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morpeh [17]
1 year ago
7

Two remote control cars with masses of 1.16 kilograms and 1.98 kilograms travel toward each other at speeds of 8.64 meters per s

econd and 3.34 meters per second, respectively. The cars collide head-on, and the less massive car recoils with a speed of 2.16 meters per second.a. What is the final speed of the second car? Include units in your answer. More information. Hint: Pay attention to directions! Must be in 3 significant digits.
Physics
1 answer:
Black_prince [1.1K]1 year ago
3 0

The initial momentum of the system can be expressed as,

p_i=m_1u_1+m_{2_{}}u_2

The final momentum of the system can be given as,

p_f=m_1v_1+m_{2_{}}v_2

According to conservation of momentum,

p_i=p_f

Plug in the known expressions,

\begin{gathered} m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ m_2v_2=m_1u_1+m_2u_2-m_1v_1 \\ v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2} \end{gathered}

Initially, the second mass move towards the first mass therefore the initial speed of second mass will be taken as negative and the recoil velocity of first mass is also taken as negative.

Plug in the known values,

\begin{gathered} v_2=\frac{(1.16\text{ kg)(8.64 m/s)+(1.98 kg)(-3.34 m/s)-(1.16 kg)(-2.16 m/s)}}{1.98\text{ kg}} \\ =\frac{10.02\text{ kgm/s-}6.61\text{ kgm/s+}2.51\text{ kgm/s}}{1.98\text{ kg}} \\ =\frac{5.92\text{ kgm/s}}{1.98\text{ kg}} \\ \approx2.99\text{ m/s} \end{gathered}

Thus, the final velocity of second mass is 2.99 m/s.

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Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

P = pressure inside the tube,

P• = 1 atm

10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

P - 101325 = 56593

P = 56593 + 101325

P = 157918 Pascal

Now, remember, in an ideal gas,

P1V1/T1 = P2V2/T2, where V is constant, then we have

P1/T1 = P2/T2, and when we substitute the values, we have

101325/(273 + 18) = 157918/ T2

101325/291 = 157918/ T2

T2 = (157918 * 291)/101325

T2 = 453 K

T2 = 453 - 273 = 180° C

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4 years ago
Who is the first person to trace the characteristics of successive generations of a living thing?
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3 years ago
suppose a child walks from the outer edge of a rotating Merry-Go-Round to the inside does angular velocity of the Merry-Go-Round
Akimi4 [234]

Okay here initially child walks on the outer edge of the disc and then started to move inside

So here as the child and Merry go round is an isolated system so there is no external Torque on this system from outside

As here we can see there is external force acting on this system by the hinge of the Merry go round as well as due to gravity so we can not use momentum conservation to solve such type of questions.

But as we can say that there is no external torque on this system about the hinge point so we will use conservation of angular momentum for this system

Here as we know that

\tau = \frac{dL}{dt}

where L = angular momentum

since here torque is ZERO

0 = \frac{dL}{dt}

L = constant

so here we can write initial angular momentum of the system as

L = (I_1 + I_2)*\omega

here we know that

I_1 = moment of inertia of merry go round

I_2 = moment of inertia of child

so here we can say

(I_1 + I_2)* \omega_1 = (I_1 + I_2')\omega_2

so here as the child moves from edge to inside the disc it moment of inertia will decrease because as we know that moment of inertia of child is given as

I_2 = mr^2

here m = mass of child

r = distance of child from center

Since child is moving inside so his distance from center is decreasing

so here moment of inertia of child is decreasing as he starts moving inside

so final angular speed of merry go round will increase as child go inside

\omega_2 = \frac{(I_1 + I_2)*\omega}{(I_1 + I_2')}

so here as

I_2' < I_2

final angular speed will be more than initial speed as child moves inside

3 0
3 years ago
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