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3 years ago

5

You are 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 of the ball appear to an observer standing at rest outside the train?
I know it appears to be traveling faster than the train, but I don't know the technicalities

1 answer:

S_A_V [24]3 years ago

4 0

I think to an observer outside the train, the speed of the ball will actually look like more than the speed of the train.

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An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity c

KiRa [710]

Acceleration can be defined as the change of speed in an instant of time, that is

$a = \frac{v_f-v_i}{\Delta t}$

Here,

$v_f =$ Final velocity

$v_i =$ Initial velocity

$\Delta t =$ Change in time

From this expression we will calculate the requested values replacing the variables in each of the given terms

PART A) The values under this condition are:

$v_f = 5.3m/s$

$v_i = 15m/s$

$\Delta t = 10.4s$

Replacing,

$a = \frac{5.3m/s-(15m/s)}{10.4s}$

$a = -0.93m/s^2$

PART B ) The values under this condition are:

$v_f = -15m/s$

$v_i = -5.3m/s$

$\Delta t = 10.4$

Replacing,

$a = \frac{-(15m/s)-(-5.3m/s)}{10.4s}$

$a = -0.93m/s^2$

Therefore the acceleration in the second time interval is $-0.93m/s^2$

PART C) The values under this condition are:

$v_f = -15m/s$

$v_i = 15m/s$

$\Delta t = 10.4$

Replacing,

$a = \frac{-15m/s-(15m/s)}{10.4s}$

$a = -2.9m/s^2$

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3 years ago

The physical (natural) laws which establish how things work can be investigated by the scientific method. True or False

UNO [17]

Answer:

True

Explanation:

It's a bit hard to explain, so you can watch a youdube video on it or something.

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2 years ago

Read 2 more answers

nevsk [136]

Answer:

The mass is $m = 3.75 \ kg$

Explanation:

From the question we are told that

The initial temperature is $T_1 = 15^oC$

The final temperature is $T_2 = 100 ^o C \ (boiling point )$

Generally the maximum heat produced by 1 Liter of natural gas is $9000 \ cal$

So the amount of heat produced by 100 L is

$E = 9000 * 100$

=> $E = 9000 00 \ cal$

Generally given that the efficiency is $\eta = 0.35$

Then actual heat received by the water is

$H = 0.35 * E$

=> $H = 0.35 * 9000 00$

=> $H = 315000 \ cal$

Converting to kcal

=> $H = 315000 \ cal = \frac{315000}{1000} = 315 \ kcal$

Generally the specific heat of water is

$c_w = 1 kcal/ kg \cdot ^oC$

Generally the heat received by the water is mathematically represented as

$H = m * c_w * (T_2 - T_1)$

=> $315 = m * 1 * ( 100 - 15 )$

=> $m = 3.75 \ kg$

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Deffense [45]

C because conduction you don’t need to touch anything

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The order of magnitude is 10⁸ .

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