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vampirchik [111]
3 years ago
15

Which of these scenarios describes circular motion?

Physics
1 answer:
Anika [276]3 years ago
6 0

A)

The moon orbiting the Earth

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A 535 kg roller coaster car began at rest at the top of a 93.0 m hill. Now it is at the top of the first loop-de-loop. This roll
Nikitich [7]
The correct answer is <span>The car has both potential and kinetic energy, and it is moving at 24.6 m/s.</span>
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2 years ago
URGENT
olga55 [171]

If the boat's speed is s, and the river's speed is r, and the boat is traveling east (0 degrees),

(0,r) + (s cos297,s sin297) = (6,0)

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2 years ago
What is the maximum weight a boat can hold if a boat can displace 60.5ml?
Valentin [98]

Answer:

a. 60.5 kg

Explanation:

Given data,

The maximum water a boat can displace is, 60.5 ml

According to the principle of buoyancy, the weight of the floating body is equal to the weight of the liquid displaced.

Under standard temperature and pressure, a unit mass of water equals one liter.

If a boat can displace a maximum of 60.5 ml of water, then it can hold a mass of a maximum of 60.5 kg of mass.

6 0
3 years ago
Liam throws a water balloon horizontally at 8.2 m/s out of a window 18 m from the ground.
Alecsey [184]

Time taken by the water balloon to reach the bottom will be given as

h = \frac{1}{2} gt^2

here we know that

h = 18 m

g = 9.8 m/s^2

now by the above formula

18 = \frac{1}{2}*9.8* t^2

18 = 4.9 t^2

t = 1.92 s

now in the same time interval we can say the distance moved by it will be

d = v_x * t

d = 8.2 * 1.92 = 15.7 m

so it will fall at a distance 15.7 m from its initial position

5 0
3 years ago
Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of
wariber [46]

Answer:

For left = 0  N/C

For right = 0  N/C

At middle = -7.6836 * 10^{-11} \vec{i}  N/C

Explanation:

Given data :-

б =6.8 * 10^{-22} C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

E = \frac{\sigma }{2\varepsilon }

1) To the left of the plate

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

2) To the right of them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

3) Between them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(-\vec{i}) = (\frac{\sigma }{\varepsilon })(-\vec{i}) = -\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} }  \vec{i} =   -7.6836 * 10^{-11} \vec{i} N/C

5 0
3 years ago
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