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omeli [17]
3 years ago
6

Write three example for types of energy​

Physics
2 answers:
AURORKA [14]3 years ago
8 0
1.) chemical
2.) nuclear
3.) thermal
Hatshy [7]3 years ago
3 0
These are some examples. The different types of energy include thermal energy, radiant energy, chemical energy, nuclear energy, electrical energy, motion energy, sound energy, elastic energy and gravitational energy.
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The manufacturer of a bulletproof vest wants the vest to be able to stop a bullet with a mass of 0.4 kg and a velocity of 1800 m
11111nata11111 [884]
We know that momentum = mass times velocity
So a. 720 kgm/s
7 0
3 years ago
A halfback on an apparent breakaway for a touchdown is tackled from behind. If the halfback has a mass of 98 kg and was moving a
uranmaximum [27]

Answer:

The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

Explanation:

Given that,

Mass of halfback = 98 kg

Speed of halfback= 4.2 m/s

Mass of corner back = 85 kg

Speed of corner back = 5.5 m/s

We need to calculate their mutual speed immediately after the touchdown-saving tackle

Using conservation of momentum

m_{h}v_{h}+m_{c}v_{c}=m_{h+c}v_{h+c}

Where, m_{h}= mass of halfback

m_{c}=mass of corner back

v_{h}= velocity of halfback

v_{c}= velocity of corner back

Put the value into the formula

98\times4.2+85\times5.5=(98+85)\times v

v=\dfrac{98\times4.2+85\times5.5}{98+85}

v=4.80\ m/s

Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

3 0
3 years ago
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0
3 years ago
Two identical charges are separated by a distance d. If the distance between them is increased to 3d, what will happen to the fo
belka [17]
A-It will be one-ninth the original force
3 0
3 years ago
Read 2 more answers
Given that:<br><br> = 2i + 9j ; and ⃗ = -i – 4j . Find . ⃗ ​
Marat540 [252]

Answer:

A.B = -38

Explanation:

A = 2i + 9j and B = -i - 4j.

So, A.B = (2i + 9j).(-i - 4j)

= 2i.(-i) + 2i.(-4j) + 9j.(-i) + 9j.(-4j)

= -2i.i - 8i.j - 9j.i - 36j.j

since i.i = 1, j.j = 1, i.j = 0 and j.i = 0, we have

A.B = -2(1) - 8(0) - 9(0) - 36(1)

A.B = -2 - 0 - 0 - 36

A.B = -38

5 0
3 years ago
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