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kvasek [131]
3 years ago
13

Akhtar, Kiran and Rahul were riding in a motorocar that was moving with a high velocity on

Physics
1 answer:
VikaD [51]3 years ago
4 0

Answer:

According to the law of conservation of momentum:

Momentum of the car and insect system before collision = Momentum of the car and insect

system after collision

Hence, the change in momentum of the car and insect system is zero.

The insect gets stuck on the windscreen. This means that the direction of the insect is

reversed. As a result, the velocity of the insect changes to a great amount. On the other hand,

the car continues moving with a constant velocity. Hence, Kiran’s suggestion that the insect

suffers a greater change in momentum as compared to the car is correct. The momentum of

the insect after collision becomes very high because the car is moving at a high speed.

Therefore, the momentum gained by the insect is equal to the momentum lost by the car.

Akhtar made a correct conclusion because the mass of the car is very large as compared to

the mass of the insect.

Rahul gave a correct explanation as both the car and the insect experienced equal forces

caused by the Newton’s action-reaction law. But, he made an incorrect statement as the

system suffers a change in momentum because the momentum before the collision is equal to

the momentum after the collision.

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Researchers found the "cosmic microwave background radiation", which is a heat imprint left over from the big bang.

The redshift of light emitted by most galaxies indicates the universe is expanding.

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3 years ago
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Which are characteristics of mammals? Check all that apply.
Neporo4naja [7]

Answer:

one of the characteristics of a mammal is their several hollow bones another is their three chambered heart and the last is highly developed nervous system

Explanation:

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6 0
3 years ago
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

\alpha = Angular acceleration

\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0
3 years ago
The electric field in a certain region of Earth’s atmo- sphere is directed vertically down. At an altitude of 300 m the field ha
I am Lyosha [343]

Answer:

Net charge contained in the cubeq= 3.536×10^-6C

Explanation:

Formular for total flux in a cube is given as:

Total flux= E300Acos(180) + E200Acos(0)

Where A is crossectional area

Total flux= A(E200-E300)

Total flux= q/Eo

q= Eo×total flux

q=(8.84×10^-12)×(100)^2×(100-60)

q= 3.536×10^-6C

6 0
4 years ago
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A 544 g ball strikes a wall at 14.3 m/s and rebounds at 14.4 m/s. The ball is in contact with the wall for 0.042 s. What is the
Nezavi [6.7K]

Answer:

F = 371.738\,N

Explanation:

Let assume that ball strikes a vertical wall in horizontal direction. The situation can be modelled by the appropriate use of the definition of Moment and Impulse Theorem, that is:

(0.544\,kg)\cdot (14.3\,\frac{m}{s})-F\cdot \Delta t = -(0.544\,kg)\cdot (14.4\,\frac{m}{s} )

F\cdot \Delta t = 15.613\,\frac{kg}{m\cdot s}

The average force acting on the ball during the collision is:

F = \frac{15.613\,\frac{kg}{m\cdot s} }{0.042\,s}

F = 371.738\,N

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3 years ago
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