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Lynna [10]
3 years ago
6

How many meters in 32 kilometers

Physics
2 answers:
Alisiya [41]3 years ago
7 0
Hello !

32 km = 32000 m
sleet_krkn [62]3 years ago
6 0
There are 32,000 meters in 32 kilometers.
You set up the problem: 32 kilometers=_____ meters
Then you times 32 by 1,000: 32 x 1,000 = 32,000
Going from kilometers to meters you times by a thousand and from meters to kilometers you divide by a thousand.
Hope this helps! :D
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An object is dropped near the earth's surface. At the end of 4.0 s, how fast is it falling in m/s? (Ignore air resistance.)
Sati [7]
For free falling bodies, the final velocity may be calculated through the equation,
                                   Vf = gt
Where g is the acceleration due to gravity (9.8 m/s²) and t is the time elapsed. Substituting the known values,
                                   Vf = (9.8 m/s²) x (4 s) = 39.2 m/s
Therefore, the object's velocity is approximately 39.2 m/s. 
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3 years ago
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Which of the following statements is true? The melting and freezing points of a substance are the same. The melting and boiling
insens350 [35]
<span>The melting and freezing points of a substance are the same.

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4 0
2 years ago
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How long will your trip take if you travel 4000 m at an average speed of 8m/s
SpyIntel [72]
  • Answer:

<em>500 sec</em>

<em>8 min 20 sec</em>

  • Explanation:

<em>Hi there !</em>

<em />

<em>8 m ................ 1 s </em>

<em>4000 m ........ x s</em>

<em>x = 4000m×1s/8m = 500 sec = 8 min 20 sec</em>

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8 0
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

3 0
2 years ago
Find the magnitude of the average force ⟨Fx⟩⟨Fx⟩ in the x direction that the particle exerts on the right-hand wall of the conta
hodyreva [135]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

The attached file has a detailed solution of the given problem.

4 0
3 years ago
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