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3 years ago

In which situation does the law of conservation of momentum apply?

Physics

1 answer:

zvonat [6]3 years ago

7 0

As per Newton's II law we know that

$F = \frac{dP}{dt}$

rate of change in momentum is Net outside force on the object

now here we will have if net outside force will be zero

then we can say

$F = 0 = \frac{dP}{dt}$

so from above equation we can say

$dP = 0$

and hence we have

$P = constant$

so the correct answer will be

<em>3.) in the absence of outside forces</em>

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What is a

LiRa [457]

Answer:

I think there six points

4 0

2 years ago

U235 + n → Xe134 + Sr100 + 2n

xenn [34]

Nuclear fission formula by the looks of it. Possibly how Professor Lisa Meitner realised that she had split the atomic nucleus. The Xenon and the Strontium (Xe and Sr) would presumably show up in a radio chemical assaying test at her university.
A few years later, Professor J Robert Oppenheimer watched a nuclear test somewhere near Los Alamos, US and lamented "I am become death, the destroyer of worlds". Shortly thereafter, Hiroshima and Nagasaki were razed to the ground and annihilated by nuclear bombs. Professor Meitner, probably inadvertently, had got the keys to the doors to "nuclear hell", and JRO ended up turning them. Something like that maybe, and a very harrowing and tumultuous period in human history.
Note in the fission equation, that out come two neutrons. They go off and produce a similar fission in another U235 nucleus into a chain reaction which, i not moderated by, say, Boron, can end up as a "mushroom cloud".

8 0

2 years ago

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago

4- by eating food only will you get energy?

Serga [27]

Answer:

yes

Explanation:

All parts of the body (muscles, brain, heart, and liver) need energy to work. This energy comes from the food we eat. Our bodies digest the food we eat by mixing it with fluids (acids and enzymes) in the stomach.

4 0

3 years ago

Given that 1 pound is equal to 4.45 newton’s what is the weight of a 500N child in pounds?

Bond [772]

Answer:

112.36 pounds

Explanation:

Since 1 pound = 4.45 Newtons, a 500N child in pounds = 500÷4.45 = 112.36 pounds (approximately).

4 0

2 years ago