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dexar [7]
3 years ago
14

Define Momentum in detail.

Physics
2 answers:
Harrizon [31]3 years ago
8 0

Explanation:

Momentum Is defined as the product of of mass and its velocity

Momentum (M) =mass *velocity

SI unit of momentum is kgm/s

The rate of change in momentum

=change in momentum / time

=(mv-mu)/t

blsea [12.9K]3 years ago
7 0

Answer:

Momentum is simply the product of mass and velocity. It is represented mathematically as

Momentum = Mass x Velocity.

The S. I unit of momentum is kilograms meter per second i.e Kg.m/s.

Momentum is a vector quantity i.e it has both magnitude and direction.

Momentum of an object is greatly affected my the mass and velocity of the object as we can clear from the equation.

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A light flashes at position x=0m. One microsecond later, a light flashes at position x=1000m. In a second reference frame, movin
padilas [110]

Answer:

To the right relative to the original frame.

Explanation:

In first reference frame <em>S</em>,

Spatial interval of the event, \rm \Delta x=1000\ m-0\ m=1000\ m.

Temporal interval of the event, \rm \Delta t = 1\ \mu s=10^{-6}\ s.

In the second reference frame <em>S'</em>, the two flashes are simultaneous, which means that the temporal interval of the event in this frame is \rm \Delta t'=0\ s.

The speed of the frame <em>S' </em>with respect to frame <em>S</em> = v.

According to the Lorentz transformation,

\rm \Delta t'=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right ).\\\\Since,\ \Delta t'=0,\\\therefore \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right )=0\\\Rightarrow \Delta t-\dfrac{v\Delta x}{c^2}=0\\\dfrac{v\Delta x}{c^2}=\Delta t\\v=\dfrac{c^2\Delta t}{\Delta x }.\\\\Also, \ \Delta t,\ \Delta x>0\ \Rightarrow v>0.

And positive v means the velocity of the second frame<em> </em><em>S'</em> is along the positive x-axis direction, i.e., to the right direction relative to the original frame <em>S</em>.

7 0
3 years ago
In an LRC circuit, why does the sum of the voltages around the circuit not equal the applied voltage as kirchhoff’s rule require
wel

Kirchhoff's loop rule is always true.

The foundation of circuit analysis is Kirchhoff's Laws for current and voltage. We have the fundamental set of tools required to begin evaluating circuits with these two principles, In an RLC circuit, the most fundamental elements of a resistor, inductor, and capacitor are connected across a voltage supply.

Learn more about Kirchhoff loop rule here-  

brainly.com/question/15003023

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5 0
2 years ago
Read 2 more answers
A gazelle leaps from a cliff 2.5 m high with a speed of 5.6m/s.
jeyben [28]

Initial speed of Gazelle is along x direction and its value will be

v_x = 5.6 m/s

also its initial height is given as

y = 2.5 m

Part a)

now from kinematics along Y direction

\Delta y = v_y t + \frac{1}{2} at^2

as we know that

\Delta y = 0

v_y = 0

a = 9.8 m/s^2

2.5 = 0 + \frac{1}{2} (9.8) t^2

t = 0.714 s

Part b)

distance moved horizontally

\Delta x = v_x t

as we know that

v_x = 5.6 m/s

now we will have

v_x = 5.6 (0.714) = 4m

so it will lend at distance of 4 m.

Part c)

final velocity in vertical direction

v_{fy} = v_y + at

v_{fy} = 0 + (9.8)(0.714) = 7 m/s

v_x = 5.6 m/s

so net speed will be

v^2 = v_x^2 + v_y^2

v^2 = 7^2 + 5.6^2

v = 8.96 m/s


7 0
3 years ago
You are explaining to your friends why astronauts feel weightless while orbiting in the space shuttle. They ask you to prove thi
meriva

Answer:

Explanation:

Force of gravity = GMm/r^2 = ma

a being the acceleration due to gravity at some distance r from the center of the Earth.  I'll use 6400 km for the radius of the Earth so r = 6734 km or 6734000 meters

a = GM/r^2

plugging in G = 6.67 x 10^-11

M is the mass of the Earth = 6 x 10^24

and r is from above

a = 8.825 m/s^2  = 0.9g

so 90% the acceleration of gravity on the surface.

5 0
3 years ago
The windshield of a speeding car hits a hovering insect. Consider the time interval from just before the car hits the insect to
HACTEHA [7]

Answer:

A. False

B True

C. False

D.False

E. True

F. False

G. False

H. False

I. True

Explanation:

A. False: The system being analyzed consists of the bug and the car.  These are the two bodies involved in the collision.

B. True: The system being analyzed consists of the bug and the car

C. False: The magnitudes of the change in velocity are different from the car and the bug. The velocity of the bug changes from 0 to the velocity of the car, while there is no noticeable change in the velocity of the car

D.False: There is barely any change in the momentum of the car since the mass of the bug is very small.

E. True: Since the mass of the bug is small, and was initially at rest, the magnitude of the change in monentum will be large because the new velocity will be that of the car.

F. False: The system being analyzed consists of the bug and the car. Those are the two bodies involved in the collision

G. False: The car barely changes in velocity since the mass of the bug is small.

H. False: The car barely changes in momentum because the collision does not affect its speed so much. on the other hand the momentum change of the bug is large since its mass is small.

I. True: The bug which was initially at rest will begin moving with the velovity of the speeding car, while the car barely changes in its velocity

5 0
3 years ago
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