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spayn [35]
3 years ago
7

HELP PLS! :/

Physics
1 answer:
zlopas [31]3 years ago
7 0

Explanation:

formula: Vi = Vf - (at)

Vi: intial velocity

Vf: final velocity

a: acceleration

t: time

fill in formula with the numbers you are given

Vi= 41.6m - ((9.81 m/s^2)(1.89s))

parenthesis first according to pemdas

Vi= 41.6m - 18.54m/s

Answer: 23.06m/s

disclaimer: I havent done physics in awhile so I have no idea if this is right. just an attempt to help steer you in the right direction hopefully. good luck

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In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo
Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of 32^{\circ} (when light enters from the air.)

Explanation:

Let v_{1} denote the speed of light in the first medium. Let v_{\text{air}} denote the speed of light in the air. Assume that the light entered the boundary at an angle of \theta_{1} to the normal and exited with an angle of \theta_{\text{air}}. By Snell's Law, the sine of \theta_{1}\! and \theta_{\text{air}}\! would be proportional to the speed of light in the corresponding medium. In other words:

\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}.

When light enters a boundary at the critical angle \theta_{c}, total internal reflection would happen. It would appear as if the angle of refraction is now 90^{\circ}. (in this case, \theta_{\text{air}} = 90^{\circ}.)

Substitute this value into the Snell's Law equation:

\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}.

Rearrange to obtain an expression for the speed of light in the first medium:

v_{1} = v_{\text{air}} \cdot \sin(\theta_{1}).

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For 0 < \theta < 90^{\circ}, \sin(\theta) is monotonically increasing with respect to \theta. In other words, for \!\theta in that range, the value of \sin(\theta)\! increases as the value of \theta\! increases.

Therefore, compared to the medium in this question with \theta_{c} = 27^{\circ}, the medium with the larger critical angle \theta_{c} = 32^{\circ} would have a larger \sin(\theta_{c}). such that light would travel faster in that medium.

4 0
3 years ago
How Many Days Would a Scientist Have To Wait For The Radioactivity To Be 12.5 The Starting Amount
kiruha [24]
They have to wait 18 days.
3 0
4 years ago
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A screen door spring is stretched
bija089 [108]

Answer:

216.39 N/m

Explanation:

F = 64.7 N

x = 0.299 m

Plug those values into Hooke's Law:

F = kx

64.7 N = k(0.299 m)

---> k = 216.39 N/m

8 0
3 years ago
If the momentum of a 1000 kg car travelling at 10 m/s was transferred completely to a 20.0 kg traffic barrier, what would the fi
denis-greek [22]

Answer:

d. 500 m/s

Explanation:

Momentum: This is the product of the mass of a body to its velocity, The S.I unit of momentum is kgm/s.

Mathematically, momentum can be expressed as,

M = mv....................... equation 1

Where M = momentum, m = mass, v = velocity

deduced from the question,

Momentum of the car = momentum of the barrier.

MV = mv ............................. Equation 1

Where M = mass of the car, V = velocity of the car, m = mass of the barrier, v = velocity of the barrier.

making v the subject of the equation,

v = MV/m........................ Equation 2

Given: M = 1000 kg, V = 10 m/s, m = 20 kg.

Substitute into equation 2

v = 1000(10)/20

v = 500 m/s.

Hence the speed of the barrier = 500 m/s

The right option is d. 500 m/s

8 0
4 years ago
Of the ball is
defon

Answer:b

Explanation:

velocity of a falling object increases.

so Kinetic energy of a falling object increases

height of a falling object from the ground decreases.

so the potential energy of a falling object decreases

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4 years ago
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