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3 years ago

What occurs when parallel rays of light hit a rough or bumby surface

1 answer:

5 0

When parallel rays of light hit a rough or bumpy surface, there will be diffused reflection. Meaning, the light that bounces after it hits a rough or bumpy surface is not parallel anymore and cannot form an image. The answer to this question is 'diffused reflection'.

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Raising 100 grams of water from 40 to 60 °C (the specific heat capacity of water is 1

faust18 [17]

Heat in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:

Heat = mC(T2-T1)
Heat = 100(1)(60-20)
Heat = 4000 calories addition to the system

Hope this answers the question. Have a nice day.

4 0

3 years ago

Which is an example of a wedge? A. B C D

Salsk061 [2.6K]

Answer:

B I believe, because the axe is wedged into the log

4 0

3 years ago

Read 2 more answers

A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

5 0

4 years ago

A small mailbag is released from a helicopter that is descending steadily at 2.00 m/s. After 5.00s what is the speed of the mail

grandymaker [24]

The acceleration of gravity (on Earth) is 9.8 m/s² downward.

This means that every falling object gains 9.8 m/s more downward speed
every second that it falls.

In 5 seconds of falling, it gains (5 x 9.8 m/s) = 49 m/s of downward speed.

If it was already descending at 2.0 m/s at the beginning of the 5 sec,
then at the end of the 5 sec it would be descending at

(2 m/s + 49 m/s) = 51 m/s .

7 0

3 years ago

The air around a pool and the water in the pool receive equal amounts of energy from the sun. Why does the air experience a grea

labwork [276]

Answer:

Explanation:

4 0

3 years ago