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ankoles [38]
3 years ago
15

The diver on the diving board is 10 meters high and has a mass of 0.050kg, what is his GPE?

Physics
1 answer:
sveta [45]3 years ago
6 0

Answer:

<h2>4.9 J</h2>

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

GPE = 10 × 9.8 × 0.05

We have the final answer as

<h3>4.9 J</h3>

Hope this helps you

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What happens when an object experiences friction?
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Answer:Whenever a moving object experiences friction, some of its kinetic energy is transformed into thermal energy. Mechanical energy is always transformed into thermal energy due to friction. Mechanical energy is always transformed into thermal energy due to friction.

Explanation:

Whenever a moving object experiences friction, some of its kinetic energy is transformed into thermal energy. Mechanical energy is always transformed into thermal energy due to friction. Mechanical energy is always transformed into thermal energy due to friction.

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How do two objects at different temperatures attain thermal equilibrium?
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C.As the two objects touch, thermal energy flows as heat from the warmer block to the colder block until particles in both blocks move at the same rate and reach the same temperature.

Explanation:

Heat is the transfer of thermal energy from an object at higher temperature to an object at colder temperature.

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Political scientists use political models to describe which of the following? a. political spectrum b. political shift c. politi
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D

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Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the othe
Colt1911 [192]

Answer:

Explanation:

First of all we shall calculate electric field near charge 2Q .

electric field due to charge Q = K x Q /  (5 x 10⁻² )²

E₁  = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis

E₁  = KQ x 10⁴  i / 25  

Similarly electric field due to charge 3Q near 2Q

=  3KQ x 10⁴  i / 25 . It is acting along y-axis

E₂ = 3KQ x 10⁴  j / 25

Similarly electric field due to charge 4Q near 2Q

=  4KQ x 10⁴  j / (25 x 2 )

= 2 KQ x 10⁴  / 25 . It is acting acting along north east direction

unit vector in north east direction = ( i + j )/ √2

So E₃ can be represented by

E₃ = 2 KQ x 10⁴  ( i + j )  / 25 x √2

Total field =  KQ x 10⁴  i / 25 + 3KQ x 10⁴  j / 25 + 2 KQ x 10⁴  ( i + j )  / ( 25 x √2 )

= KQ x 10⁴  [ i + 3 j + √2 i + √2 j ) / 25

= 400 KQ ( 2.414 i + 4.414 j )  N / C

Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j )  x 2Q  N

= 800 KQ² ( 2.414 i + 4.414 j ) N

= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N

= 108.63 ( i + 2 j ) N .

Magnitude of this force

= 108.63 x √5

= 243 N approx .

4 0
3 years ago
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