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Refer to the attachment for the graph. The shape of both functions should resemble part of a parabola. Assumption: air resistance on the car is negligible.

Explanation:

The toy car started with a large amount of (gravitational) potential energy (PE) when it is at the top of the tree. Since it wasn't moving (as it was within Michelle's grip,) its kinetic energy (KE) would be equal to zero.
As the car falls to the ground, its PE converts to KE.
When the car was about to reach the ground, its PE is almost zero, while its KE is at its maximum.

The size of gravitational PE depends on both the mass and the height of the object. In this case, assume that the mass of the car stayed the same, PE should be proportional to the height of the car.

Assume that air resistance on the car is negligible. The height $h$ of the car at time $t$ could be found with the equation:

$\displaystyle h = -\frac{1}{2}\, g\, t^2 + h_0 \, (\text{Initial height})$ ,

where

$g \approx \rm 9.81\; m \cdot s^{-2}$ near the surface of the earth, and
$h_0$ is the initial height of the car.

On the other hand, $\displaystyle \text{GPE} = m \, g \, h = -\frac{m \cdot g^2}{2}\, t^2 + \underbrace{m \cdot g \cdot h_0}_{\text{Initial GPE}}$ .

In other words, plotting the gravitational PE of the car against time would give a parabola. Since $\displaystyle -\frac{m \cdot g^2}{2} < 0$ (the quadratic coefficient is smaller than zero,) the parabola should open downwards. Besides, since at $t = 0$ the initial GPE is positive, the $y$ -intercept of this parabola should also be positive.

Assume that the air resistance on the car is negligible. The mechanical energy (ME) of the toy car should conserve (stay the same.) The mechanical energy of an object is the sum of its PE and KE. The PE of the toy car has already been found as a function of time. Therefore, simply subtract the expression of PE from mechanical energy to find an expression for KE.

To find the value of mechanical energy, consider the PE of the toy car before it was dropped. Since initially KE was equal to zero, the mechanical energy of the toy car would be equal to its initial PE. That's $m \cdot g \cdot h_0$ . If there's no air resistance, the value of ME would stay at

Subtract PE from ME to obtain an expression for KE:

$\begin{aligned} \text{KE} &= \text{ME} - \text{PE} \cr &= m \cdot g \cdot h_0 - \left(-\frac{m \cdot g^2}{2}\, t^2 + m \cdot g \cdot h_0\right)\cr &= \frac{m \cdot g^2}{2}\, t^2\end{aligned}$ .

That's also a parabola when plotted against $t$ . Note that since the quadratic coefficient $\displaystyle \frac{m \cdot g^2}{2}$ is positive, the parabola shall open upwards.

7 0

4 years ago

1. A tree is making a 12 m shadow on the ground, the angle of elevation is 30 degrees, what is the height of the tree?

makvit [3.9K]

Answer:

6.93m

Explanation:

Using the method of SOH CAH TOA

The base of the tree is 12m which is the adjacent side of the triangle to be generated, the height of the tree will be facing the angle of elevation and it is the opposite side.

Using the formula tan(theta) = Opposite/Adjacent

Given theta = 30°

Adjacent is 12m which is the distance the tree is making with the shadow on the ground

Opposite side will be the height of the tree.

Tan(theta) = height/12

Tan30° = height/12

Height = 12tan30°

Height = 6.93m

Therefore the height of the tree is 6.93m

8 0

4 years ago

Read 2 more answers

The creation and study of new and very massive elementary particles is an important part of contemporary physics. To create a pa

dimaraw [331]

Consider a collision between two protons that is perfectly inelastic: an incident proton with mass m(p), kinetic energy K, and momentum magnitude p combines with a target proton that was initially stationary to form a single product particle with mass M.

Now,

In this case, the problem is alleviated by using colliding beams.

A pair of interacting particles' combined momentum may be zero.

After the collision, the center of mass is at rest.

Therefore, all the initial kinetic energy can be used to create particles.

Let K be the kinetic energy of each of the two identical colliding particles.

Therefore, the energy released by the two colliding beams are:

E₁ = K + mc² and E₂ = K + mc²

In the final state,

E(f) = Mc² where M is the new mass of the single product after the collision.

By conservation of energy,

E₁ + E₂ =E(f)

K + mc² + K + mc² = Mc²

2K + 2mc² = Mc²

Taking 2mc² common, we get

Mc² = 2mc²[ 1 + (K/mc²)]

Learn more about collision here:

brainly.com/question/12644900

#SPJ4

4 0

1 year ago