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Fudgin [204]
3 years ago
15

What does snells law of refraction fail

Physics
1 answer:
Art [367]3 years ago
6 0
Snell's Law of Refraction fail<span> when light incidents on the surface of separation of the 2 media normally or through the normal. It is because when light incidents through the normal, the angle of incidence is equal to zero. Therefore the angle of </span>Refraction<span>is also zero. ... Therefore the </span>Snell's Law fails<span> here.</span>
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A piece of amber is charged by rubbing with a piece of fur. If the net excess charge on the fur is +8.0 nC (+8.0 10-9 C), how ma
Ann [662]

Answer:

Number of electron on the fur will be 5\times 10^{10}electron

Explanation:

We have given net charge on the fur q=8nC=8\times 10^{-9}C

Charge on one electron e=1.6\times 10^{-19}C

Let there are n number of electrons

So ne=8\times 10^{-9}

n\times 1.6\times 10^{-19}=8\times 10^{-9}

n=5\times 10^{10}electron

So number of electron on the fur will be 5\times 10^{10}electron

6 0
3 years ago
A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo
blondinia [14]

Answer:

His pitching speed is 38 m/s.

Explanation:

Hi there!

Please see the attached figure for a better understanding of the problem.

The position of the ball at any time t is given by the following vector:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector of the ball at time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Let's place the origin of the frame of reference at the throwing point so that x0  and y0 = 0.

When the ball reaches the ground, its position vector will be r1 (see figure). Using the equation of the vertical component of the position vector, we can find the time at which the ball reaches the ground. At that time, the horizontal component of the position is 30 m and the vertical component is -3.0 m (see figure):

y = y0 + 1/2 · g · t²  (y0 = 0)

y = 1/2 · g · t²

-3.0 m = 1/2 · (-9.8 m/s²) · t²

-3.0 m / -4.9 m/s² = t²

t = 0.78 s

Now, knowing that at this time x = 30 m, we can find v0:

x = x0 + v0 · t  (x0 = 0)

x = v0 · t

30 m = v0 · 0.78 s

v0 = 30 m / 0.78 s

v0 = 38 m/s

His pitching speed is 38 m/s.

3 0
4 years ago
A missile is moving 1350 m/s at a 25.0 angle
murzikaleks [220]
I will answer both versions assuming what you want to know is the distance it travels up from and over the ground. and how long until it reaches space. 540 meters per second up and over. to reach space which is 100km above sea level, it would take about 5400 minutes
4 0
3 years ago
A particle moves along the x-axis. It is initially at the position 0.170 m, moving with a velocity of 0.190 m/s and acceleration
sukhopar [10]

Answer:

(a)-0.701m

(b)-0.674m/s

Explanation:

Given a = -0.24,m/s² Xo = 0.170m, u = 0.190m/s

X = Xo + ut +1/2at²

X = 0.170 + 0.190×3.6 + 1/2×(-0.24) 3.6² = -0.701m

(b) its velocity at the end of this time interval is v

v = u + at

= 0.190 + (-0.24)× 3.6 = -0.674m/s

7 0
3 years ago
Con base en el tema ley de ohm, resistividad y resistencia resuelva los siguientes
Bumek [7]
Answer:Explanation:gfgfgfgfgfgfgfgfgfg
3 0
3 years ago
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