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Ket [755]
2 years ago
11

A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35. (a) At this instant, wha

t are the magnitude and direction of its angular momentum relative to point.(b) If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of its angular momentum at this instant?
Physics
1 answer:
zaharov [31]2 years ago
7 0

Answer:

(A) L = 115.3kgm²/s

(B) dL/dt = 94.1kgm²/s²

Explanation:

The magnitude of the angular momentum of the rock is given by the foemula

L = mvrSinθ

We have been given θ = 36.9°, m = 2.0kg, v = 12.0m/s and r = 8.0m.

Therefore L = 2.00 × 12 × 8.0 × Sin 36.9° =

115.3 kgm²/s

(B) The magnitude of the rate of angular change in momentum is given by

dL /dt = d(mvrSinθ)/dt = mgrSinθ = 2.00 × 9.8 × 8.0× Sin36.9 = 94.1kgm²/s²

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A 10-kg wagon is pulled by a force of 70 Newtons while friction provides 20 Newtons of force. How much acceleration will the wag
kkurt [141]

Answer:

5m/s²

Explanation:

Given parameters:

Mass of wagon  = 10kg

Force of pull  = 70N

Frictional force = 20N

Unknown:

Acceleration of the wagon = ?

Solution:

Frictional force is a force that opposes motion.

The net force is given as:

   Net force  = mass x acceleration

 Force of pull  - Frictional force = mass x acceleration

 Insert the parameters and solve;

     70 - 20  = 10 x acceleration

       50  = 10 x acceleration

      Acceleration  = 5m/s²

5 0
3 years ago
Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen
Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

\lambda = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}

sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m

For first dark fringe

dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

3 0
2 years ago
Identify the prefix that would be used to express<br> 2,000,000,000 bytes of computer memory?
Reika [66]

Answer:

It would be 2 GB

4 0
2 years ago
Read 2 more answers
Newton's law of gravity was inconsistent with Einstein's special relativity because
Lubov Fominskaja [6]

Answer:

Mass and thus force depends on the reference frame chosen

Explanation:

This can be explained as Newton's law of gravity provides action which are instantaneous at a distance and involves the evaluation of all the quantities at present time or at the instant they occur.

If the body undergoes a change in its mass distribution there will be an immediate change in its gravitational force without any lag.

Now, if we talk about special relativity, it would be absurd to say that an information can travel faster than light. The effect is in synchronization with the cause in one reference frame where the effect occurs after the cause for some observer in some other reference frame.

In order to observe Newton's law of gravity all the observer's in different reference frames must observe the same phenomena which could only be possible if time were absolute and in special relativity, time is not absolute.

Therefore, Newton's law of gravity was inconsistent with the Einstein's Special Relativity.

3 0
2 years ago
Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

5 0
2 years ago
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