The mass attached to the spring must be 0.72 kg
Explanation:
The frequency of vibration of a spring-mass system is given by:
(1)
where
k is the spring constant
m is the mass attached to the spring
We can find the spring constant by using Hookes' law:
where
F is the force applied on the spring
x is the stretching of the spring
When a mass of m = 2.82 kg is applied to the spring, the force applied is the weight of the mass, so we have
and using 
and 
, we find
Now we want the frequency of vibration to be
f = 7.42 Hz
So we can rearrange eq.(1) to find the mass m that we need to attach to the spring:
#LearnwithBrainly
the best option would be d.) examples of constructive erosion.
1.2 x (2.2 x 10⁵) = 264,000 Ω
0.8 x (2.2 x 10⁵) = 176,000 Ω
With a 'nominal' value of 220,000 Ω, it could actually be anywhere <em>between 176,000Ω and 264,000Ω</em> .
Answer:
D they both are the same acceleration because they are in free fall
Answer: A
Explanation: Neglecting air resistance every object will return to the ground at the same interval regardless of their weights, we can say that the objects are undergoing free falls, they are falling under the sole of gravity.
No matter the weight of the objects or the elevation, without significant air resistance, they will reach the ground at the same time.
