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Veseljchak [2.6K]
3 years ago
15

A force when you are bulling an object is ............................force​

Physics
1 answer:
dsp733 years ago
4 0
Yes it is a force omg so smart
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The primary source of evidence proposed by many scientist to support the theory of an ancient earth is ____ dating.
vova2212 [387]
It depends on your definition of “ancient.” Radiometric dating using Carbon-14 can reliably date back to about 50,000 years, uranium-lead or lead-lead dating can date back multiple millions, potassium-argon dating can reach 1.5 billion, and rubidium-strontium can reach 50 billion (nearly 4x the age of the universe). It depends on the context in which this question is being asked.
7 0
3 years ago
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

3 0
3 years ago
A miner of mass 90kg travels down a slide calculate the potential energy of the miner when he moves15m vertically downwards
Lena [83]
Answer:

Gravitational Potential Energy = mgh

Explanation:

As the miner moves down, the GPE changes because the height changes.

Gravitational Potential Energy = mgh


8 0
3 years ago
Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of
IRISSAK [1]

Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

6 0
4 years ago
Why is earth's temperature just right for life?
Rudik [331]
The answer is letter b
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