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Anastasy [175]
3 years ago
9

Questlon 7 of 10

Physics
2 answers:
katovenus [111]3 years ago
6 0

Answer:

Plasma

Explanation:

Aye pecks said so

jek_recluse [69]3 years ago
5 0

Answer:

0B. Liqiud can always conduct electricity

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A weight of 1400 pounds is suspended from two cables as shown in the figure. What is the tension in the left cable? _________ po
juin [17]

Answer:

Following are the solution to this question:

Explanation:

Law:

\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}

      = 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}

\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}

\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\

4 0
3 years ago
A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl
kondaur [170]

Answer:

Part a)

y_m = 0.157 mm

part b)

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

Explanation:

As we know that the speed of wave in string is given by

v = \sqrt{\frac{T}{m/L}}

so we have

T = 17.5 N

m/L = 5.4 g/cm = 0.54 kg/m

now we have

v = \sqrt{\frac{17.5}{0.54}}

v = 5.69 m/s

now we have

Part a)

y_m = amplitude of wave

y_m = 0.157 mm

part b)

k = \frac{\omega}{v}

here we know that

\omega = 2\pi f

\omega = 2\pi(92.2) = 579.3 rad/s

so we  have

k = \frac{579.3}{5.69}

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

4 0
3 years ago
When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 45.
SpyIntel [72]

Answer:

Torque on the rocket will be 1.11475 N -m

Explanation:

We have given that muscles generate a force of 45.5 N

So force F = 45.5 N

This force acts on the is acting on the effective lever arm of 2.45 cm

So length of the lever arm d = 2.45 cm = 0.0245 m

We have to find torque

We know that torque is given by \tau =F\times d=45.5\times 0.0245=1.11475N-m

So torque on the rocket will be 1.11475 N -m

3 0
3 years ago
A raft with the area A , thickness= h and the mass 600 kg, Floats in still water with 7 cm
elena55 [62]

<span>In this problem, we need to solve for Bubba’s mass. To do this, we let A be the area of the raft and set the weight of the displaced fluid with the raft alone as ρwAd1g and ρwAd2g with the person on the raft, </span>where ρw is the density of water, d1 = 7cm, and d2= 8.4 cm. Set the weight of displaced fluid equal to the weight of the floating objects to eliminate A and ρw then solve for m.

<span>ρwAd1g = Mg</span>

ρw<span>Ad2g = (M + m) g</span>

<span>d2∕d1 = (M + m)/g</span>

m = [(d2<span>∕d1)-1] M = [(8.4 cm/7.0 cm) - 1] (600 kg) =120 kg</span>

This means that Bubba’s mass is 120 kg.

7 0
3 years ago
A uniform string of length 0.50 m is fixed at both ends. Find the
kozerog [31]

Answer:

configuration of string:

Node - Antinode - Node    or N-A-N

This is 1/2 wavelength since a full wavelength is N-A-N-A-N

f (fundamental) = V / wavelength

F0 = 300 m/s / 1 m = 100 / sec

F1 = 300 m/s / .5 m = 600 / sec

Each increase is a multiple of the fundamental since the wavelength

increases by 1/2 wavelength to keep nodes at both ends of the string

4 0
2 years ago
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