Answer:
Explanation:
Magnetic field near current carrying wire
=
i is current , r is distance from wire
B = 10⁻⁷ x
force on second wire per unit length
B I L , I is current in second wire , L is length of wire
= 10⁻⁷ x x 33 x 1
= 3234 x
This should balance weight of second wire per unit length
3234 x = .075
r = x 10⁻⁷
= .0043 m
= .43 cm .
Answer:
A = m³/s³ = [L]³/[T]³ = [L³T⁻³]
B = m³s = [L³T]
Explanation:
We have the equation:
V = At³ + B/t
where, the dimensions of each variable are as follows:
V = m³ = [L]³
t = s = [T]
substituting these in equation, we get:
m³ = A(s)³ + B/s
for the homogeneity of the equation:
A(s)³ = m³
<u>A = m³/s³ = [L]³/[T]³ = [L³T⁻³]</u>
Also,
B/s = m³
<u>B = m³s = [L³T]</u>
Answer:
the answer is 1.35. Have a nice day!!
Answer:
the answer is A.
Explanation:
Using the laws of newton:
∑F = ma
where ∑F is the sumatory of forces acting in the system, m the mass and a the acelertion of the system.
Then, if the block is moving with constant velocity, its aceleration is equal to 0, so:
∑F = m(0)
∑F = 0
It means that:
F - = 0
where F is the force applied and is the friction force. Replacing the value of F, we get:
310N - = 0
Finally, solving for :