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klio [65]
1 year ago
5

The deepest section of ocean in the world is the Marianas Trench, located in the Pacific Ocean. Here the ocean floor is as low a

s 10,918 m below the surface. The index of refraction of water is 1.33. Calculate the time for the laser beam to reach the bottom of the trench
Physics
1 answer:
Levart [38]1 year ago
3 0
B I just took the test
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An object of mass 8kg is attached to massless string of length 2m and swum with a tangential velocity of 3 what is the tension o
Paul [167]

Answer:

36 N

Explanation:

If the object of mass, m = 8 kg is swung in a horizontal circle of radius, r = 2m = length of string with tangential velocity v = 3 m/s, the tension in the string is the centripetal force which is T = mv²/r

= 8 kg × (3 m/s)²/2 m

= 4 kg × 9 m/s²

= 36 N

6 0
3 years ago
How does the density of fluid affect the magnitude of buoyancy acting on an object immersed in it
Oksi-84 [34.3K]

Answer:

Explanation:

more the density , more will be the buoyant force acting on it , less the density less will be the buoyant force acting on it. This is why people float in dead sea and sink in other seas

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3 years ago
The blood pressure in the human body is greater at the feet than at the brain
GarryVolchara [31]
Blood pressure is greater in feet because of gravity
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3 years ago
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A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch
Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

x=V_{o}cos \theta t----------------------(1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y is the final height of the ball

y_{o} is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}

t=1.597s

Substituting (6) in (1):

67.1 =V_{o} cos(0\°) 1.597-------------------(4)

Step 2:  Finding V_{o}:

From equation(4)

67.1 =V_{o}(1) 1.597

V_0 = \frac{6.71}{1.597}

V_{o}=42.01 m/s (8)  

Substituting V_{o} in (3):

V=42.01 -(9.8)(1.597)

v =42 .01 - 15.3566  

V=26.359 m/s

5 0
2 years ago
Which telescope would be better viewing a faint, distant star? Why?
grin007 [14]
Reflecting telescope. Reflecting telescopes tend to have larger objective (due to the use of mirrors, mirrors are a lot cheaper than lenses) and have the ability to collect more light, while refracting telescopes are limited to objective lenses with smaller diameters due to their structural limitations (chromatic abbreviation, for example). Therefore, reflecting telescopes should be better at viewing faint distant stars
4 0
2 years ago
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