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2 years ago

11

The law of reflection applies to ONLY totally reflected waves. Is it true or false

1 answer:

saw5 [17]2 years ago

8 0

The answer is T. B.

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HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!pls

barxatty [35]

I think this is the solution:

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5 0

2 years ago

If the resistance of a circuit increases, that means that the current has to

Gnoma [55]

Answer:

If resistance increases current decreases.

Explanation:

Current is <em>inversely proportional</em> to the resistance.
from the relation given below, we can clearly see the relation between current and resistance;

V=IR

I ∝ 1/R

This relation shows that when resistance increases,current decreases.

4 0

2 years ago

** URGENT** The voltage across the primary winding is 350,000 V, and the voltage across the secondary winding is 17,500 V. If th

Vlad [161]

As we know that in transformers we have

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

here we know that

$V_s = 17,500 Volts$

$V_p = 350,000 Volts$

$N_s = 600 coils$

now from above equation we will have

$\frac{17500}{350000} = \frac{600}{N_p}$

$N_p = 600\times \frac{350000}{17500}$

$N_p = 12000 coils$

6 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

2 years ago

Helppp plssss I am needing help for science and there is a photo pls answer

disa [49]

Answer:

sorry I don't really know :P

Explanation:

7 0

2 years ago

Read 2 more answers