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tekilochka [14]
2 years ago
8

What energy change takes place when the wrecking ball strikes the wall?

Physics
1 answer:
Ilia_Sergeevich [38]2 years ago
3 0

Answer:

D

Explanation:

The wrecking ball transfers kinetic energy to the wall.

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3 years ago
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Why are some forces considered to be noncontact forces? A. Objects must be far apart in order to exert a force. B. Objects do no
kompoz [17]

Answer:

B. Objects do not have to touch each other to experience a force.

Explanation:

For example ..One of the noncontact forces is magnetic force whereby a magnetic object will be attracted to another magnetic object of oppsite charged particles, through waves called electromagnetic waves. On the other hand, the two magnetic objects of similar charged particles can repel through electromagnetic waves..

8 0
3 years ago
A fan spins at 6.0 rev/s. You turn it off, and it slows at 1.0 rev/s2. What is the angular displacement before it stops
Komok [63]

Answer:

Angular displacement before it stops = 18 rev

Explanation:

Given:

Speed of fan w(i) = 6 rev/s

Speed of fan (Slow) ∝ = 1 rev/s

Final speed of fan w(f) = 0 rev/s

Find:

Angular displacement before it stops

Computation:

w(f)² = w(i) + 2∝θ

0² = 6² + 2(1)θ

0 = 36 + 2θ

2θ = -36

Angular displacement before it stops = -36 / 2

θ = -18

Angular displacement before it stops = 18 rev

4 0
3 years ago
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s344n2d4d5 [400]

Answer:

Yes

Explanation:

Element D will also burn in oxygen, because elements classified under same group has similar physical or chemical properties.

5 0
4 years ago
If 5.4 J of work is needed to stretch a spring from 15 cm to 21 cm and another 9 J is needed to stretch it from 21 cm to 27 cm,
qaws [65]

Answer:

the natural length of the spring is 9 cm

Explanation:

let the natural length of the spring = L

For each of the work done, we set up an integral equation;

5.4 = \int\limits^{21-l}_{15-l} {kx} \, dx \\\\5.4 = [\frac{1}{2}kx^2 ]^{21-l}_{15-l}\\\\5.4 = \frac{k}{2} [(21-l)^2 - (15-l)^2]\\\\k = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}  \ \ \ -----(1)

The second equation of work done is set up as follows;

9 = \int\limits^{27-l}_{21-l} {kx} \, dx \\\\9 = [\frac{1}{2}kx^2 ]^{27-l}_{21-l}\\\\9 = \frac{k}{2} [(27-l)^2 - (21-l)^2] \\\\k = \frac{2(9)}{(27-l)^2 - (21-l)^2} \ \ \ -----(2)

solve equation (1) and equation (2) together;

\frac{2(9)}{(27-l)^2 - (21-l)^2} = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}\\\\\frac{2(9)}{2(5.4)} = \frac{(27-l)^2 - (21-l)^2}{(21-l)^2 - (15-l)^2}\\\\\frac{9}{5.4} = \frac{(729 - 54l+ l^2) - (441-42l+ l^2)}{(441-42l+ l^2) - (225 -30l+ l^2)} \\\\\frac{9}{5.4 } = \frac{288-12l}{216-12l} \\\\\frac{9}{5.4 } =\frac{12}{12}  (\frac{24-l}{18 -l})\\\\\frac{9}{5.4 } = \frac{24-l}{18 -l}\\\\9(18-l) = 5.4(24-l)\\\\162-9l = 129.6-5.4l\\\\162-129.6 = 9l - 5.4 l\\\\32.4 = 3.6 l\\\\l = \frac{32.4}{3.6} \\\\

l = 9 \ cm

Therefore, the natural length of the spring is 9 cm

4 0
3 years ago
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