Magnitude of normal force acting on the block is 7 N
Explanation:
10N = 1.02kg
Mass of the block = m = 1.02 kg
Angle of incline Θ
= 30°
Normal force acting on the block = N
From the free body diagram,
N = mgCos Θ
N = (1.02)(9.81)Cos(30)
N = 8.66 N
Rounding off to nearest whole number,
N = 7 N
Magnitude of normal force acting on the block = 7 N
Answer:
Explanation:
When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:
Where 
is the length of the string and 
the velocity of propagation. Use this expression to find the value of .
The velocity of propagation is given by the expression:
Where 
is the desirable variable of the problem, the linear mass density, and 
is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:
With the value of the tension and the velocity you can find the mass density:
since both components, length and time, are measurable
<span>since Rate = length ÷ time </span>
<span>∴ rate is also measurable and ∴ quantitative.
</span>
Answer:
The speed is equals to 22.49 m/s
Explanation:
Given Data:
Required:
Speed of Traverse wave = c =?
Solution:
As we know that
Now the equation for speed of traverse wave is calculated through:
=
Substituting the values
=22.49 m/s
