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Yakvenalex [24]
3 years ago
13

SOMEBODY PLEASE HELP!! 20 PTS!!!

Physics
2 answers:
Neporo4naja [7]3 years ago
4 0
1. D
2. C
3. D
4. B
5. D
aliya0001 [1]3 years ago
3 0

Answer:

1. Reflection

2. travel from one medium to another

3. Same waves to travel in opposite direction.  

Explanation:

1. When a wave strikes a solid barrier, it bounces back in the same medium. This wave behavior of bouncing back is known as reflection. Its like a basketball hitting a backboard. The ball bounces back at the same angle as it was incident. ∠i = ∠r

2. For refraction to occur in a wave, the wave must travel from one medium to another. When light travels from through mediums of different optical densities, it bends. The wave bends away normal when it enters from denser medium to rarer medium. The wave bends towards the normal when it enters from rarer to denser medium. The angle of refraction and angle of incidence are related by Snell's law.

\frac{sin(i)}{sin(r)} = \frac{\mu_2}{\mu_1}

3. The formation of standing wave requires two same waves to travel in the opposite direction and interfere. The incident wave and reflected wave when interfere, form standing waves. There waves are also resonances or harmonics. A standing wave oscillates at one place and does not transfers any energy.

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Answer:14 s

Explanation:

Given

Velocity of ant is 0.1 m/s in a direction 45^{\circ}

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i.e. from diagram

\sin 45=\frac{1}{L}

L=\sqrt{2}

time=\frac{Distance}{speed}

t=\frac{\sqrt{2}}{0.1}

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A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, v_{o} = 20.0 m/s

Angle made by the ramp, \theta = 22.0^{\circ}

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH

where

H = dsin22^{\circ} = 5sin22^{\circ}

g = 9.8 m/s^{2}

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}

v = \sqrt{400 - 19.6\times 5sin22^{\circ}} = 19.06 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m

(b) now, considering the coefficient of friction bhetween ramp and the ball, \mu = 0.150:

velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH - \mu gd

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5

v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5} = 18.7 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m

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Answer:

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Answer:

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(Note that either negative or positive values go to show the positioning and hence, they are not a strong necessity in your final answer.)

So happy that i could help you!

Now this question could turn out to be easy for you!!

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