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A 1.5m long string weighs 0.0020 kg. It is tensioned to 100N. A disturbance travels along it with a wavelength of 1.5m, find:a)

Zigmanuir [339]

Answer:

the propagation velocity of the wave is 274.2 m/s

Explanation:

Given;

length of the string, L = 1.5 m

mass of the string, m = 0.002 kg

Tension of the string, T = 100 N

wavelength, λ = 1.5 m

The propagation velocity of the wave is calculated as;

$v = \sqrt{\frac{T}{\mu} } \\\\\mu \ is \ mass \ per \ unit \ length \ of \ the \ string\\\\\mu = \frac{0.002 \ kg}{1.5 \ m} = 0.00133 \ kg/m\\\\v = \sqrt{\frac{100}{0.00133} } \\\\v = 274.2 \ m/s$

Therefore, the propagation velocity of the wave is 274.2 m/s

7 0

2 years ago

A long solenoid that has 1,130 turns uniformly distributed over a length of 0.430 m produces a magnetic field of magnitude 1.00

sweet-ann [11.9K]

Given Information:

Number of turns = N = 1130 turns

Length of solenoid = L = 0.430 m

Magnetic field = B = 1.0x10⁻⁴ T

Required Information:

Current = I = ?

Answer:

I = 0.0302 A

Explanation:

The current flowing in the solenoid winding can be found using

I = BL/μ₀N

Where μ₀ is the permeability of free space, N is the number of turns, B is the magnetic field and L is the length of solenoid

I = 1.0x10⁻⁴*0.430/4πx10⁻⁷ *1130

I = 0.0302 A

or

I = 30.28 mA

6 0

2 years ago

It is friction that provides the force for a car to accelerate, so for high-performance cars the factor that limits acceleration

bearhunter [10]

Answer:

About 12 seconds

Explanation:

6 0

3 years ago

A 15-turn circular wire loop with a radius of 3.0 cm is initially in a uniform magnetic field with a strength of 0.5 T. The fiel

lana66690 [7]

To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that

$\epsilon = - d\frac{\phi_B}{dt}$

$\epsilon = -NA\frac{dB}{dt}$

Where,

N = Number of loops

A = Cross sectional Area

B = Magnetic Field

$\epsilon = (15)(\pi(0.03)^2)\frac{0-0.5}{0.1}$

$\epsilon = 0.212V$

$\epsilon = 0.21V$

Therefore the correct answer is A.

6 0

3 years ago

Two identical resistors are connected in parallel across a 26-V battery, which supplies them with a total power of 7.1 W. While

Gnom [1K]

Answer:

A) R = 190.42 Ω

B) P = 5.325 W

Explanation:

We are given;

Total power;P_tot = 7.1 W

Voltage;V = 26 V

A)We are told that while the battery is still connected, one of the resistors is heated, so that its resistance doubles.

Thus, the power is doubled.

Now, formula for power is;

P = IV

Thus, since power is doubled, we have;

P = 2(IV)

Now, formula for current is; I = V/R

So, P = 2V²/R

Making R the subject, we have;

R = 2V²/P

In this question, P is p_total = 7.1 W

Thus;

R = (2 × 26²)/7.1

R = 190.42 Ω

B) Now, the resistance of the resistors are R and 2R.

Formula for power in this context is;

P = V²/R

Thus,

Total power delivered to the resistors is;

P = V²/R + V²/2R

P = 3V²/2R

P = (3 × 26²)/(2 × 190.42)

P = 5.325 W

8 0

3 years ago

Are these right? Just need you to confirm or tell we what I got wrong. Thanks​

Are these right? Just need you to confirm or tell we what I got wrong. Thanks