Answer:
A. The force between the wires Is attractive, this is because there are two wires of equal current flowing in the same direction thus using the right hand rule the force vectors of the two wires point to each other.
B: see attached file for explanation
Explanation:
Answer:
LT⁻¹
Explanation:
Assuming the given expression is
x = A t³ + B t.........(1)
x is the distance
we have to calculate dimension of dx/dt
from expression (1)
x = A t³
A = LT⁻³
x = B t
B = LT⁻¹
now,
dx/dt = 3At^2 + B
from rule of dimension
dimension of dx/dt is equal to dimension of At^2
dx/dt = A t²
dx/dt = LT⁻³ x T²
dx/dt = LT⁻¹
hence, dimension of dx/dt is equal to LT⁻¹
Hi, thank you for posting your question here at Brainly.
First, you have to draw the system for better understanding. Pls refer to the attached image.
Since the system is not moving, the summation of forces are zero since they are at equilibrium
Let's base on the 2-kg object first to find the tension (T) of the rope.
Forces in the y-direction = 0 = T - W
0 = T - 2 kg (9.81 m/s2)
T = 19.62 N
Let's base on the 2.41 kg load.
Forces in the y-direction = 0 = Fn - Wy
0 = Fn - (2.41kg)(9.81m/s^2)(cos47)
Fn = 16.12 N
Forces in the x-direction = 0 = Ff + T - Wx
0 = Ff + 19.62 - (2.41kg)(9.81m/s^2)(sin47)
Ff = -2.33 N
The magnitude of the frictional force is 2.33 N.
PART A)
Impulse is defined as change in momentum
here finally after hitting the wall car comes to rest
So final momentum of the car will be ZERO
Initial momentum of the car is given as
here m = 2000 kg
v = 77 m/s
now we have
so here impulse will be
PART B)
As we know that force of impact is defined as
so here it is ratio of impulse and time
so if we increase the time of impact by 3 times
so the force will be decrease by factor of 1/3
so here force is decreased by 1/3