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3 years ago

What happens in terms of energy when a moving car hit a parked car causing the park card to move?

Physics

2 answers:

olga_2 [115]3 years ago

5 0

Its like newtons 3rd law that once in motion a outer force has to stop it

ad-work [718]3 years ago

5 0

Answer:

The moving car transfers Kinetic Energy to the parked car

Explanation:

Options are not supplied here but I have previously answered this question in a test and these options were supplied.

A) The moving car transfers kinetic energy to the parked car.

B) Kinetic energy in the moving car disappears.

C) Kinetic energy in the parked car is created.

D) The parked car transfers kinetic energy to the moving car.

A) This option provides us with the true outcome, the moving car transfers Kinetic energy to the parked car. This is in agreement with the law of conservation of energy. A transference of energy is the outcome of this event.

B) Kinetic energy cannot disappear, this is because according to the law of conservation of energy, energy can neither be created nor destroyed.

If energy disappears, this would imply that energy is destroyed which is not agreeable.

C) Kinetic energy in the parked car is created... Just like the option B above, and according to the above stated law, energy can neither be created nor destroyed. It is also not agreeable here that energy is created.

D) Energy can be transferred but, "a parked car indicates that the car only has potential energy" and as such cannot transfer kinetic energy to the moving car since it does not possess kinetic energy.

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Quick puzzle question. If god created the universe who created god?

mrs_skeptik [129]

Answer:

My mom always told me he was just there

Explanation:

4 0

3 years ago

Read 2 more answers

what is the energy (in eV units) carried by one photon violet light that has a wavelength of 4.5e-7?

DaniilM [7]

The energy of a photon is given by
$E=hf$
where h is the Planck constant and f is the photon frequency.

We can find the photon's frequency by using the following relationship:
$f= \frac{c}{\lambda}$
where c is the speed of light and $\lambda$ is the photon's wavelength. By plugging numbers into the equation, we find
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{4.5 \cdot 10^{-7} m}=6.67 \cdot 10^{14}Hz$

And so now we can find the photon energy
$E=hf=(6.6 \cdot 10^{-34} Js)(6.67 \cdot 10^{14}Hz )=4.4 \cdot 10^{-19} J$

We know that 1 Joule corresponds to
$1 J = 1.6 \cdot 10^{-19} eV$
So we can convert the photon's energy into electronvolts:
$E= \frac{4.4 \cdot 10^{-19} J }{1.6 \cdot 10^{-19} J/eV}=2.75 eV$

4 0

4 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago

the power of a physicians eyes is 57.1 D while examinging a patient. how far from her eyes (in m) is the feature being examined.

yanalaym [24]

Answer:

14 cm

Explanation:

Power of eye = 57.1 D

The relation between the focal length and the power is

f = 1 / P = 1 / 57.1 = 0.0175 m = 1.75 cm

The distance between the image and the lens is, v = 2 cm

Let the distance between the object and the eye is u

Use the lens equation

1/ f = 1 / v - 1 / u

1 / 1.75 = 1 / 2 - 1 / u

1 / u = 1 / 2 - 1 / 1.75

1 / u = (1.75 - 2) / 3.5

u = - 14 cm

Thus, the distance between the feature and eye is 14 cm .

8 0

3 years ago

During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 * t^2 to s

Mila [183]

Answer:

70560 m

Explanation:

The formula to calculate the distance travelled during a free fall motion is

$d=-\frac{1}{2}gt^2$

where

d is the distance travelled

g = -9.8 m/s^2 is the acceleration due to gravity

t is the time

In this situation,

t = 120 s

Therefore the distance travelled after 120 s is

$d=-\frac{1}{2}(-9.8)(120)^2=70560 m$

6 0

4 years ago