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attashe74 [19]
2 years ago
7

Imagine holding two identical bricks under water. Brick A is just beneath the surface of the water, while brick B is at a greate

r depth. What is the force needed to hold brick B in place? (Assume the density of water doesn't change with height.) a) the same as b) larger c) smaller than the force required to hold brick A in place.
Physics
1 answer:
FromTheMoon [43]2 years ago
4 0

Answer:

I think it's c

Explanation:

The brick's mass, helps to hold it under the water but brick A is being held up. If you dropped the brick, it would sink to the bottom. You need a stronger force to hold it up.

Hope this helps you! x

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A lab cart is loaded with different masses and moved at various constant velocities? the anser should be

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Four students made a graphic organizer describing the parts of the atom.
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I believe the answer is D.

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What type of motion occurs when an object spends around and axis without altering its linear position?
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A cartoon shows two friends watching an unoccupied car in free fall after it has rolled off a diff. One friend says to the other
olga55 [171]

Answer:

The statement is not correct.

Explanation:

To know if the statement is correct, we shall determine the velocity of the car after 3 s. This is illustrated below.

Data obtained from the question include:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 3 s

Final velocity (v) =?

v = u + gt

v = 0 + (9.8 × 3)

v = 0 + 29.4

v = 29.4 m/s

Thus, the velocity of the car after 3 s is 29.4 m/s.

Hence, the statement made by the friend is not correct as the car has a falling velocity of 29.4 m/s after 3 s.

8 0
2 years ago
¿Cuál es la aceleración centrípeta de un móvil que recorre una
Lady bird [3.3K]

Answer:

a)  a = 4.57 m/s², b)  a = 6.48 m / s² , c)  a = 1.42 m / s²,d)   r = 82.3 m

 

Explanation:

The centripetal acceleration is the acceleration responsible for the change of direction of the acceleration vector and occurs in circular movements, the expression is

           a = v² / r

let's apply this precaution to our cases

a) let's calculate

          a = 8²/14

         a = 4.57 m/s²

b) an automobile at v = 65 km / h (1000 m / 1km) (1 h / 3600 s) =18,055 m/s

let's reduce feet to meters

          1 ft = 0.3048 m

           r = 165 ft (0.3048 m / 1 ft) = 50.292 m

          a = 18,055 2 / 50,292

           a = 6.48 m / s²

c) we calculate

          a = 1.25²2 / 1.1

          a = 1.42 m / s²

d) we look for the radius

          a = v² / r

          r = v² / a

we reduce

          v = 80 km / h (1000 m / 1km) (1h / 3600s) = 22.22  ms

          r = 22.22²/6

          r = 82.3 m

e) the cenripeta acceleration is used to take the curves on the highway,

    Used in centrifuges to separate compounds

       It is used in the games of the park of atraccio

     Used in CD players and computer hard drives

5 0
2 years ago
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