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baherus [9]
3 years ago
6

Ohh by asks zh a. Mans jks shall snnkkksajsiaoksoo

Physics
2 answers:
Ahat [919]3 years ago
5 0

Answer:

han

Explanation:

jabxhxuxjx

zhannawk [14.2K]3 years ago
4 0

Answer: fREe pOINtS

Explanation:

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The critical angle between a medium and air is 430. What is the speed of light in that medium?
vagabundo [1.1K]

Answer:

The speed of light is that medium is 281907786.2 m/s.

Explanation:

since the critical angle is Фc = 430, we know that the refractive index is given by:

n = 1/sin(Фc)

  = 1/sin(430)

  = 1.06

then if n is the refractive index of the medium and c is the speed of light, then the speed of light in the medium is given by:

v = c/n

  = (3×10^8)/(1.06)

  = 281907786.2 m/s

Therefore, the speed of light is that medium is 281907786.2 m/s.

8 0
2 years ago
QUESTION 7
ryzh [129]

Answer:

<em>The force required is 3,104 N</em>

Explanation:

<u>Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = ma

Where a is the acceleration of the object.

On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

v_f=v_o+at

Where:

vf is the final speed

vo is the initial speed

a is the acceleration

t is the time

Solving for a:

\displaystyle a=\frac{v_f-v_o}{t}

We are given the initial speed as vo=20.4 m/s, the final speed as vf=0 (at rest), and the time taken to stop the car as t=7.4 s. The acceleration is:

\displaystyle a=\frac{0-20.4}{7.4}

a=-2.757\ m/s^2

The acceleration is negative because the car is braking (losing speed). Now compute the force exerted on the car of mass m=1,126 kg:

F = 1,126\ kg * 2.757\ m/s^2

F= 3,104 N

The force required is 3,104 N

6 0
2 years ago
The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?
Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight =   60 .3 (9.81) = 591.54 N

Accel of lift changes this to    60.3 ( 9.81 - L)     where L - accel of lift

                                           60.3 ( 9.81 - L ) = 564

                                               solve for L = .457 m/s^2  DOWNWARD

                                                        so L = - .457 m/s^2

4 0
1 year ago
Solve the problem.
gulaghasi [49]
As the shock waves travel in concentric outward circles from the epicenter, and the diameter is measured 120 miles,
area of a circle =<span>π</span><span>r*r</span>

d=120
<span>r=<span>120/2</span></span><span>r=60</span><span><span>60*60</span>=3600</span><span>3600*π=11309.734</span>
<span>11309.734 square miles</span>
5 0
2 years ago
A current of 12 amps is measured in a circuit with a total resistance of 9.0 ohms. What is the size of the voltage source that s
Serjik [45]
Data:
i (current) = 12 A
R (resistance) = 9.0 Ω
V (voltage) = ? (volts)

Formula:
V = R*i

Solving:
V = R*i
V = 9.0*12
\boxed{\boxed{V = 108\:volts}}\end{array}}\qquad\quad\checkmark
6 0
2 years ago
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