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Firdavs [7]
1 year ago
9

in case one, a car speeds up from zero m/s to 15 m/s. in case two, the same car speeds up from 15 m/s to 30 m/s. the mass of the

car is 1000 kg. compare the energy needed to provide the increase in speed in each case. give your answers in joules. (a) in case one, how much energy is needed? j (b) in case two, how much energy is needed? j (c) in case two, how many more times the energy was needed than in case one? three times as much energy was needed. four times as much energy was needed. the same amount of energy was needed in both cases. twice as much energy was needed.
Physics
1 answer:
GrogVix [38]1 year ago
7 0

Three time  the energy was needed than in case one.

What is energy?

In physics, energy is the quantifiable property that is transmitted to a body or a physical system and is visible in the form of heat and light. The law of conservation of energy states that energy can be converted in form but cannot be created or destroyed.

Case 1:

Vo = 0 m/s

V1 = 15 m/s

Case 2:

Vo = 15 m/s

V1 = 30 m/s

Change in KE = 1/2 × m × (V1^2 - Vo^2)

KE1 = 1/2 × 1000 × (15^2 - 0)

= 112.5 kJ

KE2 = 1/2 × 1000 × (30^2 - 15^2)

= 1/2 × 1000 × 675

= 337.5kJ

Case 1 has an increase of 112.5 kJ energy while Case 2 has an increase of 337.5kJ.

To know more about energy click

brainly.com/question/582060

#SPJ4

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eimsori [14]

Answer:

The car will travel 30 miles during the 30-minutes period of acceleration.

Explanation:

Given data :

Initial velocity = v₁ = 50 miles/hour

Final velocity = v₂ = 70 miles/hour

Time = t = 30 min = 0.5 hour

Using the definition of acceleration, we find the acceleration (a)

                   a = (v₂ - v₁) ÷ t

                   a = (70 - 50) ÷ 0.5

                   a = 20 ÷ 0.5

                   a = 40 miles/hour²

Using 3rd equation of motion, we find the distance travel (s)

                    2as = v₂² -  v₁²

                    2(40)s = 70² - 50²

                    80 × s = 4900 - 2500

                         s = 2400 ÷ 80

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3 years ago
An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Exp
Artemon [7]

Explanation:

It is given that,

Mass of the football player, m = 92 kg

Velocity of player, v = 5 m/s

Time taken, t = 10 s

(1) We need to find the original kinetic energy of the player. It is given by :

k=\dfrac{1}{2}mv^2

k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2

k = 1150  J

In two significant figure, k=1.2\times 10^3\ J

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

i.e. P=\dfrac{W}{t}=\dfrac{\Delta K}{t}

P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}

P = 115 watts

In two significant figures, P=1.2\times 10^2\ Watts

Hence, this is the required solution.  

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