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nexus9112 [7]
2 years ago
12

What is an incident wave,and a normal

Physics
1 answer:
Dafna1 [17]2 years ago
5 0

Answer:

An incident wave is a current or voltage wave that travels through a transmission line from the generating source towards the load. It becomes incident when it arrives at a discontinuity or another medium with different propagation characteristics. Wave normal. A unit vector which is perpendicular to an Equiphase surface of a wave, and has its positive direction on the same side of the surface as the direction of propagation.

Explanation:

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What is the direction of the net force that acts on an object undergoing uniform circular motion?
8_murik_8 [283]
D. The direction of the force is toward the center of the object's circular path.
5 0
3 years ago
Read 2 more answers
On a sky coaster (human pendulum) that reaches 10 meters from it's equilibrium position, a man of 120 kg is able to reach a maxi
dimaraw [331]

Answer:

14 m/s

Explanation:

Using the principle of conservation of energy, the potential energy is converted to kinetic energy, assuming any losses.

Kinetic energy is given by ½mv²

Potential energy is given by mgh

Where m is the mass, v is the velocity, g is acceleration due to gravity and h is the height.

Equating kinetic energy to be equal to potential energy then

½mv²=mgh

V

Making v the subject of the formula

v=√(2gh)

Substituting 9.81 m/s² for g and 10 m for h then

v=√(2*9.81*10)=14.0071410359145 m/s

Rounding off, v is approximately 14 m/s

6 0
3 years ago
Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of
slamgirl [31]

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

<u>Explanation:</u>

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2

Substituting the velocities in the equation

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0
2 years ago
19,792,000,000 in scientific notation will have how many significant figures
xz_007 [3.2K]

Answer:

= 1.9792 × 10^10

Significant Figures= 5

Explanation:

Look at the attachment below

Hope this helps (:

8 0
3 years ago
A drunken sailor stumbles 580 meters north, 530 meters northeast, then 480 meters northwest. What is the total displacement and
kvv77 [185]

Answer:

(a)  1294.66 m

(b) 88.44°

Explanation:

d1 = 580 m North

d2 = 530 m North east

d3 = 480 m North west

(a) Write the displacements in vector forms

\overrightarrow{d_{1}}=580\widehat{j}

\overrightarrow{d_{2}}=530\left ( Cos45\widehat{i}+Sin45\widehat{j} \right )

\overrightarrow{d_{2}}=374.77\widehat{i}+374.77\widehat{j}

\overrightarrow{d_{3}}=480\left ( - Cos45\widehat{i}+Sin45\widehat{j} \right )

\overrightarrow{d_{3}}=-339.41\widehat{i}+339.41widehat{j}

The resultant displacement is given by

\overrightarrow{d}\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}

\overrightarrow{d}=\left ( 374.77-339.41 \right )\widehat{i}+\left ( 580+374.77+339.41 \right )\widehat{j}

\overrightarrow{d}=35.36\widehat{i}+1294.18\widehat{j}

magnitude of the displacement

d ={\sqrt{35.36^{2}+1294.18^{2}}}=1294.66 m

d = 1294.66 m

(b) Let θ be the angle from + X axis direction in counter clockwise

tan\theta =\frac{1294.18}{35.36}=36.6

θ = 88.44°

4 0
3 years ago
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