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ZanzabumX [31]
3 years ago
13

An older-model car accelerates from 0 to speed v in 9.0 s. A newer, more powerful sports car of the same mass accelerates from 0

to 3v in the same time period. Assuming the energy coming from the engine appears only as kinetic energy of the cars, compare the power of the two cars. (Take Pnew and Pold as the powers of the new car and old car, respectively.)
Physics
1 answer:
Sergio [31]3 years ago
6 0

Answer:The new car has 9 times the power of the old car

Explanation:

Given data

old car

Mass =m

Velocity =v

Time to attain the velocity t= 9secs

New car

Mass =m (same as old car)

Velocity =3v

Time to attain the velocity t= 9secs

We know that the kinetic energy expression is given as

KE=1/2mv²

KE for old car =1/2mv²J

KE for new car =1/2m(3v)²

=1/2m9v²J

Now the expression for power is

P= energy *time

Pold = 1/2mv²*9

4.5(mv²)Hp

Pnew = 1/2m*9v²*9

1/2m*81*v²

40.5(m*v²)Hp

Now comparing the powers of both cars

Pnew/Pold =40.5mv²/4.5mv²

Pnew/Pold = 9

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oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

d=\frac{1,280,000,000km}{37,000}=34,594.59km

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}

hence, the speed of the Hubble is approximately 384km/min

5 0
3 years ago
True or false the refraction of a wave is how many wavelengths pass a fixed point each second
Ksenya-84 [330]
False. That description fits the wave's 'frequency'. 
It has nothing to do with refraction.
8 0
3 years ago
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Charge A and charge B are 3.00 m apart, and charge A is +1.44 C and charge B is +3.10 C. Charge C is located between them at a c
Sidana [21]

Answer:

the distance from charge A to C is r₁₃= 1.216 m

Explanation:

following Coulomb's law , the force exerted by 2 point charges between themselves is:

F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant

since C ( denoted as 3) is at equilibrium

F₁₃=F₂₃

k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²

q₁/r₁₃²=q₂/r₂₃²

r₁₃²/q₁=r₂₃²/q₂

r₂₃=r₁₃*√(q₂/q₁)

since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have

r₁₃+r₂₃=d=r₁₂

r₁₃+r₁₃*√(q₂/q₁)=d

r₁₃*(1+√(q₂/q₁))=d

r₁₃=d/(1+√(q₂/q₁))

replacing values

r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m

thus the distance from charge A to C is r₁₃= 1.216 m

7 0
3 years ago
A rocket is launched from the launch pad at a speed of 700 km/hr.
Juliette [100K]

Answer:

Explanation:

Vm = Δs/Δt

700km/h = Δs/1.5h

700 = Δs/1.5

\frac{700}{1} = \frac{S}{1.5}

S = 700 x 1.5

S = 1050 Km

*(S = Δs)

Answer: The rocket will have traveled 1050 Km

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5 0
3 years ago
Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the re
lapo4ka [179]

Answer:

K_{system} = \frac{k}{10}

Explanation:

When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})

Here, n = 10

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\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})

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\frac{1}{K_{system}} = \frac{10}{k}

K_{system} = \frac{k}{10}

7 0
3 years ago
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