Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
Answer:
Explanation:
If E₀ is the electric field outside the smaller sphere and r is the radius of larger sphere.
E₀ = kQ/r²
The radius of the larger sphere is 3r and the charge on both sphere is same then the electric field outside the larger sphere is given as
E = kQ/(3r)² = kQ/9r² = 1/9 (kQ/r²)= 1/9 x E₀
hence the correct option is e.
The force of thrust is greater than the force if gravity !
Answer found on quizlet !
-- Momentum is (mass) x (speed).
Object B has 1.5 times as much momentum as Object A has.
-- Kinetic energy is (1/2) x (mass) x (speed) .
Object B has 1.5 times as much kinetic energy as Object A has.
-- If they would both stop long enough to get on the scale,
Object B would weigh 1.5 times as much as Object A does.
Explanation:
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