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Hatshy [7]
3 years ago
15

The tension of a guitar string is increased by 40%. By what factor odes the fundamental frequency of vibration change? a. 1.13 b

. 1.18 c. 1.06 d. 1.09 e. 1.24
Physics
1 answer:
bogdanovich [222]3 years ago
8 0

Answer:

<h3> b. 1.18</h3>

Explanation:

The fundamental frequency in string is expressed as;

F1 = 1/2L√T/m .... 1

L is the length of the string

T is the tension

m is the mass per unit length

If the tension is increased by 40%, the new tension will be;

T2 = T + 40%T

T2 = T + 0.4T

T2 = 1.4T

The new fundamental frequency will be;

F2 = 1/2L√1.4T/m ..... 2

Divide 1 by 2;

F2/F = (1/2L√1.4T/m)/1/2L√T/m)+

F2/F = √1.4T/m ÷ √T/m

F2/F = √1.4T/√m ×√m/√T

F2/F = √1.4T/√T

F2/F = 1.18√T/√T

F2/F = 1.18

F2 = 1.18F

Hence the fundamental frequency of vibration changes by a factor of 1.18

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Answer:

12N

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Answer:

(a) \frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

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Explanation:

(a)

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P = \sigma AT^4

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P = Energy Radiated per Second = ?

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So the ratio of power at 250 K to the power at 2000 K is given as:

\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b)

Now, for 90% radiator blackbody at 2000 K:

P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4

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7 0
3 years ago
8TH GRADE SCIENCE I NEED NOW DO NOT SKIP
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Explanation:

1. Force=mass*acceleration

acceleration=force/mass

=100/50

=2m/s^2

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Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If
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Let F_{1}=F_{2}=F.

Normal force equals (using Newton's third law) N=mg+F\sin30^{o}-F\sin37^{o}.

F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o}), but F_{f}\leq F(\cos30^o+\cos37^o) for all F_{f} (in order to start moving the break). Therefore F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o}), solving for F: F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}

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