When a car traveling at 30m/s hits the gas pedal up to 65m/s in 3.5 sec what is the cars acceleration during that time

A 2000 kg car is moving at 15 m/sec when it collides with a 1200 kg car sitting still. Briefly compare the impulse imparted on t

Roman55 [17]

Answer:

The 1200 kg car.

Explanation:

The 1200 kg car had bo momentum at the initial stage of the journey because it wasn't on motion. But after collision with the 2000 kg car it gained alot of momentum.

4 0

3 years ago

Where is the centre of mass of a system of two particles is situated?

Sever21 [200]

Answer:

In a two particle system, the center of mass lies on the center of the line joining the two particles.

4 0

3 years ago

What question can a student BEST answer when comparing and contrasting the models?

OverLord2011 [107]

Answer:

A

Explanation:

4 0

2 years ago

An object has more momentum if it has which of the following?

maks197457 [2]

Answer:

C.) A high velocity and Large mass.

Explanation:

Momentum of any object is defined by following formula

Here : m = mass of object

v = velocity of object

now we know that since momentum is product of mass and velocity

So in order to have more momentum we need the value of this product to be more. So this product will me large is both the physical quantity will be more in magnitude. So if mass is large and velocity will be more then the product of them will be large and hence the momentum of object will be more. Btw I had that question too.

6 0

3 years ago

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$