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Ganezh [65]
3 years ago
5

Another engine reaches its top speed in 7.5s. It is able to perform 250,000 J of wok in that time. How much power this engine ha

ve in that time?
Physics
1 answer:
EastWind [94]3 years ago
3 0

Answer:

33,333.3J/s

Explanation:

Energy expended which is the work done in Joule

=250,000J

Time taken to expend the energy =7.5s

Power =Unknown

Formula

Power = <u> </u><u> </u><u> </u><u> </u><u> </u><u>Work</u><u> </u><u> </u><u> </u>

Time

Hence, Power = <u>250,000J</u>

7.5s

Power = 33,333.3J/s

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Answer:

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Explanation:

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Astronauts aboard the ISS move at about 8000 m/s, relative to us when we look upward.How long does an astronaut need to stay abo
Luba_88 [7]

Answer:

#_time = 7.5 10⁴ s

Explanation:

In order for the astronaut to be younger than the people on earth, it follows that the speed of light has a constant speed in vacuum (c = 3 108 m / s), therefore with the expressions of special relativity we have.

            t = \frac{t_p}{ \sqrt{1-  (v/c)^2} }

where t_p is the person's own time in an immobile reference frame,

           t_{p} = t \sqrt{1 - (\frac{v}{c})^2 }

let's calculate

we assume that the speed of the space station is constant

              t_p = 1 \sqrt{1 - \frac{8 \ 10^3}{3 \ 10^8} }

             t_p = 1 \sqrt{1- 2.6666 \ 10^{-5}}

             t_ =  0.99998666657   s

             

therefore the time change is

             Δt = t - t_p

             Δt = 1 - 0.9998666657                  

              Δt = 1.3333 10⁻⁵ s

this is the delay in each second, therefore we can use a direct rule of proportions. If Δt was delayed every second, how much second (#_time) is needed for a total delay of Δt = 1 s

               #_time = 1 / Δt

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               #_time = 7.5 10⁴ s

5 0
3 years ago
By method of dimension show that the following equation are homogenous.
il63 [147K]

Answer:

Proof in explanataion

Explanation:

The basic dimensions are as follows:

MASS = M

LENGTH = L

TIME = T

i)

Given equation is:

H = \frac{u^2Sin^2\phi}{2g}

where,

H = height (meters)

u = speed (m/s)

g = acceleration due to gravity (m/s²)

Sin Ф = constant (no unit)

So there dimensions will be:

H = [L]

u = [LT⁻¹]

g = [LT⁻²]

Sin Ф = no dimension

Therefore,

[L] = \frac{[LT^{-1}]^2}{[LT^{-2}]}\\\\\ [L] = [L^{(2-1)}T^{(-2+2)}]

<u>[L] = [L]</u>

Hence, the equation is proven to be homogenous.

ii)

F = \frac{Gm_1m_2}{r^2}\\\\

where,

F = Force = Newton = kg.m/s² = [MLT⁻²]

G = Gravitational Constant = N.m²/kg² = (kg.m/s²)m²/kg² = m³/kg.s²

G = [M⁻¹L³T⁻²]

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Therefore,

[MLT^{-2}] = \frac{[M^{-1}L^{3}T^{-2}][M][M]}{[L]^2}\\\\\ [MLT^{-2}] = [M^{(-1+1+1)}L^{(3-2)}T^{-2}]\\\\

<u>[MLT⁻²] = [MLT⁻²]</u>

Hence, the equation is proven to be homogenous.

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