Explanation:
Period of a mass on a spring is:
T = 2π√(m/k)
T is inversely proportional with the square root of k. So as the spring constant increases, the period decreases.
Answer:
The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s
Explanation:
Given that,
Mass of halfback = 98 kg
Speed of halfback= 4.2 m/s
Mass of corner back = 85 kg
Speed of corner back = 5.5 m/s
We need to calculate their mutual speed immediately after the touchdown-saving tackle
Using conservation of momentum
Where, 
= mass of halfback
=mass of corner back
= velocity of halfback
= velocity of corner back
Put the value into the formula
Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s
Hello,
<u>Solution for A:</u>
Force = 3.00N
Mass = 0.50 Kgs
Time = 1.50 Seconds
According to newton's second law of motion;
Force = Mass times Acceleration(a)
3.00 = 0.50 * a
a = 3.00/0.50 = 6.00 m/s^2
We know that acceleration = Velocity / time
So Velocity = time * acceleration = 1.50 * 6 = 9.00 m/s^2
<u>Solution for B:</u>
The net force = 4.00N - 3.00N = 1.00N to the left
Force = 1.00N
Mass = 0.50Kg
Time = 3.00 Seconds
Again; F = MA (Where F is force, M is mass and A is acceleration)
1.00N = 0.5 * A
A = 1/0.5 = 2 m/s^2
Velocity = Acceleration * Time = 2 * 3 = 6 m/s
Answer:
See explanation
Explanation:
We have to convert to angular velocity in rads-1 as follows;
Angular velocity in rad/s = 2π/60 × 1900 rpm = 199 rad/s
Given that
angular velocity =angle turned /time taken
Time taken = angle turned/angular velocity
Converting 35° to radians we have;
35 × π/180 = 0.61 radians
Time taken = 0.61 radians/199 rad/s
Time taken = 0.0031 seconds
