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3 years ago

Two identical cars, each traveling at the same velocity, and having the same mass come to a stop

Physics

1 answer:

olya-2409 [2.1K]3 years ago

4 0

Answer:

The second option - same change in momentum, different force

Explanation:

They experience the same change in momentum, but the force on car A is greater.

p=m*v

Impulse, or change in p=F*t

We know the cars have the same change in momentum, because they both go from the same initial momentum (p=mv), to zero momentum. Therefore, both experience the same change in momentum. Since the time is different, but the change in momentum is the same, the force has to be different.

For example: if both cars had 200 kg*m/s of momentum at the start, and 0 at rest, the impulse is 200 kg*m/s

So,

p(a)= 200=F(0.3)

p(b)= 200=F(5)

Which force has to be greater to make the equations true? The force for Car A, of course!

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Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t

salantis [7]

Explanation:

Mass of bumper cars, $m_1=m_2=110\ kg$

Initial speed of car A, $u_1=8\ m/s$

Initial speed of car Z, $u_2=-10\ m/s$

Final speed of car A after the collision, $v_1=-10\ m/s$

We need to find the velocity of car Z after the collision. Let it is equal to $v_2$ . Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$110\times 8+110\times (-10)=110\times (-10)+110v_2$

$v_2=\dfrac{-1320}{110}\ m/s$

$v_2=-12\ m/s$

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

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3 years ago

What are organelles?<br>Which part of a cell holds organelles?

DanielleElmas [232]

Answer:

An organelle is a structure in a cell that has specific jobs to do. Organelles are held in the cytoplasm

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3 years ago

Read 2 more answers

The typical unit for a period used with Kepler's third law is

ollegr [7]

Well, if you're using the law to work with periods of Earth satellites,
then the most convenient unit is going to be 'hours' for the largest
orbits, or 'minutes' for the LEOs.

But if you're using it to work with periods of planets, asteroids, or
comets, then you'd be working in days or years.

6 0

3 years ago

Two positive charges of same magnitude are separated by some distance if we bring a unit positive charge from one charge to anot

Studentka2010 [4]

Answer:

Increases.

Explanation:

The electric potential increases when the two positive charges of same magnitude bring close to one charge to another because there is repulsive force between them due to same charge and when the two opposite charges move away from each other, the potential energy decreases. When two opposite charges are brought closer together, electric potential energy decreases while on the other hand, when we move opposite charges apart from each other than the work done against the attractive force that leads to an increase in electric potential energy.

3 0

3 years ago

On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const

IceJOKER [234]

Answer:

Explanation:

Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} (1)$

Since the car starts from rest, v₀ =0.
We know the value of t = 5 sec., but we need to find the value of a.
Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

$a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2 (2)$

Replacing a and t in (1):

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m. (3)$

Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

$a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2 (4)$

Replacing v₀, at and t in (1), we have:

$x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m (5)$

Therefore, as the truck travels twice as far as the car, the right answer is a).

7 0

3 years ago