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Anestetic [448]
3 years ago
12

(96 Points )A bus and a bicycle have a head-on collision. Compare the force of impact between the bus and the bicycle; compare t

he accelerations of the bus and bicycle.
Physics
1 answer:
lidiya [134]3 years ago
6 0

Answer:

Small acceleration for the bus and major for the bicycle

Explanation:

In a head-on collision of this type, i order to find force involve in a collision, one has to study the change in linear momentum of each object. Because Force s defined as the change of momentum with time.

Recall that momentum is the product of the mass of the object times its velocity. The mass of the bus (M) is much larger than that of the bicycle (m), and even if their speeds are similar, the momentum of the bus will be much larger than that of the bike due to this mass difference. We could assume an inelastic collision (where the two objects stay together after the collision),

The initial total momentum of the system (imagining that the bus is pointing to the right (positive convention), while the bike is pointing in the opposite (to the left), can be written as: Pi=Mv-mv (we are assuming that both vehicles go at the same speed for simplicity)

The final total momentum of the system that underwent an inelastic collision will be: P_f=(M+m)v_f

Therefore, since the mass M is much larger than mass m, the final momentum will be pretty similar to the initial momentum of the bus.

The change in momentum for the bus in the fraction of time of the collision will be almost unnoticeable for the bus (minimal change in its momentum, and therefore minimal acceleration accounting for the change).

For the bicycle, the change in momentum involves change in the direction of motion, (going to the left before the collision, and to the right afterwards) which will include a mayor acceleration backwards.

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In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 1850 J of work is done on the gas
Oliga [24]

Answer:

The value of change in internal l energy of the gas = 1850 J

Explanation:

Work done on the gas (W) =  - 1850 J

Negative sign is due to work done on the system.

From the first law  we know that Q = Δ U + W ------------- (1)

Where Q = Heat transfer to the gas

Δ U = Change in internal energy of the gas

W = work done on the gas

Since it is adiabatic compression of the gas so heat transfer to the gas is zero.

⇒ Q = 0

So from equation (1)

⇒ Δ U = - W ----------------- (2)

⇒ W = - 1850 J (Given)

⇒ Δ U = - (- 1850)

⇒ Δ U = + 1850 J

This is the value of change in internal energy of the gas.

6 0
3 years ago
If the distance between two charges is doubled, by what factor is the magnitude of the electric force changed? F_e final/F_e, in
motikmotik

To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,

F_{initial} = \frac{kq_1q_2}{r^2}

Here,

k = Coulomb's constant

q_{1,2} = Charge at each object

r = Distance between them

As the distance is doubled so,

F_{final} = \frac{kq_1q_2}{( 2r )^2}

F_{final} = \frac{ kq_1q_2}{ 4r^2}

F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}

F_{final} = \frac{1}{4} F_{initial}

\frac{F_{final}}{ F_{initial}} = \frac{1}{4}

Therefore the factor is 1/4

6 0
3 years ago
Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.
rjkz [21]

Answer:

Momentum of block B after collision =-50\ kg\ ms^{-1}

Explanation:

Given

Before collision:

Momentum of block A = p_{A1}= -100\ kg\ ms^{-1}

Momentum of block B = p_{B1}= -150\ kg\ ms^{-1}

After collision:

Momentum of block A = p_{A2}= -200\ kg\ ms^{-1}

Applying law of conservation of momentum to find momentum of block B after collision p_{B2}.

p_{A1}+p_{B1}=p_{A2}+p_{B2}

Plugging in the given values and simplifying.

-100-150=-200+p_{B2}

-250=-200+p_{B2}

Adding 200 to both sides.

200-250=-200+p_{B2}+200

-50=p_{B2}

∴ p_{B2}=-50\ kg\ ms^{-1}

Momentum of block B after collision =-50\ kg\ ms^{-1}

6 0
2 years ago
Please help ASAP. How is heat related to work?
earnstyle [38]
Energy is the ability to do work so I would say that thermal or heat energy is a type of work. Don’t know if this will work but that’s what I would put.
7 0
3 years ago
Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are tripled and their separation
stira [4]

Answer:

Remains the same

Explanation:

5 0
2 years ago
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