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lukranit [14]
3 years ago
9

Train A is moving at 100 kmh–1 through a station. Train B is also travelling at 100 kmh–1 through the station but in the opposit

e direction to train A. What is the speed of a seated passenger on? (a) train A relative to an observer on the station? (b) train A relative to another seated passenger on the same train? (c) train B relative to a passenger on train A?
Physics
1 answer:
lisov135 [29]3 years ago
7 0

Explanation:

(a) An observer on the station has a speed of 0 km/h.  The speed of  a passenger on Train A is 100 km/h.  The relative speed is 100 km/h − 0 km/h = 100 km/h.

(b) The speed of both passengers is 100 km/h, in the same direction.  The relative speed is 100 km/h − 100 km/h = 0 km/h.

(c) The speed of both passengers is 100 km/h, in opposite directions.  The relative speed is 100 km/h − (-100 km/h) = 200 km/h.

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They attract and stick together
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Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon'
kramer
Chameleon's tongue is more fast than thought. Its long sticky tongue moves at an amazing ballistic speed which lashes out unsuspecting insects and bugs. Now let us see how fast it is.

GIven:

acceleration of the chameleon's tongue- 260 m/s
2 for 20 ms
constant speed 30 ms
50 ms total time
1/20 of a second

solution:

<u>260</u> = <u>  n</u><u>  </u>
20       50
<u>20 n </u>= <u>13, 000</u>
20            20
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4 0
2 years ago
You suspect that a power supply is faulty, but you use a power supply tester to measure its voltage output and find it to be acc
ValentinkaMS [17]

Answer:

Load

Explanation:

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3 0
3 years ago
If a bullet loses 1/nth of its velocity while passing through a plank,then no of planks required to stop the bullet is?
Novay_Z [31]
Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing <span><span>1/nth</span><span>1/nth</span></span><span> of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous).

</span><span>With this assumption, the energy loss becomes
</span><span>
ΔE = <span>1/2 </span>m<span>v2 </span>− <span>1/2 </span>m <span><span>(<span>v−<span>v/n</span></span>) </span><span>2
</span></span></span>
and the number of planks <span>NN</span><span> becomes
</span>
N = <span><span><span>1/2</span>m<span>v2 /</span></span><span>ΔE </span></span>= <span><span>n2/ </span><span>2n−1
</span></span>
Otherwise, if you assume that the bullet loses <span><span>1/<span>nth</span></span><span>1/<span>nth</span></span></span><span> of its velocity per plank, then the answer is </span><span><span>N=∞</span></span><span><span>

</span>
</span>
8 0
3 years ago
Tommy was walking at a rate of 4 miles hour at noon and at 12:30 pm he was walking at a brisk rate of 6 miles hour . Two hours l
maks197457 [2]

Answer:

B) Tommy had a positive acceleration between noon and 12:30 pm.

Explanation:

Acceleration is defined as the rate of change of velocity:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time

In the problem,

- At noon, Tommy is walking at a velocity of 4 mi/h

- At 12.30 pm, Tommy is walking at a velocity of 6 mi/h

- A time of half an hour (0.5 h) passed between the two moments

So Tommy's acceleration is

a=\frac{6 mi/h-4 mi/h}{0.5 h}=4 mi/h^2

and the acceleration is positive, since the velocity has increased.

3 0
3 years ago
Read 2 more answers
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