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3 years ago

A bicyclist, traveling with a constant speed, covers 61.Om in 7.80s.

Physics

1 answer:

dybincka [34]3 years ago

5 0

Answer:

<h2>The velocity is 7.82 m/s</h2>

Explanation:

In this problem, we are going to solve for the velocity since what we are given as parameters are:

1. Distance covered s= 61m

2. time taken to cover the distance t= 7.8 secs

we know that the expression for velocity is given as

velocity= distance/time

We then substitute the given data into the formula to solve for velocity.

velocity= 61/7.8

velocity= 7.82 m/s

<h2>The velocity is 7.82 m/s</h2>

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The mass of a string is 7.7 × 10-3 kg, and it is stretched so that the tension in it is 190 N. A transverse wave traveling on th

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3 years ago

calculate the efficiency of a light bulb that had an input of 400 j transfers 100j as a light and 300j as heat.

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6 0

3 years ago

Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g

ollegr [7]

Answer:

The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

<u>velocity of wave on the string with greater tension;</u>

$v_1 = \sqrt{\frac{T_1}{\mu }$

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

$v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}$

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

$v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s$

<u>Fundamental frequency of wave on the string with greater tension;</u>

Beat frequency = F₂ - F₁

= 270.6 - 258

= 12.6 Hz

Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0

3 years ago