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liberstina [14]
4 years ago
10

A vertical scale on a spring balance reads from 0 to 200 \rm{N}. The scale has a length of 10.0 \rm{cm} from the 0 to 200 \rm{N}

reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.05 \rm{Hz}.
Ignoring the mass of the spring, what is the mass m of the fish?
Physics
1 answer:
pav-90 [236]4 years ago
8 0

Answer:

m = 12.05 kg

Explanation:

Spring constant in K, N/m

K = 200/10* 100

K = 2000 N/m

Angular Frequency = sqrt (Spring constant / (Mass )

ω = 2 π f

ω =  2π* 2.05 Hz = 12.8805 rad/s

ω^2 = Spring constant / Mass

Mass= Spring constant / ω^2

ω^2 = 165.907 rad^2/s^2

m = 2000 (N/m)/165.907 (rad^2/s^2)

m = 12.05 kg

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5 0
4 years ago
When a battery, a resistor, a switch, and an inductor form a circuit and the switch is closed, the inductor acts to oppose the c
GaryK [48]

Answer:

The time constant becomes twice.

Explanation:

T = Time constant of the L-R circuit

L = Inductance of the inductor

R = Resistance of the resistor

Time constant of the L-R circuit is given as

T = \frac{L}{R}\\

T_{1} = initial time constant of the L-R circuit = T

T_{2} = final time constant of the L-R circuit

L_{1} = Initial inductance of the inductor =  L

L_{2} = Initial inductance of the inductor =  2L

For the same resistance, the time constant depend directly on the inductance, hence

\frac{T_{1}}{T_{2}} = \frac{L_{1}}{L_{2}}\\\frac{T}{T_{2}} = \frac{L}{2L}\\\frac{T}{T_{2}} = \frac{1}{2}\\T_{2} = 2T

7 0
4 years ago
6. A car accelerates from rest to 5m/s in 20 s. The force from the engine is 2000N. The force of air resistance and friction act
otez555 [7]

Answer:

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4 0
3 years ago
If the ac peak voltage across a 100-ohm resistor is 120 V, then the average power dissipated by the resistor is ________
Kazeer [188]

Answer:

The average power dissipated is 72 W.

Explanation:

Given;

peak voltage of the AC circuit, V₀ = 120 V

resistance of the resistor, R = 100 -ohm

The average power dissipated by the resistor is given by;

P_{avg} = \frac{1}{2} I_oV_o= I_{rms}V_{rms} = \frac{V_{rms}^2}{R}

where;

V_{rms} is the root-mean-square-voltage

V_{rms} = \frac{V_o}{\sqrt{2}} \\\\V_{rms} = \frac{120}{\sqrt{2}}\\\\V_{rms} = 84.853 \ V

The average power dissipated by the resistor is calculated as;

P_{avg} = \frac{V_{rms}^2}{R}\\\\P_{avg} = \frac{84.853^2}{100}\\\\P_{avg} = 72 \ W

Therefore, the average power dissipated is 72 W.

5 0
3 years ago
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