Answer:
a= 1.59 m/s² : Magnitude of the acceleration
β = 65.22° (north of east) : Direction of the acceleration
Explanation:
Conceptual analysis
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Problem development
The acceleration is presented in the direction of the resultant force applied .
Calculation of the resultant forces (R)
R= 429.5 N
We apply the formula (1) to calculate the magnitude of the acceleration(a) :
∑F = m*a , m= 270 kg
R= m*a
429.5 =270*a
a= 1.59 m/s²
Calculation of the direction of the acceleration (β)
β = 65.22° (north of east)
B would be first cause they are tall then
c cause they are
short then D can they are tall then A
This is what I would say
For the purpose of the exercise, we can assume that the Earth is at perihelion (closest point to the Sun) on December 21st (in reality, it happens around January 4th) and that the Earth is at aphelion (farthest point from the Sun) on June 21st (in reality, this happens around July 4th). The distance Earth-Sun is the following:
- Perihelion: 147.1 milion km
- Aphelion: 152.1 milion km
- Average distance: 149.6 milion km
So, we can calculate the percentage change with respect to the average distance as:
Answer:
Explanation:
(a) the angle between the coil and the coil's magnetic dipole moment is same that means zero degree.
(b) L = 17.6 cm = 0.176 m
i = 5.95 mA = 5.95 x 106-3 A
B = 4.74 mT = 4.74 x 10^-3 T
(b) Let r be the radius of the loop is r
Circumference = 2 x 3.14 x r
0.176 = 2 x 3.14 x r
r = 0.028 m
B = μo/4π x (N x 2 π i) / r
Where, N be the number of turns
4.74 x 10^-3 = (10^-7 x N x 2 x 3.14 x 5.95 x 10^-3) / 0.028
N = 3.55 x 10^4
(c) Maximum torque = N i A B
τ = 3.55 x 10^4 x 5.95 x 10^-3 x 3.14 x 0.028 x 0.028 x 4.74 x 10^-3
τ = 2.465 x 10^-3 Nm