All categories

3 years ago

What is the kinetic energy of a an 80kg football player running at 8 m/s?

1 answer:

4 0

In the question it is already given that the football player is 80 kg.
Then the mass of the football player = 80 kg
Velocity at which the football player is running = 8 m/s
<span>Kinetic Energy = 0.5 • mass • square of velocity
Now we have to put the known data in this equation to find the actual velocity of the footballer.
</span> <span></span>So
Kinetic Energy of the footballer = 0.5 * 80 * (8 * 8)
= 0.5 * 80 * 64
= 2560
So the Kinetic energy of the footballer is 2560 joules

You might be interested in

An object of mass 2kg raised to a height 10m possess potential energy of 200J. What is the kinetic energy and potential energy a

Shtirlitz [24]

Explanation:

${\bold{\sf{\underline{Understanding \: the \: concept}}}}$

✠ This question says that there is an object and its mass is 2 kg ; it's raised to a height 10 m and possess potential energy of 200 J. Now this question ask us to find the kinetic energy and the potential energy at a height 4 metre.

$\bold{↬{ }}$ ${\bold{\sf{\underline{Given \: that}}}}$

✰ Mass = 2 kilograms

✰ Raised height = 10 metres

✰ Posses potential energy = 200 Joules

$\bold{↬{ }}$ ${\bold{\sf{\underline{To \: find}}}}$

✰ Kinetic energy at a height 4 metre

✰ Potential energy at a height 4 metre

${\bold{\sf{\underline{Solution}}}}$

✰ Kinetic energy at a height 4 metre = 120 J

✰ Potential energy at a height 4 metre = 80 J

${\bold{\sf{\underline{Using \: concepts}}}}$

✰ Potential energy formula.

${\bold{\sf{\underline{Using \: formula}}}}$

✰ Potential energy = mgh

${\bold{\sf{\underline{We \: also \: write \: these \: as}}}}$

✰ Potential energy as P.E

✰ Mass as m

✰ Joules as J

✰ Height as h

✰ Raised height as g

${\bold{\sf{\underline{Full \: solution}}}}$

<h3>✠ Let us find the Potential energy.</h3>

↦ Potential energy = mgh

↦ Potential energy = 2 × 10 × 4

↦ Potential energy = 20 × 4

↦ Potential energy = 80 J

<h3>✠ Now according to the question let us find the kinetic energy</h3>

↦ Kinetic energy = Posses potential energy - Finded potential energy

↦ Kinetic energy = 200 J - 80 J

↦ Kinetic energy = 120 Joules

4 0

3 years ago

Read 2 more answers

A hiker, caught in a rainstorm might absorb 1 liter of water in her clothing. if it is windy so that this amount of water is eva

NeTakaya

Water evaporates at 100⁰C

So change in temperature = 100-20 = 80⁰C

Amount of water to be evaporated = 1 liter = 1L*1kg/liter = 1 kg

Specific heat of water is 1 calorie/gram ⁰C = 4.186 joule/gram =4186 J/kg

So heat required E = mcΔT = 1 * 4186 *80= 334880 J =334.88 kJ

So amount of heat require to evaporate water = 334.88 kJ

4 0

3 years ago

How could you increase the number of snowmen destroyed by the sled sliding down the hill? SledAQ1 A. Increase the starting heigh

Dima020 [189]

To increase the number of snowmen destroyed, you would want more energy and all of the above would increase the energy of the system so D.

8 0

3 years ago

Read 2 more answers

9. Calculate the distance (in km) that Charlie runs if he maintains the average

Karo-lina-s [1.5K]

<u>Correct Question:</u>

Calculate the distance (in km) charlie runs if he maintains an average speed of 8 km/hr for 1 hour

<u>Answer:</u>

The total distance covered by Charlie is 8 km in 1 hour.

<u>Explanation:</u>

The average velocity as given in the question is,

v = 8 km/hr

Total time taken,

$$t=1 hour$

As we know the formula to evaluate the total distance d when the average velocity and time is given;

$v=\frac{d}{t}$

$d=v \times t$

$d=8 \times 1$

$d=8 k m$

Hence, the total distance covered by Charlie in 1 hour will be 8 km.

5 0

3 years ago

Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains

OLEGan [10]

Answer:

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)

C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J

C = 252000 J/326.7 kg°C

5 0

3 years ago