All categories

3 years ago

What happens in terms of energy when a moving car hits a parked car, causing the parked car to move?

Physics

2 answers:

padilas [110]3 years ago

8 0

Answer:

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

Explanation:

In terms of energy when a moving car hits a parked car, causing the parked car to move, the moving car transfers kinetic energy to the parked car.

Arlecino [84]3 years ago

5 0

Answer:

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

Explanation:

As we know that energy is neither be created nor it is destroyed but we can transfer energy from one form to another form.

As we know that here a moving car will collide with a parked car

So initially the moving car is having some kinetic energy while parked car has no energy.

Then we know that after collision parked car start moving so here we can say that parked car is getting kinetic energy after collision.

So we can say that here energy is transferred from moving car to parked car in the form of kinetic energy due to which it starts moving.

So correct answer is

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

You might be interested in

I need C,D,E,F,H please

VARVARA [1.3K]

Answer:

cant see blury. but wanna TEXT?

Explanation:

7 0

3 years ago

It dont have the subject science so it science

WINSTONCH [101]

Answer:

The human activity that uses the most water worldwide is for agriculture. In other words, watering crops.

Explanation:

pls mark brainliest

7 0

3 years ago

Consider a day in the physics lab room with room temperature of 20 0C and pressure of 1.0 atm. A container of 3.5 L is left open

svetlana [45]

Answer:

The number of moles of air that escape is approximately 0.027 moles

Explanation:

The room temperature, T₁ = 20 °C = 298.15 K

Atmospheric pressure, P₁ = 1.0 atm

The volume of the container, V₁ = 3.5 L

The final temperature of the air in the container after heating on the Bunsen burner, T₂ = 95 °C = 368.15 K

The container opened finally

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

Where;

R = The universal gas constant = 0.08205 L·atm/(mol·K)

Therefore, we get;

n₁ = 1.0 × 3.5/(0.08205 × 298.15) ≈ 0.143

The number of moles, n₁ ≈ 0.143 moles

When the gas is heated to 95 °C, the number of moles becomes

n₂ = P₂·V₂/(R·T₂)

P₂ = 1.0 atm atmospheric pressure, V₂ = 3.5 L, the volume of the container, T₂ =

∴ n₂ = 1.0 × 3.5/(0.08205 × 368.15) ≈ 0.116

The number of moles of air remaining in the container, n₂ ≈ 0.116 moles

The number of moles of air that escape, n = n₁ - n₂

∴ n = 0.143 - 0.116 = 0.027

The number of moles of air that escape, n ≈ 0.027 moles

3 0

3 years ago

A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to

daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s

Explanation:

It is given that,

Mass of cart 1, $m_1=2\ kg$

Mass of cart 2, $m_2=4\ kg$

Initial speed of cart 1, $u_1=3\ m/s$

Initial speed of cart 2, $u_2=0$ (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{2\times 3}{(2+4)}$

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

$J=F\times t=m(V-u)$

$J=m(V-u)$

$J=2\times (1-3)$

J = -4 kg-m/s

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

8 0

4 years ago

A current of 0.86 a flows through a copper wire 0.47 mm in diameter when it is connected to a potential difference of 15 v. how

likoan [24]

To calculate the length of the wire, we use formulas,

$R = \frac{V}{I}$ (A)

$R= \rho \frac{l}{A}$ (B)

Here, R is the resistance of the wire, I is the current flows through wire and V is potential difference. A is cross sectional area of wire and $\rho$ is the density of copper wire and is value, $\rho = 1.7\times 10^{-8} \Omega m$ .

Given $I = 0.86 A,V=15 V and r = \frac{0.47 mm}{2} =2.35 \times 10^{-4} m, V= 15 V.$

Substituting the values of I and V in equation (A ) we get,

$R=\frac{15V}{0.86A} = 17.44 \Omega$

Now from equation (B),

$l=\frac{R A}{\rho }$

Therefore,

$l= \frac{17.44\times\pi \times r^2 }{1.7\times 10^{8} \Omega m} \\\\ l= \frac{17.44\times 3.14 \times(2.35\times10^{-4}m)^2 }{1.7\times 10^{-8} \Omega m} = 177.9 m$

Thus the length of the copper wire is 177.9 m.

8 0

3 years ago