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alekssr [168]
3 years ago
13

A ball is rolling along a straight and level path. The ball is traveling at a constant speed of 2.0 m/s. The mass of the ball is

3.0 kg. What is the kinetic energy of the ball?
A. 6.0 J
B. 9.0 J
C. 12.0 J
D. 3.0 J
Physics
1 answer:
antoniya [11.8K]3 years ago
7 0
Hope this is correct A. 6.0J
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Who is the father of the humans?
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Adam (Arabic:, romanized:dam) is said to be the first human person on Earth, as well as the first prophet (Arabic:, nab) of Islam. Muslims see Adam's role as the father of the human race with veneration.

Explanation:

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A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,00
Savatey [412]

Answer:

The answer is "Option B".

Explanation:

r=15\times 10^{7}\ km\ = 15\times 10^{10}\ m\\\\w=\frac{2\pi}{1\ year}\\\\=\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec}\\\\a=w^2r\\\\=(\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec})^2 \times 15 \times 10^{10}\ \frac{m}{s^2}\\\\

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7 0
3 years ago
You used a telescope and other mathematics to discover that Jupiter is 5.20 au from the sun. Use the equation to find its orbita
meriva

Answer:

11.9 years

Explanation:

We can find the orbital period by using Kepler's third law, which states that the ratio between the square of the orbital period and the cube of the average distance of a planet from the Sun is constant for every planet orbiting aroudn the Sun:

\frac{T^2}{r^3}=const.

Using the Earth as reference, we can re-write the law as

\frac{T_e^2}{r_e^2}=\frac{T_j^2}{r_j^3}

where

Te = 1 year is the orbital period of the Earth

re = 1 AU is the average distance of the Earth from the Sun

Tj = ? is the orbital period of Jupiter

rj = 5.20 AU is the average distance of Jupiter from the Sun

Substituting the numbers and re-arranging the equation, we find:

T_j=\sqrt{\frac{T_e^2 r_j^3}{T_j^2}}=\sqrt{\frac{(1 y)^2 (5.2 AU)^3}{(1 AU)^3}}=11.9 y


4 0
3 years ago
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A square coil of wire with 15 turns and an area of 0.40 m2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped
Drupady [299]

Answer:

The magnitude of the average induced emf is 90V

Explanation:

Given;

area of the square coil, A = 0.4 m²

number of turns, N = 15 turns

magnitude of the magnetic field, B = 0.75 T

time of change of magnetic field, t = 0.05 s

The magnitude of the average induced emf is given by;

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E = -(15 x 0.4 x 0.75) / 0.05

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|E| = 90 V

Therefore, the magnitude of the average induced emf is 90V

6 0
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