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svetlana [45]
3 years ago
10

A car with a mass of 1.50x10^3 kg starts from rest and accelerates to a speed of 18.0m/s in 12.0 s. assume that the force of res

istance remains constant 400.0n during this time. what is the average power developed by the car's engine.
Physics
1 answer:
Luden [163]3 years ago
8 0
The first thing you should know for this case is the definition of distance.
 d = v * t
 Where,
 v = speed
 t = time
 We have then:
 d = v * t
 d = 9 * 12 = 108 m
 The kinetic energy is:
 K = ½mv²
 Where,
 m: mass
 v: speed
 K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
 The work due to friction is
 w = F * d
 Where,
 F = Force
 d = distance:
 w = 400 * 108 = 4.32 * 10 ^ 4
 The power will be:
 P = (K + work) / t
 Where,
 t: time
 P = 2.86 * 10 ^ 5/12 = 23.9 kW
 answer:
 the average power developed by the engine is 23.9 kW
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4 years ago
A person who weighs 69kg rides with an acceleration of +1.45 m/s2. What is the force exerted by the legs on the person's upper b
trapecia [35]

Answer:

The force exerted by the legs on the person's upper body 32.106 N.

Given:

Mass = 69 kg

Acceleration = 1.45 \frac{m}{s^{2} }

To find:

Force exerted by legs on the person's upper body = ?

Formula used:

Force = mass × acceleration

Solution:

Mass of legs = \frac{32}{100} mass of body

Mass of legs = 22.08 kg

According to Newton's second law of motion,

Force = mass × acceleration

Force = 22.08 × 1.45

Force = 32.106 N

The force exerted by the legs on the person's upper body 32.106 N.

4 0
4 years ago
Two wires with equal lengths are made of pure copper. The diameter of wire A is three times the diameter of wire B. When 8 kg ma
noname [10]

Answer: c. YA < YB

Explanation:

The formula for Young’s modulus is = Tensile stress / Tensile strain

Tensile stress = Force x Length

Force = mass x acceleration due to gravity

 = 8kg x 10m/s

 = 80kgm/s

Tensile stress  = 80kgm/s x 2m = 160kgm2/s

Tensile strain = Area x change in length

Area = pi x D2 / 4 ; Pi = 3.14

Change in length = L2 – L1 (New length – Initial length)

Given parameters:

Length of wire A = Length of wire B, (let’s use 2meters for the calculation)

For wire A, Diameter = 3 x Wire B diameter

Assuming Diameter of wire B = 1meter

Therefore, diameter of wire A = 1 x 3 = 3meters

It is said that wire B stretches more than wire A when the man of 8kg is placed on both

For wire B, let’s assume new length is = 4m

For wire A let’s assume new length is = 3m.

(i) Tensile strain of wire A =  

Area of wire A = 3.14 x (32)/4 = 7.065m2

Change in length = 3m - 2m = 1m.

Therefore, tensile strain = 7.065m2 x 1m = 7.065m3

Young’s modulus for wire A (YA) = 160kgm2/s divided by 7.065m3  

   = 22.64Pa.

(ii) Tensile strain of wire B =

Area of wire B = 3.14 x (12)/4 = 0.785m2

Change in length = 4m – 2m = 2m

Therefore, tensile strain = 0.785m2 x 2m = 1.57m3

Young’s modulus for wire B (YB) = 160kgm2/s divided by 1.57m3

   = 101.91Pa.

From the calculations above, we see that YA is less than YB (YA < YB). This is true given that wire A has a greater diameter than wire B which in turn impacts the Area of the wire since the diameter is directly proportional to area and the area is inversely proportional to the young’s modulus.

5 0
3 years ago
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