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Flura [38]
3 years ago
7

A student shakes a rope such that 20 complete vibrations are made in 4.00 seconds. Determine the vibrational frequency of the ro

pe, along with the corresponding unit.
Physics
1 answer:
fgiga [73]3 years ago
8 0

Answer:

The vibrational frequency of the rope is 5 Hz.

Explanation:

Given;

number of complete oscillation of the rope, n = 20

time taken to make the oscillations, t = 4.00 s

The vibrational frequency of the rope is calculated as follows;

Frequency = \frac{number \ of \ complete \ vibrations}{time \ taken} \\\\Frequency = \frac{20 }{4 \ s} \\\\Frequency = 5 \ Hz

Therefore, the vibrational frequency of the rope is 5 Hz.

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Which factor does not affect the resistance of a material?
34kurt

the best answer is b

5 0
3 years ago
Read 2 more answers
URGENT <br> Find the total<br> equivalent<br> resistance for the<br> circuit.
Nesterboy [21]

Answer:

20 + 10 = 30 \\ 30and50 in \: parallel \\  =  \frac{50 \times 30}{50 + 30}  \\  =  \frac{1500}{80}  \\ 18.75 \\ 30nd40parallel \\  \frac{30 \times 40}{70}  \\  \frac{1200}{70}  = 17.143 \\ total = 18.75 + 17.143 = 35.893ohm \\ thank \: u

3 0
3 years ago
100 points!!! Please help!!!
Greeley [361]

Answer:

41°

Explanation:

Kinetic energy at bottom = potential energy at top

½ mv² = mgh

½ v² = gh

h = v²/(2g)

h = (2.4 m/s)² / (2 × 9.8 m/s²)

h = 0.294 m

The pendulum rises to a height of  above the bottom.  To determine the angle, we need to use trigonometry (see attached diagram).

L − h = L cos θ

cos θ = (L − h) / L

cos θ = (1.2 − 0.294) / 1.2

θ = 41.0°

Rounded to two significant figures, the pendulum makes a maximum angle of 41° with the vertical.

7 0
4 years ago
Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

8 0
3 years ago
PLEASE PLEASE PLEASE A mad scientist places massive amounts of charge on basketball sized aluminum balls. The charge on the ball
bazaltina [42]

Answer:

4.2 x 10⁷N

Explanation:

Given parameters:

Charge on ball:

             q₁  = 3C

              q₂ = 14C

Distance between balls  = 9000m

Unknown:

Force acting on the two balls

Solution:

The force experienced by the two charges is given by coulombs law. It is mathematically expressed as;

                      F  = \frac{k q_{1} q_{2} }{r^{2} }

where k  = 9 x 10⁹Nm²/C²

           q is the charges

             r is the distance

Input the variables and solve;

                 

        F  = \frac{9 x 10^{9} x 3 x 14 }{9000}  = 4.2 x 10⁷N

8 0
3 years ago
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