All categories

2 years ago

A car moving at a speed of 20 m/s has a kinetic energy of 200,000 J from that motion. What is the car's mass?

Physics

1 answer:

zhenek [66]2 years ago

8 0

Answer:

i think the answer is 1000 kg but i might be wrong

You might be interested in

A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and directi

Alex_Xolod [135]

Magnitude: 12.1 N.
Direction: 17.0° to the 8 N force.

<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

$\displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$ .
$\displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$ .

The sum of forces on each direction will be the resultant force on that direction:

Resultant force parallel to the 8 N force: $(8 + \dfrac{5\sqrt{2}}{2})\;\text{N}$ .
Resultant force normal to the 8 N force: $\dfrac{5\sqrt{2}}{2}\;\text{N}$ .

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

$\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N}$ (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

$\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491$ .

Find the size of the angle using inverse tangent:

$\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree$ .

In other words, the resultant force is 17.0° relative to the 8 N force.

4 0

3 years ago

Quick Quiz 40.2 While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: y

Firlakuza [10]

The order of decreasing photon energy is FM radio, AM radio, yellow light, and microwaves.

Electromagnetic radiation: The electromagnetic field's waves, which are conveying electromagnetic radiant energy through space, are what make up electromagnetic radiation. It consists of X-rays, gamma rays, microwaves, infrared, light, and radio waves. These waves are all a component of the electromagnetic spectrum.

Microwave radiation has wavelengths between one meter and one millimeter, which correspond to frequencies between 300 MHz and 300 GHz, respectively.

Radio waves: The electromagnetic spectrum's longest wavelengths, which are found in radio waves, are normally found at frequencies of 300 gigahertz and below.

The radio waves have the highest photon energy and the lowest is microwaves.

So, the highest to lowest order is FM radio, AM radio, yellow light and microwaves.

Learn more about waves here:

brainly.com/question/1968356

#SPJ4

4 0

2 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .