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VashaNatasha [74]
1 year ago
10

A 12 kg block is accelerating at the rate of 9.0 m/s? while being acted on by two forces.

Physics
1 answer:
Aliun [14]1 year ago
8 0

Answer:

See below

Explanation:

F = ma

F = 12 * 9 = 108 N

  108 N  needed      <u> add  30 N more east </u>

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Hold the paper up to a mirror so
Ne4ueva [31]

Answer:

The image of everything in front of the mirror is reflected backward, retracing the path it traveled to get there. Nothing is switching left to right or up-down. Instead, it's being inverted front to back. ... That reflection represents the photons of light, bouncing back in the same direction from which they came

Explanation:

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2 years ago
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To stretch a certain nonlinear spring by an amount x requires a force F given by F = 40 x − 6 x 2 , where F is in Newtons and x
strojnjashka [21]

Answer:

64 J

Explanation:

The potential energy change of the spring ∆U = -W where W = work done by force, F.

Now W = ∫F.dx

So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)

∆U = - ∫-Fdx

=  ∫F.dx

Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U =  ∫₀²F.dx

=  ∫₀²(40x - 6x²).dx

=  ∫₀²(40xdx - 6x²dx)

=  ∫₀²(40x²/2 - 6x³/3)

=  ∫₀²(20x² - 2x³)

= [20x² - 2x³]₀²

= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)

= [(20(4) - 2(8)) - (0 - 0))

= [80 - 16 - 0]

= 64 J

5 0
3 years ago
what is the largest and smallest possible resultant force of two force with magnitude of 41N and 14N​
MArishka [77]

Explanation:

6gtvctyvyvyvyvyfcfufy

8 0
3 years ago
Can lamp that works on a 2.5 v work on a 1.12 v ?​
12345 [234]

Answer:

Explanation:

Thinking about the logics it can but it may be dim because 1.12 is lower than 2,5v so this will mean u lamp may not work or may work very dimely due to the low voltage it is receiving.

5 0
2 years ago
How much voltage is in the primary coil if there are 3200 windings in the
Lesechka [4]

Answer:

Voltage in primary coil is 3.91 V

Explanation:

For transformer we know that the working principle is given as

\frac{V_1}{V_2} = \frac{N_1}{N_2}

here we know that

V_1 [tex] = voltage in primary coil[tex]V_2 = 25 V

N_1 = 500

N_2 = 3200

Now we have

\frac{V_1}{25} = \frac{500}{3200}

V_1 = 3.91 V

8 0
3 years ago
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