Answer:
a) α = 0.0334 rad / s² , b) w = 2.14 rad/s see that the angular velocity doubles.
Explanation:
This is a magular kinematics exercise
Let's reduce the magnitudes to the SI system
w = 0.17 rev /s (2π rad / 1rev) = 1.07 rad / s
a) as part of rest its initial velocity is zero w or = 0
w = w₀ + α t
α =
α =
α = 0.0334 rad / s²
b) If we double the angular relation what will be the final velocity
w = w₀ + (2α) t
w = 0 + 2 0.0334 32
w = 2.14 rad/s
We see that the angular velocity doubles.
Answer:
The time taken by missile's clock is
Solution:
As per the question:
Speed of the missile,
Now,
If 'T' be the time of the frame at rest then the dilated time as per the question is given as:
T' = T + 1
Now, using the time dilation eqn:
(1)
Using binomial theorem in the above eqn:
We know that:
Thus eqn (1) becomes:
Now, putting appropriate values in the above eqn:
Answer:
The direction of the B-field is in the +y-direction.
Explanation:
The corresponding formula is
This means, we should use right-hand rule.
Our index finger is pointed towards +x-direction (direction of velocity),
our middle finger should point towards the direction of the B-field,
and our thumb should point towards the +z-direction (direction of the force).
Since our middle finger in this situation points towards +y-direction, the B-field should be in +y-direction.
Answer:
The arrow is at a height of 500 feet at time t = 2.35 seconds.
Explanation:
It is given that,
An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s
The projectile formula is given by :
We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,
On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds
So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.