At 4 m/s?
How do the two kinetic energies compare to one another? QUADRUPLES !
#3 What is the kinetic energy of a 2,000 kg bus that is moving at 30 m/s?
Potential energy
9 × 10²¹ electrons flow through a cross section of the wire in one hour.
<h3>What is the relation between current and charge?</h3>
- Mathematically, current = charge / time
- In S.I. unit, Charge is written in Coulomb and time in second.
<h3>What is the amount of charge flown through a wire for one hour if it carries 0.4 A current?</h3>
- Charge= current × time
- Current= 0.4 A, time = 1 hour= 3600 s
- Charge= 0.4× 3600
= 1440 C
<h3>How many numbers of electrons present in 1440C of charge?</h3>
- One electron= 1.6 × 10^(-19) C
- So, 1440 C = 1440/1.6 × 10^(-19)
= 9 × 10²¹ electrons
Thus, we can conclude that the 9 × 10²¹ electrons flow through a cross section of the wire in one hour.
Learn more about current here:
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Answer:
Being an elastic object, rubber ball will be an ideal choice as it will bounce off the bowling pit and will experience a large change in momentum in comparison with the beanbag which will either slow down or come to a halt upon hitting a bowling pit. That is why rubber ball will experience a greater impulse and the bowling pin will experience the negative impulse of the rubber ball.
For Rubber Ball
Upon elastic collision it will reverses the direction and move with velocity equal or less then original
change in momentum = P
For Beanbag
value of impulse will large if velocity is zero.
Explanation:
Answer:
The magnitude of magnetic field at given point = 
× 
T
Explanation:
Given :
Current passing through both wires = 5.0 A
Separation between both wires = 8.0 cm
We have to find magnetic field at a point which is 5 cm from any of wires.
From biot savert law,
We know the magnetic field due to long parallel wires.
⇒ 
Where 
magnetic field due to long wires, 
, 
perpendicular distance from wire to given point
From any one wire 
5 cm, 
3 cm
so we write,
∴ 
![B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]](https://tex.z-dn.net/?f=B%20%3D%5Cfrac%7B%204%5Cpi%20%5Ctimes10%5E%7B-7%7D%20%5Ctimes5%7D%7B2%5Cpi%20%7D%20%5B%5Cfrac%7B1%7D%7B0.03%7D%20%2B%20%5Cfrac%7B1%7D%7B0.05%7D%20%5D)
Therefore, the magnitude of magnetic field at given point = 
consider the motion along the horizontal direction :
v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s
X = horizontal displacement of the ball = 2.0 m
a = acceleration along the horizontal direction = 0 m/s²
t = time taken to land = ?
using the kinematics equation
X = v₀ t + (0.5) a t²
2.0 = 3.0 t + (0.5) (0) t²
t = 2/3
consider the motion of the ball along the vertical direction
v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s
Y = vertical displacement of the ball = height of the table = h
a = acceleration along the vertical direction = 9.8 m/s²
t = time taken to land = 2/3
using the kinematics equation
Y = v₀ t + (0.5) a t²
h = 0 t + (0.5) (9.8) (2/3)²
h = 2.2 m
C 2.2 m
