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mixer [17]
3 years ago
13

A car accelerates uniformly from rest to 20 m/sec in 5.6 sec along a level stretch of road. Ignoring friction, determine the ave

rage power required to accelerate the car if (a) the weight of the car is 9,000 N, and (b) the weight of the car is 14,000 N.
Physics
1 answer:
bonufazy [111]3 years ago
3 0

Answer:

(a) P=33000W

(b) P=51000W

Explanation:

The average power is defined as the amount of work done during a time interval:

P=\frac{W}{t}(1)

According to work-energy theorem, the work done is equal to the change in kinetic energy. So, we have:

W=\Delta K\\W=K_f-K_0\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\(2)

Recall that the weight is given by:

w=mg\\m=\frac{w}{g}(3)

The car accelerates uniformly from rest (v_0=0). Replacing (3) in (2), we have:

W=\frac{wv_f^2}{2g}

(a) Finally, we replace this in (1):

P=\frac{wv_f^2}{2gt}\\P=\frac{9000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=33000W

(b)

P=\frac{14000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=51000W

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A car moves with the speed of 120m/s for 4 minutes ,calculate the distance covered by the car​
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Answer:

960 m

Explanation:

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A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric
Liula [17]

The electric field produced by a single-point charge is given by

E(r)=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge


To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.


1) The first charge is q=-3.00 nC=-3.00 \cdot 10^{-9} C, and it is located at x=0, so its distance from the point x=0.200 m is

r=0.200 m-0=0.2 m

Therefore, the electric field is

E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.


2) The second charge is q=-5.50 nC=-5.5 \cdot 10^{-9}C and it is located at x=0.800 m, so its distance from the point is

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Therefore, the electric field is

E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N

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