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
3.5m is ur answer ask for more questions anytime
Answer: rolling friction
Explanation: I think it is the answer
Answer:
The dimension of the nullspace of T = 4
Explanation:
The rank/dimension theorem is explains that:
Suppose V and W are vector spaces over F, and T:V → W is linear. If V is finite dimensional, then
nullity(T) + rank(T) = dim(V).
rank(T) = dimension of T = dim(T) = dim(W) = 7
nullity(T) = dimension of the nullspace of T = dim(T) = ?
dim(V) = 11
nullity(T) = dim(V) - dim(T) = 11 - 7
nullity(T) = 4.
