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Licemer1 [7]
2 years ago
13

ASAP

Physics
1 answer:
nikitadnepr [17]2 years ago
6 0

Answer:

A. 59.4

Explanation:

The refractive index of the glass, n₁ = 1.50

The angle of incidence of the light, θ₁ = 35°

The refractive index of air, n₂ = 1.0

Snell's law states that n₁·sin(θ₁) = n₂·sin(θ₂)

Where;

θ₂ = The angle of refraction of the light, which is the angle the light will have when it passes from the glass into the air

Therefore;

θ₂ = arcsin(n₁·sin(θ₁)/n₂)

Plugging in the values of n₁, n₂ and θ₁ gives;

θ₂ = arcsin(1.50 × sin(35°)/1.0) ≈ 59.357551° ≈ 59.4°

The angle the light will have when it passes from the glass into the air, θ₂ ≈ 59.4°.

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A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k
Veronika [31]

Answer:

14.98\ \text{kg m/s}

45.26^{\circ}

Explanation:

P_1 = Initial momentum of the pin = 13 kg m/s

P_i = Initial momentum of the ball = 18 kg m/s

P_2 = Momentum of the ball after hit

55^{\circ} = Angle ball makes with the horizontal after hitting the pin

\theta = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}

Momentum in the y direction

P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}

(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}

The pin's resultant velocity is 14.98\ \text{kg m/s}

P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}

The pin's resultant direction is 45.26^{\circ} below the horizontal or to the right.

4 0
2 years ago
A 1000kg car is rolling slowly across a level surface at 1 m/s heading twoards a group o fsmall innocent children. The doors are
Degger [83]

Answer:

The force required to push to stop the car is 288.67 N

Explanation:

Given that

Mass of the car, m = 1000 kg

Initial speed of the car, u = 1 m/s

The car and push on the hood at an angle of 30° below horizontal, \theta=30^{\circ}

Distance, d = 2 m

Let F is the force must you push to stop the car.

According work energy theorem theorem, the work done is equal to the change in kinetic energy as :

W=\dfrac{1}{2}m(v^2-u^2)F\times d=\dfrac{1}{2}m(v^2-u^2)

v = 0

Fd\ cos\theta=\dfrac{1}{2}m(u^2)      F=\dfrac{\dfrac{1}{2}m(u^2)}{d\ cos\theta}F=\dfrac{\dfrac{1}{2}\times 1000\times (1)^2}{2\ cos(30)}F = -288.67 N

The force required to push to stop the car is 288.67 N

3 0
2 years ago
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