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3 years ago

An object, initially at rest, is subject to an acceleration of 34 m/s2. How long will it take for that

Physics

2 answers:

SIZIF [17.4K]3 years ago

6 0

Answer:

If it is moving 34 m/s it will take 100 seconds, or 1:40 to reach 3400 meters.

Explanation:

I found this answer by dividing 3400 by 34 and converting seconds to minutes

nevsk [136]3 years ago

6 0

Answer:

14 s

Explanation:

Given:

v₀ = 0 m/s

a = 34 m/s²

Δx = 3400 m

Find: t

Δx = v₀ t + ½ at²

(3400 m) = (0 m/s) t + ½ (34 m/s²) t²

t = 14.1 s

Rounded to two significant figures, it takes 14 seconds.

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3 years ago

Convert 123,453 to a scientific notation

Natalija [7]

Answer:

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2 years ago

Nova, whose mass is 50kg drops 2 K, what is the amount of heat lost from Nova's body? (specific heat of the human body is 3470 J

Alex17521 [72]

Use the equation q=mc/\T, where q is the heat lost, m is mass, and /\T is the change in temperature, and c is the specific heat.
q=50kg(3470J/kg K)(2K)
q=<span>347000 J

Any other questions, just ask.
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6 0

3 years ago

A muon is traveling at 0.988

AlexFokin [52]

As per Einstein's relation of relativity

$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$

here we know that

$m_0 = 207 m_e = 207 \times 9.11 \times 10^{-31} kg$

$m_0 = 1.88 \times 10^{-28} kg$

now here we know that

$v = 0.988 c$

now from above equation mass of the muon is given as

$m = \frac{1.88 \times 10^{-28}}{\sqrt{1 - 0.988^2}}$

$m = 1.22 \times 10^{-27} kg$

now for the momentum of muon we can use

$P = mv$

$P = 1.22 \times 10^{-27} \times 0.988(3 \times 10^8)$

$P = 3.62 \times 10^{-19} kg m/s$

so above is the momentum of muon

6 0

3 years ago

A falling object accelerates from -10.00 to-30.00m/s. How much time does that take??????

meriva

With the equation Vf = Vi + at, you can determine the unknown variable. I presume that gravity is the acceleration, so 30 = 10 + 9.8(t) —> 20 = 9.8(t) —> 2.040816 = t

7 0

3 years ago