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Doss [256]
3 years ago
13

According to the law of reflection, _____ in the diagram below.

Physics
1 answer:
inessss [21]3 years ago
4 0

Answer:

C. B=C

Explanation:

According to the law;

<em>The</em><em> </em><em>angle</em><em> </em><em>of</em><em> </em><em>incidence</em><em>(</em><em>i</em><em>)</em><em> </em><em>is</em><em> </em><em>equal</em><em> </em><em>to</em><em> </em><em>the </em><em>angle</em><em> </em><em>of</em><em> </em><em>reflect</em><em>ion</em><em> (</em><em>r</em><em>)</em>

Let B be the incident Ray, i.

And C be the reflected Ray, r.

Peace.

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Usain Bolt accelerates at a rate of 3.7
MissTica

Answer:

Explanation:

I think that you to run more than 12 miles

4 0
4 years ago
All Physicsts over here plz help in these questions!!!!!!!!!
mrs_skeptik [129]
Hello!

First one we can use that PE=mgh so we have

4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m

Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have

g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2

Hope this helps. Any questions please ask. Thank you.
6 0
4 years ago
Read 2 more answers
A diode for which the forward voltage drop is 0.7 V at 1.0 mA is operated at 0.5 V. What is the value of the current
Vesnalui [34]

Answer:

the current value is 0.335 \mu A

Explanation:

The computation of the value of the current is given below:

z_i = I_s e^{\frac{0.7}{ut} }= 10^{-3}\\\\Z_z = I_s e^{\frac{0.5}{ut} }\\\\\frac{Z_z}{Z_i}= \frac{Z_z}{10^{-3}}  = e^{\frac{0.5\times 0.7}{0.025} }\\\\= 0.335 \mu A

Hence, the current value is 0.335 \mu A

4 0
3 years ago
A catapult with a spring constant of 10,000 N/m is used to launch a target from the deck of a ship. The spring is compressed a d
Mnenie [13.5K]

Answer:

(C) 40m/s

Explanation:

Given;

spring constant of the catapult, k = 10,000 N/m

compression of the spring, x = 0.5 m

mass of the launched object, m = 1.56 kg

Apply the principle of conservation of energy;

Elastic potential energy of the catapult = kinetic energy of the target launched.

¹/₂kx² = ¹/₂mv²

where;

v is the target's  velocity as it leaves the catapult

kx² = mv²

v² = kx² / m

v² = (10000 x 0.5²) / (1.56)

v² = 1602.56

v = √1602.56

v = 40.03 m/s

v ≅ 40 m/s

Therefore, the target's velocity as it leaves the spring is 40 m/s

6 0
3 years ago
PLEASE HELPPPP!!!!
Tresset [83]

you will hear a higher pitch due to a higher frequency.

3 0
3 years ago
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