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3 years ago

1. The energy an object has while it is moving is called kinetic energy, or the energy of

Physics

1 answer:

Anestetic [448]3 years ago

4 0

Answer:

Explanation:

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A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calc

balandron [24]

Answer:

Approximately $28^{\circ}$ .

Explanation:

The refractive index of the air $n_{\text{air}}$ is approximately $1.00$ .

Let $n_\text{glass}$ denote the refractive index of the glass block, and let $\theta _{\text{glass}}$ denote the angle of refraction in the glass. Let $\theta_\text{air}$ denote the angle at which the light enters the glass block from the air.

By Snell's Law:

$n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}})$ .

Rearrange the Snell's Law equation to obtain:

$\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}$ .

Hence:

$\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}$ .

In other words, the angle of refraction in the glass would be approximately $28^{\circ}$ .

7 0

2 years ago

ball A of mass 5.0 kg moving at 20 meters per second collides with ball B of unknown mass moving at 10. meters per second in the

antiseptic1488 [7]

Answer:The mass of ball B is 10 kg.

Explanation;

Mass of ball A = $M_A=5 kg$

Velocity of the ball A before collision: $U_A=20 m/s$

Velocity of ball A after collision= $V_A=10 m/s$

Mass of ball B= $M_B$

Velocity of the ball B before collision: $U_B=10 m/s$

Velocity of ball B after collision= $V_B=15 m/s$

$M_AV_A+M_BV_B=M_AU_A+M_BU_B$

$5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s$

$M_B=10kg$

The mass of ball B is 10 kg.

8 0

3 years ago

block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8

dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: $\sqrt{800} \,\,\,N$

Explanation:

First try to write all forces in vector component form:

The force F1 acting at 45 degrees would have multiplication factors of $\frac{\sqrt{2} }{2}$ on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is $F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

y-component of F1 is $F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

y-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is: $v_x=4\,*\,2=8\,m/s$