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AlekseyPX
3 years ago
8

What is true about coal as an energy source

Physics
1 answer:
seropon [69]3 years ago
6 0
Nonrenewable............................
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A 75 kg astronaut floating in space throws a 5 kg rock at 5 m/sec. How fast does the astronaut move backwards?
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The astronaut will get a velocity 0.064ms−1 opposite to the direction of the object.

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Which formula can be used to calculate the horizontal displacement of a horizontally launched projectile?
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At which points would the river be traveling the slowest?
Vesnalui [34]

Answer:

This would be traveling at the lower reaches.

Explanation:

A river would be traveling the fastest at the upper reaches and it becomes slower at the middle reaches and the slowest at the lower reaches. A place where water flows fast in a river is where the width is narrow and the bottom is steep. (This is just examples incase you would like to keep notes).

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Are dimensionless quantities always unitless​
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3 years ago
Read 2 more answers
A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the
lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

l^2 = x^2 + y^2-----(1)

Differentiating with respect to t ( time ),

0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}  ( l = 26 feet = constant )

\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}

\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

We have,

y = 10, \frac{dx}{dt}= -3\text{ feet per min}

\frac{dy}{dt}=\frac{3x}{10}-----(X)

(i) From equation (1),

26^2 = x^2 + 10^2

676=x^2 + 100

576 = x^2

\implies x = 24\text{ feet}

From equation (X),

\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}

(ii) From equation (X),

\frac{dy}{dt}\propto x

Thus, for different value of x the value of \frac{dy}{dt} would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

\frac{dy}{dt}=0\text{ feet per min}

3 0
3 years ago
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