Answer:
75.7238461542 kg
20.4664723031 m
Explanation:
F = Force = 82 N
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Force is given by
The mass of the block is 75.7238461542 kg
Distance moved in the next 4.2 seconds
The distance moved is 20.4664723031 m
True, if you move something forward at 100 miles an hour but your on something moving backwards 100 miles an hour you up staying in the same location, aka zero velocity.
Answer:
0.0979 N/c
Explanation:
Electric field, E is given as a product of resistivity and current density
E=jP where P is resistivity and j is current density
But the current density is given as
where I is current and A is area and
Substituting this into the first equation then
Given diameter of 0.259 cm= 0.00259 m and the radius will be half of it which is 0.001295 m
Answer:
A) τ = 1,222 10⁻⁶ N m
, B) w = 0.24 rad / sec
, v = 2.88 10⁻³ m / s
Explanation:
Part A
We can get the torque
τ= F x r
bold are vector
τ = F r sin θ
Let's use according to Newton's law
F - W = 0
F = mg
τ = mg r sin θ
Let's reduce the magnitudes to the SI system
m = 12 ug = 12 10⁻⁶ kg
r = 12 mm = 12 10⁻³ m
Let's calculate
τ = 12 10⁻⁶ 9.8 12 10⁻³ sin 60
τ = 1,222 10⁻⁶ N m
Part B
Let's use Newton's law for rotational movement
τ = I α
The moment of inertia of the antero that we approximate as a particle is
τ = m r² α
α = τ / m r²
α = 1,222 10⁻⁶ / (12 10⁻⁶ (12 10⁻³)²)
α = 0.70718 10³ rad / s²
Angular velocity is
w = w₀ + α t
w = 0 + 0.70718 10³ 0.34 10⁻³
w = 0.24 rad / sec
Angular and linear variables are related.
v = w r
v = 0.24 12 10⁻³
v = 2.88 10⁻³ m / s
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.