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Snowcat [4.5K]
3 years ago
7

The calculation of kinetic energy uses the formula, KE = 1 2 mv2. Identify the unit that is required for mass and velocity, resp

ectively, to determine the kinetic energy in Joules. A) mass in g, velocity in m s B) mass in kg, velocity in m s C) mass in g, velocity in km hr D) mass in kg, velocity in km hr
Physics
2 answers:
shusha [124]3 years ago
8 0

So the answer  is B.  because the mass have  Kg as a international unit and velocity is m/s, they are  international units in physics.

kompoz [17]3 years ago
4 0

Answer:

Mass : kg

Velocity : m/s

Explanation:

Kinetic energy is due to the motion of an object. The mathematical expression for the kinetic energy is given by :

E_k=\dfrac{1}{2}mv^2

Where

m is the mass of the object

v is the velocity of object  

The SI unit of kinetic energy is joule (J). One joule equals :

E_k=\dfrac{1}{2}\times kg\times m/s=Joule

Every physical quantity must be in SI units So, mass must be in kilogram (kg) and velocity must be in meter/second (m/s).

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dem82 [27]

Answer:

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

T = 2\pi \sqrt{\frac{r^3}{GM}}

now for the time period of moon around the earth we can say

T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}

here we know that

T_1 = 0.08 year

r_1 = 0.0027 AU

M_e = mass of earth

Now if the same formula is used for revolution of Earth around the sun

T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}

here we know that

r_2 = 1 AU

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M_s = mass of Sun

now we have

\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}

\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}

12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

4 0
3 years ago
A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic
defon

Answer:

0.52 Nm

Explanation:

A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T

Angle between the plane of loop and magnetic field = 30 Degree

Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree

θ = 60°

Torque = N i A B Sinθ

Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60

Torque = 0.52 Nm

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3 years ago
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Explanation:

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Answer:

true

Explanation:

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