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3 years ago

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A laser beam enters a 12.0 cm thick glass window at an angle of 33.0° from the normal. The index of refraction of the glass is 1

.41. At what angle from the normal does the beam travel through the glass? How long does it take the beam to pass through the plate?

1 answer:

Alenkinab [10]3 years ago

6 0

Answer:

The time is 0.563 ns.

Explanation:

Given that,

Index of refraction of glass = 1.41

Distance = 12.0 cm

Angle = 33.0°

We need to calculate the refraction angle

Using Snell's law

$n_{1}\sin\theta=n_{r}\sin\theta$

put the value into the formula

$1\times\sin33=1.41\sin\theta$

$\sin\theta=\dfrac{1\times\sin33}{1.41}$

$\theta=22.71^{\circ}$

We need to calculate the velocity of beam in glass

Using formula of velocity

$v=\dfrac{c}{n}$

Put the value into the formula

$v=\dfrac{3\times10^{8}}{1.41}$

$v=2.13\times10^{8}\ m/s$

We need to calculate the time

Using formula of distance

$v=\dfrac{d}{t}$

$t=\dfrac{12.0\times10^{-2}}{2.13\times10^{8}}$

$t=5.63\times10^{-10}\ sec$

$t=0.563\times10^{-9}\ sec$

$t=0.563\ ns$

Hence, The time is 0.563 ns.

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A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s

strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

$m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v$

Where:

$m_{M}$ , $m_{S}$ - Masses of the small meteorite and the communication satellite, measured in kilograms.

$v_{M}$ , $v_{S}$ - Speeds of the small meteorite and the communication satellite, measured in meters per second.

$v$ - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

$v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}$

If $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ and $v_{S} = 0\,\frac{m}{s}$ , the final speed is now calculated:

$v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}$

$v = 0.1\,\frac{m}{s}$

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

$K_{S} + K_{M} - K - Q_{disp} = 0$

$Q_{disp} = K_{S}+K_{M}-K$

Where:

$K_{S}$ , $K_{M}$ - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

$K$ - Kinetic energy of the satellite-meteorite system, measured in joules.

$Q_{disp}$ - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

Given that $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ , $v_{S} = 0\,\frac{m}{s}$ and $v = 0.1\,\frac{m}{s}$ , the dissipated heat is:

$Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]$ $Q_{disp} = 200000\,J$

$Q_{disp} = 200\,kJ$

The energy coverted to heat is 200 kilojoules.

4 0

3 years ago

A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M a

aleksklad [387]

Answer:

a) y = 0.98 t², t=1s y= 0.98 m,

b) he two blocks must move the same distance

c) v = 1.96 m / s, d) a = -1.96 m / s², e) x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

N-W = 0

N = M g

X axis

T- fr = Ma

the friction force has the expression

fr = μ N

fr = μ Mg

small block

w- T = m a

we write the system of equations

T - fr = M a

mg - T = m a

we add and resolved

mg- μ Mg = (M + m) a

a = $g \ \frac{m - \mu M}{m+M}$

a = $9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}$

a = 9.8 (6/30)

a = 1.96 m / s²

a) now we can use the kinematic relations

y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

y = ½ a t²

y = ½ 1.96 t²

y = 0.98 t²

for t = 1s y = 0.98 m

t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

v = vo + a t

v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

-fr = M a

- μ M g = M a

a = - μ g

a = - 0.2 9.8

a = -1.96 m / s²

e) the distance to stop the block is

v² = vo² - 2 a x

0 = vo² - 2a x

x = vo² / 2a

x = 1.96² / 2 1.96

x = 0.98 m

the time to travel this distance is

v = vo - a t

t = vo / a

t = 1.96 /1.96

t = 1 s

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3 years ago

A busy chipmunk runs back and forth along a straight line of acorns that has been set out between his burrow and a nearby tree.

saveliy_v [14]

Answer: a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

Explanation:

Acceleration is the rate of change in the velocity per time

a = change in velocity/time

a = ∆v/t

average acceleration a = (v2 -v1)/t. ....1

Given;

Final velocity v2 = 1.63m/s

Initial velocity v1 = -1.15ms

time taken t = 2.11s

Substituting into eqn 1

a = [1.63 - (-1.15)]/2.11

a = (1.63+1.15)/2.11

a = 2.78/2.11

a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

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I believe the answer would be mass. Low mass stars and medium mass stars often become white dwarfs when they die while high mass stars explode in violent explosions called supernovas and usually leave behind a black hole or a neutron star.

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