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chubhunter [2.5K]
3 years ago
9

A 2.51 kg ball is attached to a ceiling by a 1.19 m long string. The height of the room is 3.45 m. The acceleration of gravity i

s 9.81 m/s 2 . What is the gravitational potential energy associated with the ball relative to a) the ceiling?
Physics
1 answer:
ANTONII [103]3 years ago
6 0

Answer:

55,42 J

Explanation:

Since the height of the room is 3.45 m (distance between the floor and the ceiling) the difference between this value and the length of the rope 1.19 m; it will be equal to (3.45-1.19) =2.26 m. If we take as a reference point (Ep=0) the floor of the room, then the potential energy will be equal to Ep = M * g * h, replacing values in this equation (2.5 kg * 9.81 m/s2 * 2.26 m) will be 55,42 (N * m) or Jules.

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vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{2}  +  \frac{1}{3}  \\  \frac{1}{R}  =  \frac{3 + 2}{6}  =  \frac{5}{6}  \\ R =  \frac{6}{5}  = 1.2 \:  \: ohm

5.2) Ans;

\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{8}  +  \frac{1}{10}  \\  \frac{1}{R}  =  \frac{5 + 4}{40}  =  \frac{9}{40}  \\ R =  \frac{40}{9}  = 4.4 \:  \: ohm

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2 years ago
A police car parked on the side of the highway emits a 1200 Hz sound that bounces off a vehicle farther down the highway and ret
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f' = frequency observed by the police car after sound reflected from the vehicle and comes back to police car = 1250 Hz

f = frequency emitted by the police car  = 1200 Hz

V = speed of sound = 340 m/s

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frequency observed by the police car is given as

f' = f (V + v)/(V - v)

inserting the values in the above equation

1250 = 1200 (340 + v)/(340 - v)

v = 6.9 m/s

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3 years ago
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Answer:

...do

Explanation:

24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5

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Answer: Conduction

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