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svet-max [94.6K]
2 years ago
11

Unit Test Review

Physics
1 answer:
PolarNik [594]2 years ago
4 0

Answer:

option B...

they represent different concept...

i hope this will answer your question

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A heat pump absorbs heat from the atmosphere at a rate of 30 kW. If work is being done to run this heat pump at a rate of 7.7 kW
Volgvan

Answer:

Option D 3.9

Explanation:

First, you need to use the correct equation which is the following:

COP = Q/W

Where:

Q = heat absorbed

W = work done by the pump

COP = coefficient of perfomance

We have all the data, so, all you need to do is replace in the above expression and you shoould get the correct result:

COP = 30 / 7.7

COP = 3.896

This result you can round it to 3.9. option D.

8 0
3 years ago
If Johnny won a 300 meter race in 40 seconds, his speed would be ?
irina1246 [14]

Formula: s = d/t

s = speed

d = distance

t = time

Solve using the values we are given.

s = 300/40

s = 7.5m/s

Best of Luck!

4 0
3 years ago
Read 2 more answers
A 30-g bullet is fired with a horizontal velocity of 460 m/s and becomes embedded in block B which has a mass of 3 kg. After the
ICE Princess25 [194]

Answer:

energy loss due to friction and the impacts = 2.97 J; The impact loss due to AB impacting the carrier is =25.72J; The impact loss at first impact is 6,316.64J

Explanation:

First find the velocity of the bullet after the first impact using

M1V1 + 0 = (M1 + M2)v'

Where M1 is the mass of the bullet

M2 is the mass of the block B

M3 is the mass of the carrier

v' is the velocity

v' = M1V1/(M1 + M2)

v'= (30 × 10^-3 kg)(460 m/s) / (30 × 10^-3 kg + 3 kg)

v' = 13.8/3.03

v'= 4.55m/s

Also calculate final velocity of the carrier v2'

v2' = M1V1/(M1 + M2 + M3)

v2'= (30 × 10 kg)(460 m/s) / (30 × 10 kg + 3 kg + 30kg)

v2' =0.42m/s

Now to calculate energy loss due to friction

Normal force

N= W1 + W2 = (m1 + m2)g

Where W1 and W2 is the weight of the bullet and block respectively

g is gravitational acceleration for taken as 9.81m/s

= (0.030 kg + 3 kg)(9.81 m/s) 29.724 N

Friction force = coefficient of kinetic× normal force

Where coefficient of kinetic = 0.2

Ff = (0.2)(29.724)= 5.945 N

Now

Energy loss due to friction = frictional force × distance

Assume distance is 0.5 m.

Energy loss due to friction = 5.945 N × 0.5 m

= 2.97J

Kinetic energy of block with embedded bullet immediately after first impact:

1/2 × (m1 + m2)(v')^2

1/2 × (30 × 10^-3 kg + 3 kg)(4.55m/s)^2

= 1/2 × (3.03kg) × (4.55m/s)^2

= 31.36 J

Final kinetic energy of bullet, Block, and Carrier together

1/2 × (m1 + m2 + m3)(v2')^2

1/2 × (30 × 10^-3 kg + 3 kg + 30kg) (0.42m/s)^2

1/2 × (33.03kg) × (0.42m/s)^2

= 2.91 J

Therefore

Loss due to friction and stopping impact = Kinetic energy of block with embedded bullet immediately after first impact - Final kinetic energy of bullet, Block, and Carrier together

= 31.36 J - 2.91 J

= 28.69 J

Impact loss due to AB impacting the carrier = loss due to friction- energy due to friction

28.69J - 2.97J

=25.72J

Initial kinetic energy of system ABC = 1/2(m1vo)

=1/2(0.030 kg)(460 m/s)^2 = 6,348J

Therefore

Impact loss at first impact = Initial kinetic energy of system ABC - Kinetic energy of block with embedded bullet immediately after first impact:

= 6,348J - 31.36 J

= 6,316.64J

3 0
3 years ago
One of the great triumphs of spectroscopy was when astronomers identified a new element in the sun (one that was only later foun
Nezavi [6.7K]

Answer:

When Helium is identified by astronomers is one of the great triumphs of spectroscopy.

Explanation:

Janssen managed to do this great triumphs on August 18, 1868 . Janssen was the first person to introduce the helium, an element that never seen before on Earth, in the solar spectrum. At that time, he didn’t know that what he’d seen—he just think that it was something new. In the mid of 1800, the spectroscope instrument is introduced in astronomy.

Later on we heard that all helium in Universe has been created by the fusion of hydrogen nuclei.

To know more about Spectroscopy:

brainly.com/question/14677550

#SPJ4

6 0
2 years ago
1. Why is only a small portion of the solar energy that strikes Earth's atmosphere stored by primary producers?
DanielleElmas [232]

Explanation:

only a fraction of radiation strikes plants or algae and only a portion of that fraction is of wavelengths suitable for photosynthesis and much energy is lost as a result of reflection or heating of plant tissue

5 0
2 years ago
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