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3 years ago

Is a lump of coal chemical energy

Physics

1 answer:

AVprozaik [17]3 years ago

5 0

Answer:

well, not unless the coal is reacting with something. if it's just sitting there, then it can't power anything.

Explanation:

You might be interested in

Which is the wavelength of a wave that travels at a speed of 3.0 × 10^8 m/s and has a frequency of 1.5 × 10^16 Hz?

k0ka [10]

Answer:

2 x 10^-8

Explanation:

the formula of wavelength is

the speed divided by frequency

so you have the speed given = 3.0x10^8m/s

and frequency = 1.5×10^16 Hz

so wavelength = 3.0x10^8m/s / 1.5x10^16 Hz

7 0

3 years ago

How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and

aev [14]

Answer:

h= 46.66 m

Explanation:

Given that

Initial speed of the car ,u = 110 km/h

We know that

1 km/h= 0.277 m/s

u= 30.55 m/s

lets height gain by car is h.

The final speed of the car will be zero at height h.

v²=u²- 2 g h

v= 0 m/s

0²=30.55²- 2 x 10 x h ( g = 10 m/s²)

h= 46.66 m

4 0

3 years ago

Give a real-world example of how energy is transformed from electrical energy to thermal energy. Describe how the heat can be tr

timofeeve [1]

Answer:

A microwave

Explanation:

Car

Lightbulb is a really good one

Same with the sun. That one has Chemical as well

3 0

2 years ago

How much force does it take to accelerate a 2000 kg car at 1m/s^2

Dmitry [639]

Answer:

2000 N

Explanation:

F=ma

m=2000 kg

a=1m/s^2

F=(2000 kg)(1m/s^2)

F=2000 N

4 0

3 years ago

Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa

tester [92]

Answer:

Part a)

$t = 2.83 s$

Part b)

Ball thrown downwards = $v_f = 23.8 m/s$

Ball thrown upwards = $v_f = 23.8 m/s$

Part c)

$d = 22.24 m$

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

$\Delta y = 0 = v_y t + \frac{1}{2}at^2$

$0 = 13.9 t - \frac{1}{2}(9.81) t^2$

$t = 2.83 s$

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

$v_f^2 - v_i^2 = 2 a \Delta y$

$v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)$

$v_f^2 = 567.9$

$v_f = 23.8 m/s$

Part c)

Relative speed of two balls is given as

$v_{12} = v_1 - v_2$

$v_{12} = (13.9) - (-13.9) = 27.8 m/s$

now the distance between two balls in 0.8 s is given as

$d = v_{12} t$

$d = 27.8 \times 0.8$

$d = 22.24 m$

7 0

3 years ago