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grigory [225]
2 years ago
9

What is the kinetie energy of a 3-kilogram ball that is rolling at 2 meters per second?

Physics
1 answer:
Sedaia [141]2 years ago
7 0
KE=1/2m v2= 1/2 (3kg) (2m/s)2 = 6J (6joules)
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An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2 . The coil rotates in a uniform magnetic
mezya [45]

Answer:

d. 332 V

Explanation:

Given;

number of turns in the wire, N = 40 turns

area of the coil, A = 0.06 m²

magnitude of the magnetic field, B = 0.4 T

frequency of the wave, f = 55 Hz

The maximum emf induced in the coil is given by;

E = NBAω

Where;

ω is angular velocity = 2πf

E = NBA(2πf)

E = 40 x 0.4 x 0.06 x (2 x π x 55)

E = 332 V

Therefore, the maximum induced emf in the coil is 332 V.

The correct option is "D"

d. 332 V

7 0
2 years ago
Which statement describes a scientific law
Olegator [25]

The Earth is a sphere is a statement to describe a scientific law

8 0
2 years ago
The pulse site located at the point where the upper leg bends is called the
VLD [36.1K]
The pulse site located at the point where the upper leg bends is called the femoral. It is an artery found in the thigh. It is large and is deemed as the main arterial supply for the lower part of the body. It is known as the second artery that is the largest. It is being used as the catheter access artery. From it, diagnostics for the heart, brain, arms, kidney and other parts can be directed to the other arterial system. It can also be used as a source to draw blood that is from the arteries when there is low blood pressure.
3 0
3 years ago
A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has c
tatuchka [14]

We have that the values for F north, F east, F up are

  • F_N=1.09090909*10^{-5}
  • F_E=5.18181818*10^{-6}
  • F_E=2*10^{-6}

From the Question we are told that

electric force F_1 = 1.2 x 10^{-3} N(N)

electric force , F_2=5.7 x 10^{-4} N(E)

electric force , F_3=2.2 x 10^{-4} N (U)

charge on this ball one q_1= 110 nC.

charge on this ball two q_2= -50 nC.

Generally the equation for the F north  is mathematically given as

F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}

F_N=1.09090909*10^{-5}

For F East

F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}

F_E=5.18181818*10^{-6}

For F UP

F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}

F_E=2*10^{-6}

For more information on this visit

brainly.com/question/21811998

5 0
2 years ago
A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 4 kilograms is tied to the middle of the cloth
Olegator [25]

Answer:

T=26.03 N

Explanation:

Given that

Distance between poles = 12 m

Mass of block m= 4 kg

Sag distance = 5 m

Lets take tension in the clothesline is T.

The component of tension in vertical direction will be T cosθ.

By force balancing

2 T cosθ = 40

here tan\theta =\dfrac{5}{6}

θ=39.80°

2 T cos39.8 = 40

T=26.03 N

6 0
2 years ago
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