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nirvana33 [79]
2 years ago
15

Please help. This question is urgent.​

Physics
1 answer:
Alex_Xolod [135]2 years ago
8 0

Answer:

A

Explanation:

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What length of a certain metal wire of diameter 0.15 mm is needed for the wire to have a resistance of 15 ω? the resistivity of
Murljashka [212]

The resistance of the cylindrical wire is R=\frac{\rho l}{A}.

Here R is the resistance, l is the length of the wire and A is the area of cross section. Since the wire is cylindrical A=\frac{\pi d^2}{4} . Rearranging the above equation,

A=\frac{\rho l}{R}\\  \frac{\pi d^2}{4}=\frac{\rho l}{R}\\  l=\frac{\pi d^2 R}{4 \rho}

Here d=0.15, R=15, \rho=1.68(10^{-8}).

Substituting numerical values,

l=\frac{\pi 0.15^2(10^{-6}) (15)}{4 (1.68)(10^{-8})}  \\ l=15.78

The length of the wire is 15.78 \;\;m

5 0
3 years ago
An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic m
Elanso [62]

Answer:

Part a)

A = 0.066 m

Part b)

maximum speed = 0.58 m/s

Explanation:

As we know that angular frequency of spring block system is given as

\omega = \sqrt{\frac{k}{m}}

here we know

m = 3.5 kg

k = 270 N/m

now we have

\omega = \sqrt{\frac{270}{3.5}}

\omega = 8.78 rad/s

Part a)

Speed of SHM at distance x = 0.020 m from its equilibrium position is given as

v = \omega \sqrt{A^2 - x^2}

0.55 = 8.78 \sqrt{A^2 - 0.020^2}

A = 0.066 m

Part b)

Maximum speed of SHM at its mean position is given as

v_{max} = A\omega

v_{max} = 0.066(8.78) = 0.58 m/s

4 0
3 years ago
A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calc
balandron [24]

Answer:

Approximately 28^{\circ}.

Explanation:

The refractive index of the air n_{\text{air}} is approximately 1.00.

Let n_\text{glass} denote the refractive index of the glass block, and let \theta _{\text{glass}} denote the angle of refraction in the glass. Let \theta_\text{air} denote the angle at which the light enters the glass block from the air.

By Snell's Law:

n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}}).

Rearrange the Snell's Law equation to obtain:

\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}.

Hence:

\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}.

In other words, the angle of refraction in the glass would be approximately 28^{\circ}.

7 0
2 years ago
A pail in a water well is hoisted by means of a frictionless winch, which consists of a spool and a hand crank. When Jill turns
Sveta_85 [38]

Answer:

182.28 W

Explanation:

Here ,

m = 7.30 Kg

distance , d= 28.0 m

time , t = 11.0 s

average power supplied = change in potential energy/time

average power supplied = m×g×d/time

average power supplied = 7.30×9.81×28/11

average power supplied = 182.28 W

the average power supplied is  182.28 W

6 0
2 years ago
620 J of heat is added to the cylinder of an engine, which causes the gas inside to expand. As a result, the piston of the engin
Maru [420]

Answer:

<u>400</u> J work is done BY the engine.

The internal energy of the gas is <u>620</u> J

Explanation:

The given information are;

The heat added to the cylinder = 620 J

The force applied by the piston of the engine = 8.0 kN = 8,000 N

The distance over which the force moves (the piston) = 5.0 cm = 0.05 m

The work done (by the engine) = Force × Distance = 8,000 N × 0.05 m = 400 J

The internal energy is the sum of the kinetic and potential energy of the system

Therefore, given that the internal energy, U, is the sum total of the energy in the system

∴ U = The heat supplied to the system = 620 J

Which gives;

<u>400</u> J work is done BY the engine.

The internal energy of the gas is <u>620</u> J.

6 0
2 years ago
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