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Ilia_Sergeevich [38]
3 years ago
7

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

Physics
2 answers:
Alexus [3.1K]3 years ago
5 0

Answer:

6 Joules

Explanation:

KE= ½ m v2 = ½ (3kg) (2 m/s)2 = 6 Joules 8.

vitfil [10]3 years ago
3 0

Answer:

6Newton meter is the answer.

Explanation:

Mass=m=3kg

Velocity=v=2m/s

Kinetic energy =K.E.=?

As we know that

K.E.=1/2mv^2

Putting the values

K.E.=1/2✖️3kg✖️(2m/s)^2

K.E.=1/2✖️3kg✖️4m^2/s^2

K.E=1/2✖️12Nm

K.E.=6Nm is your answer

Hope it will help you :)

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3 years ago
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Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in
Karo-lina-s [1.5K]

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

(a) Self inductance is calculated as;

L = \frac{N^2 \mu_0 A}{l}

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2

L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH

(b) The energy stored in the inductor when 21 A current ;

E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s

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Sodium (Na) is a Alkali metal so that's the answer

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