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ValentinkaMS [17]
3 years ago
11

An older-model car accelerates from 0 to speed v in a time interval of Δt. A newer, more powerful sports car accelerates from 0

to 4v 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.
Physics
1 answer:
dexar [7]3 years ago
7 0

Answer:

Ratio of power produced by new to old car is 16:1

Explanation:

Assuming both cars have the same mass m

Let KE1 and KE2 be the kinetic energies of the old and new car respectively.

v1 = v

and v2 = 4v

KE1 = 1/2mv1² = 1/2mv²

KE2 = 1/2mv2² = 1/2m(4v)² = 1/2×m×16v² = 8mv²

Ratio of the power produced by the new car to the old one is simply KE2/KE1 =8mv²/1/2mv² = 16

Since the same time interval is given.

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When is a secondary source more helpful than a primary source?
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Answer:

I think the answer is C.

Explanation:

A primary source is a first hand account of an event while a secondary source is a retelling or second hand account meaning as many details will be prevalent.

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2 years ago
Every substance has a specific value of heat required to change the temperature of 1 gram of the substance by 1 degree Celsius.
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3 years ago
At what angle should the roadway on a curve with a 50m radius be banked to allow cars to negotiate the curve at 12 m/s even if t
kari74 [83]

Answer:

The road bank angle is 16.38⁰.

Explanation:

radius of curvature of the road, r = 50 m

allowable speed of car on the road, v = 12 m/s

The bank angle is calculated as;

\theta = tan^{-1} (\frac{v^2}{gr} )

where;

θ is the road bank angle

g is acceleration due to gravity = 9.8 m/s²

\theta = tan^{-1} (\frac{v^2}{gr} )\\\\\theta = tan^{-1} (\frac{12^2}{9.8 \times 50} )\\\\\theta = tan^{-1} ( 0.2939)\\\\\theta = 16.38 ^0

Therefore, the road bank angle is 16.38⁰.

8 0
3 years ago
A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the
Gnoma [55]

Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating  system = 6.74 J

We know that energy is given by

(a) Ke=\frac{1}{2}mv_{max}^2

6.74=\frac{1}{2}\times 0.267\times v_{max}^2

v_{max}=7.1m/sec

(b) Now \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec

We know that \omega =\sqrt{\frac{k}{m}}

35.68=\sqrt{\frac{k}{0.267}}

k=339.9N/m

(c) We know that energy is given by

E=\frac{1}{2}KA^2

6.74=\frac{1}{2}\times 339.9\times A^2

A=19.91cm

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