mph-126mph

Finally:

V_{A-S}=-194mph This is the velocity of the asteroid relative to the Sun and its negative sign indicates it is <u>moving towards the Sun</u>.

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What component of a longitudinal sound wave is analogous to a trough of a transverse wave?
anzhelika [568]

Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.

While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.  

6 0
3 years ago
0.125 C of charge flow out of a
zimovet [89]

Answer:

Resistance = 252.53 Ohms

Explanation:

Given the following data;

Charge = 0.125 C

Voltage = 5 V

Time = 6.3 seconds

To find the resistance;

First of all, we would determine the current flowing through the battery;

Quantity of charge, Q = current * time

0.125 = current * 6.3

Current = 0.125/6.3

Current = 0.0198 A

Next, we find the resistance;

Resistance = voltage/current

Resistance = 5/0.0198

Resistance = 252.53 Ohms

3 0
3 years ago
A machine does 500 j of work in 20 sec. What is the power of this machine?
NNADVOKAT [17]

Explanation:

P=W/t

P=500/20

P=25 W

6 0
3 years ago
Jonathan must lift a 3000n rock using a lever. jonathan uses 300n of force to push down on the lever to move the rock. what is t
-Dominant- [34]
The load is the weight of the rock that Jonathan lifts:
L=3000 N
The effort instead is the force applied in input to the lever in order to lift the rock:
E=300 N

So, the ratio between load and effort for this exercise is
\frac{L}{E}= \frac{3000 N}{300 N}=10
So, the ratio is 10:1.
3 0
3 years ago
Compare the catching of two different water balloons.
Stels [109]

Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

Δv₁ = Change in velocity = Final velocity - Initial velocity

Mass = Density × Volume

Density of water = 0.997 g/mL

Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

The momentum change is;

Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s

Case B:  Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s

Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

c. The momentum change = impulse therefore Case B involves the greatest impulse

d. Here we have;

Impulse = Momentum change = F_{average} × Δt = mΔV

∴ F_{average} = m·ΔV/Δt

∴ For Case A F_{average} = 149.55×8/Δt =  1196.4/Δt N

For Case B  F_{average} = 598.2×8/Δt =  4785.6/Δt

Where Δt is the same for Case A and Case B,  F_{average}  for Case B >>  F_{average}  for Case B

Therefore, Case B involves the greatest force.

4 0
3 years ago
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