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Allisa [31]
3 years ago
6

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

end of the 19 s? (b) what is the average power required of the car during the 19 s interval? (c) what is the instantaneous power at the end of the 19 s interval, assuming that the acceleration is constant?
Physics
1 answer:
lana [24]3 years ago
5 0
(a) Let's convert the final speed of the car in m/s:
v_f = 61 km/h = 16.9 m/s
The kinetic energy of the car at t=19 s is
K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
P= \frac{W}{\Delta t}
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W

(c) The instantaneous power is given by
P_i = Fv_f
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
a= \frac{v_f-v_i}{\Delta t}=  \frac{16.9 m/s}{19 s}=0.89 m/s^2
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W
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The focal length of the lens is 25cm

given:

power,p=4 diopters

what is focal length?

Focal length is the distance between the point of convergence of your lens and the sensor or film recording the image.

what is diopter?

The unit of optical power of lens is called diopter.It is the optical power of the lens.

we know,

p=1/f

where,

p= power

f= focal length

f=1/p

f=1/4

=0.25m

=25cm

Thus,the focal length of the lens is 25cm

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8 0
2 years ago
1.
Shalnov [3]

Answer:

No.

Explanation:

Given the following :

Velocity (V) of ball = 5m/s

Radius = 1m

Can the ball reach the highest point of the circular track

of radius 1.0 m?

The highest point in the track could be considered as the diameter of the circle :

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Maximum height which the ball can reach :

Using the relation :

Kinetic Energy = Potential Energy

0.5mv^2 = mgh

0.5v^2 = gh

0.5(5^2) = 9.8h

0.5 * 25 = 9.8h

12.5 = 9.8h

h = 12.5 / 9.8

h = 1.2755

h = 1.26m

Therefore maximum height which can be reached is 1.26m.

Since h < Diameter

7 0
3 years ago
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