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Anarel [89]
3 years ago
14

Most of the energy consumed in the U.S. is stored in which form of energy

Physics
1 answer:
kykrilka [37]3 years ago
8 0
It’s stored in Quads.
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Which statement about work and power correctly describes an automobile race?
miv72 [106K]
Well this question looks like it makes some assumptions.  So assuming that both cars have the same mass and experience the same wind resistance regardless of speed and same internal frictions, then we could say "The car that finishes last has the lowest power".  The reason is that for a given race the cars must overcome losses associated with motion.  Since they all travel the same distance, the amount of work will be the same for both.  This is because work is force times distance.  If the force applied is the same in both cases (identical cars with constant wind resistance) and the distance is the same for both (a fair race track) then W=F·d will be the same.
Power, however, is the work done divided by the time over which it is done.  So for a slower car, time t will be larger.  The power ratio W/t will be smaller for the longer time (slower car).
7 0
4 years ago
Read 2 more answers
The products of one turn of dark reaction cycle are called
harina [27]
Light Independent Reactions
5 0
3 years ago
A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction
yawa3891 [41]

Answer:

F'=\dfrac{F}{4}

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

For the slower car on the road,

F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

8 0
4 years ago
A level test track has a coefficient of road adhesion of 0.80, and a car being tested has a coefficient of rolling friction that
pav-90 [236]

Answer:

the unloaded braking efficiency is 84.6 %

Explanation:

Given the data in the question;

by Ignoring aerodynamic resistance; we can find the theoretical stopping distance using the following formula

S = (Y_{b}( V₁² - V₂²)) / ( 2g( ηbμ + f_{rl} ± sin∅_{g}))

now given that the tracked is levelled, ∅_{g} = 0, also Y_{b} = 1.04 for level or flat road

Speed V₁ = 60mil/hr = (60×5280)/(1×60×60) = 316800ft/3600s = 88ft/s

now, we substitute in our values to get the braking efficiency;

180ft = (1.04( (88ft/s)² - 0²)) / ( 2(32.2( (ηb/100)(0.80) + (0.018) ± sin(0°)))

180ft = 8053.76 / ( 64.4)(0.008ηb + 0.018)

180ft = 8053.76 / ( 0.5152ηb + 1.1592)

180( 0.5152ηb + 1.1592)  = 8053.76

( 0.5152ηb + 1.1592) = 8053.76 /180

0.5152ηb + 1.1592 = 44.7431

0.5152ηb = 44.7431 - 1.1592

0.5152ηb = 43.5839

ηb = 43.5839 / 0.5152

ηb = 84.596 ≈ 84.6 %

Therefore,  the unloaded braking efficiency is 84.6 %

7 0
3 years ago
Why would an orange roll off your cafeteria tray if you stopped suddenly
Leviafan [203]
Because the momentum would still be applied to the orange if the tray stopped moving
3 0
3 years ago
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