A is a length or distance. The sin and all the other trig functions have no units. They're just numerical ratios.
Answer:
Option C. is correct
Explanation:
The magnetic field is the area around a magnet in which there is magnetic force. When an electric current flows through a wire a magnetic field is created. A single wire does not produce a strong magnetic field. So, to increase the power of the magnetic field, increase the number of coils in the wire.
Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
the tendency of an object to resist a change in motion is called inertia
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
