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3 years ago

How high would you have to lift a 1000kg car to give it a potential energy of:

Physics

1 answer:

Elza [17]3 years ago

5 0

Given parameters:

Mass of the car = 1000kg

Unknown:

Height = ?

To find the heights for the different amount potential energy given, we need to understand what potential energy is.

Potential energy is the energy at rest due to the position of a body.

It is mathematically expressed as:

P.E = mgh

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height of the car

Now the unknown is h, height and we make it the subject of the expression to make for easy calculation.

h = $\frac{P.E }{mg}$

For 2.0 x 10³ J;

h = $\frac{2000}{1000 x 9.8}$ = 0.204m

For 2.0 x 10⁵ J;

h = $\frac{200000}{9.8 x 1000}$ = 20.4m

For 1.0kJ = 1 x 10³J;

h = $\frac{1000}{9.8 x 1000}$ = 0.102m

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A certain car can accelerate from 0 to 60 mph in 7.9 s. What is the car's average acceleration in mph/s?

Anna007 [38]

Answer:

a = 7.6\ mph/s

Explanation:

Motion With Constant Acceleration

It's a type of motion in which the velocity of an object changes uniformly in time.

The equation that describes the change of velocities is:

$v_f=v_o+at$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

Solving the equation [for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

The car accelerates from vo=0 to vf=60 mph in t=7.9 s, thus the acceleration is:

$\displaystyle a=\frac{60 \ mph-0}{7.9}$

a = 7.6\ mph/s

7 0

3 years ago

A train on a straight track goes in the positive direction for 5.9 km, and then backs up for 3.8 km

kenny6666 [7]

Answer:

2.1km

Explanation:

Ill take it as u are talking about the displacement

Since displacement has negatives and positves

5.9 - 3.8 = 2.1km

6 0

3 years ago

Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person

sertanlavr [38]

Answer:

v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

f ’= f (v + v₀) / (v- $v_{s}$ )

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

v₀ = v_{s} = v ’

we substitute

f ’= f (v + v’) / (v - v ’)

f ’/ f (v-v’) = v + v ’

v (f ’/ f -1) = v’ (1 + f ’/ f)

v ’= (f’ / f-1) / (1 + f ’/ f) v

v ’= (f’-f) / (f + f’) v

let's calculate

v ’= (3400 -3000) / (3000 +3400) 343

v ’= 400/6400 343

v ’= 21.44 m / s

3 0

3 years ago

The work accomplished (produced) by a machine is called Work _____

Nonamiya [84]

I think it’s output because output work is work done by a machine

6 0

3 years ago

A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the

yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0 (initial KE = 0 because the ball starts from rest) (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0

4 years ago