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Stels [109]
3 years ago
8

How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 n/m so that it wil

l oscillate at 1.00 hz?
Physics
1 answer:
mrs_skeptik [129]3 years ago
7 0
The frequency of a simple harmonic oscillator such as a spring-mass system is given by
f= \frac{1}{2 \pi}   \sqrt{ \frac{k}{m} }
where 
k is the spring constant
m is the mass attached to the spring.

Re-arranging the formula, we get:
m= \frac{k}{4 \pi^2 f^2}
and since we know the constant of the spring:
k=39.5 N/m
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
m= \frac{39.5 Hz}{4 \pi^2 (1.00 Hz)^2} = 1.00 kg
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Answer:

r = 5.94 10⁻² m

Explanation:

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        F =q v x B

The bold indicate vectors and the vector product determines that the force is perpendicular to the other two vectors, so the modulus of the velocity does not change, but its direction, therefore the acceleration in Newton's second law is centripetal

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The magnitude of the force is

      F = q v B sin θ

The centripetal acceleration is

      a = v² / r

Let's replace

       q v B sin θ = m v² / r

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Let's calculate

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3 years ago
You are standing on a bathroom scale in an elevator in a tall building. Your mass is
Oksanka [162]

Answer:

94 kg or 921.6 N

Explanation:

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a_N = 9.8 + 4.6 = 14.4\text{ m/s}{}^2

The resultant weight is

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Since the bathroom scale is graduated in kg, it's reading is

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The correct answer is

b. low frequencies and low energy.

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E=hf

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