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nevsk [136]
3 years ago
15

Can someone help me with the exercise 3 only c,d,e please!!!

Physics
2 answers:
saul85 [17]3 years ago
4 0
I'm not sure...
I feel ya.
fgiga [73]3 years ago
3 0
Im not sure, but i think you should try C
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Cars A and B have the same mass, but Car A's speed is 15 mph, whereas Car B is moving at 60 mph. The kinetic energy
kupik [55]

Answer:

16 times that of car A

Explanation:

Moving objects have kinetic energy which is calculated using the formula:

E=(1/2)mv² where m is the mass of the object and v is its velocity.

For car A, kinetic energy =1/2×m×15²

=112.5m........................................i

For car B,  the kinetic energy =1/2×m×60²

=1800m......................................ii

dividing ii by i we get

1800m/112.5m= 16.07

Therefore The kinetic energy of car B is 16.07 times that of car A

7 0
3 years ago
Increasing which of the following would increase the magnetic force between the permanent magnet and the coil? A. Transformer B.
Gnoma [55]

For this one, all you really need to do is eliminate any answers
that are absurd or meaningless.

You can't increase a transformer.
You can't increase a circuit.
You can't increase a generator.

When the <em><u>current</u></em> through a coil of wire increases, the magnetic field
around the coil increases, so there would be more magnetic force
between the coil and a permanent magnet.


7 0
3 years ago
&lt;
goblinko [34]

Answer:

Explanation:

initial velocity u = 32.7 m /s

final velocity v = 50.3 m /s

displacement s = 44500 m

acceleration a = ?

v² = u² + 2 a s

50.3² = 32.7² + 2 x a x 44500

2530.09 = 1069.29 + 89000a

a .016 m /s²

time taken t = ?

v = u + at

50.3 = 32.7 + .016 t

t = 1100 s

6 0
2 years ago
A car was at a 30km marker and began to drive. The car passed a 90km marker and the time had changed 30
dolphi86 [110]

Answer:

i think it's 2km pm

Explanation:

2km x 30 60.. start was 30, and now your at 90.. we had to determine how much time it took.. so 2 is the average.. or atleast per minute and sorry it i still didnt answer ur question lol im just trynna help

7 0
2 years ago
When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero fil
Anika [276]

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}

We are given the second smallest nonzero thickness at which destructive interference occurs.

This corresponds to, m = 2, therefore

2t = 2\lambda_{film}

t = \lambda_{film}

The index of refraction of soap is given, then

\lambda_{film} = \frac{\lambda_{vacuum}}{n}

Combining the results of all steps we get

t = \frac{\lambda_{vacuum}}{n}

Rearranging, we find

\lambda_{vacuum} = tn

\lambda_{vacuum} = (278)(1.33)

\lambda_{vacuum} = 369.74nm

4 0
3 years ago
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