Answer:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. The total momentum of the trolleys after separation is zero
c. The momentum of the 2 kg trolley after separation is 12 kg·m/s
d. The momentum of the 3 kg trolley is -12 kg·m/s
e. The velocity of the 3 kg trolley = -4 m/s
Explanation:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0
c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s
d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s
e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley
∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s
The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s
Answer:
Explanation:
Force = q ( v x B)
- 5.6 x 10⁻⁹ (v x - 1.25 k )
- 3.4x 10⁻⁷i + 7.4 x 10⁻⁷j
Let v = ai+bj +ck
Force = - 5.6 x 10⁻⁹ [(ai+bj +ck) x - 1.25 k )]
= - 5.6 x 10⁻⁹ ( 1.25aj - 1.25bi )
= - 7 a j + 7 b i
( 7bi - 7aj ) x 10⁻⁹
Comparing with given force
7b x 10⁻⁹ b = - 3.4 x 10⁻⁷
b = - 48.57
- 7 a x 10⁻⁹ = 7.4 x 10⁻⁷
a = - 105.7
velocity
= -105.7 i - 48.57 j + ck
b ) Component along k can not be obtained .
c ) v . F = ( -105.7 i - 48.57 j + ck ) . −(3.40×10−7N) ˆı +(7.40×10−7N) ˆȷ
= 105.7 x 3.4 x 10⁻⁷ - 48.57 x 7.4 x 10⁻⁷
= 359.38 x 10⁻⁷ - 359.38 x 10⁻⁷
=0
angle between v and F = 90 degree
Answer:
no
Explanation:
this is because its valency shell is full so it wont want any other electrons in its valence shell.
Answer:
1) electric current - ampere
2) force - joule
3 frequency - hertz
4) density - kilogram per cubic metre
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