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ohaa [14]
3 years ago
5

It's nighttime, and you've dropped your goggles into a 3.2-mm-deep swimming pool. If you hold a laser pointer 1.1 mm above the e

dge of the pool, you can illuminate the goggles if the laser beam enters the water 2.2 mm from the edge. How far are the goggles from the edge of the pool?
Physics
1 answer:
Savatey [412]3 years ago
5 0

Answer: 5.30m

Explanation:

depth of pool = 3.2 m

i = 67.75°

Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

n₁ = 1, n₂ =1.33, r= 44.09°

Hence,

Distance of Google from edge if pool is:

2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) =5.30m

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Two UFPD are patrolling the campus on foot. To cover more ground, they split up and begin walking in different directions. Offic
miss Akunina [59]

Answer:

0.256 hours

Explanation:

<u>Vectors in the plane </u>

We know Office A is walking at 5 mph directly south. Let X_A be its distance. In t hours he has walked

X_A=5t\ \text{miles}

Office B is walking at 6 mph directly west. In t hours his distance is

X_B=6t\ \text{miles}

Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office

D=\sqrt{X_A^2+X_B^2}

D=\sqrt{(5t)^2+(6t)^2}

D=\sqrt{61}t

This distance is known to be 2 miles, so

\sqrt{61}t=2

t =\frac{2}{\sqrt{61}}=0.256\ hours

t is approximately 15 minutes

3 0
3 years ago
The kinetic energies of particles in a sample of matter are increasing. This sample is most likely
Sedaia [141]

Answer:

The sample is most likely gaining thermal energy

Explanation:

4 0
3 years ago
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. Một con lắc đơn có chiều dài dây treo 1m dao động điều hoà treo trong một xe chạy trên mặt phẳng nghiêng
lyudmila [28]

Answer:

C

Explanation:

hope for help ....im expert

6 0
3 years ago
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A ball is thrown up into the air with an initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed
kaheart [24]

Answer:

B) t = 1.83 [s]

A) y = 16.51 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f} =v_{o} -g*t

where:

Vf = final velocity = 0

Vo = initial velocity = 18 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.

A) The maximum height is reached when the final velocity of the ball is zero.

0 = 18 - (9.81*t)

9.81*t = 18

t = 18/9.81

t = 1.83 [s], we found the answer for B.

Now using the following equation.

y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\

where:

y = elevation [m]

Yo = initial elevation = 0

y = 18*(1.83) - 0.5*9.81*(1.83)²

y = 16.51 [m]

7 0
3 years ago
A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2 , and 1.85 ms (1 ms = 10−3
ankoles [38]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 =  u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

4 0
3 years ago
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