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mihalych1998 [28]
2 years ago
12

Playing catch with a friend in a moving train. When you toss the ball in the direction the train is moving, how does the speed o

f the ball appear to an observer standing at rest outside the train?
I understand basic relative motion, just the ball moving on another moving object confuses me!
Physics
2 answers:
likoan [24]2 years ago
5 0
Well it would simply appear faster because the speed of the ball and the speed of the train.
telo118 [61]2 years ago
5 0

Answer:

The speed of the ball appear faster when we observe it from ground frame at rest.

Explanation:

As we know that by relate motion concept

v_{AB} = v_A - v_B

here when ball is thrown in the direction of train motion with velocity "v" relative to the train then we will assume here that let say the train at that moment is running at speed v_T

now by above concept of relative velocity we will have

v = v_{ball} - v_T

now for the speed of ball

v_{ball} = v + v_T

so from this equation we can see that the actual speed of ball in ground frame will be more than the speed observed in train frame

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Answer:

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5) Motion of rocket due to velocity of expelled gas

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What is power, and what is its relationship to voltage and amperage?
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3 years ago
Correct the following statement:
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It's the 3rd option

Explanation:

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3 years ago
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Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

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Explanation:

We have given radius of mars R_{mars}=6.9\times 10^3km=6.9\times 10^6m and radius of earth R_{E}=1.3\times 10^4km=1.3\times 10^7m

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So mass of mars M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg

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So density of mars d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3

Volume of earth  V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3

So density of earth d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3

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(B) Value of g on mars

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(c) Escape velocity is given by

v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec

5 0
3 years ago
Read 2 more answers
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gtnhenbr [62]
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For rectilinear motion, the distance traveled is equal to the initial velocity times the time, plus one-half of the acceleration times the square of the time. For projectile motion, the maximum distance is equal to the square of the initial velocity multiplied with the square of the sine of the launch angle, all over twice the gravity.
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3 years ago
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