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3 years ago

PLEASE TRY TO ANSWER AS MANY QUESTIONS AS YOU CAN !

Physics

2 answers:

Vladimir [108]3 years ago

8 0

Answer:

0.546

Explanation:

(∑Fy=0; 41.5x9.8+138Sin 28 =0; Fn = 341.9N

Ff= 341.9x0.29=99.1

∑Fx =ma; 138Cos28- 99.1 = 41.5xa;

a = .546 m/s2

suter [353]3 years ago

3 0

Good luck with solving this

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4vir4ik [10]

It brings cold water from the bottom of the ocean.

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3 years ago

Read 2 more answers

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

velocity of wave on the string with greater tension;

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

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

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

Fundamental frequency of wave on the string with greater tension;

$f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz$

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

A charge of 90 C passes through a wire in 1 hour 15 minutes . what is the current in the wire

timurjin [86]

We calculate current from the formula:
$I= \frac{q}{t}$ , where q is a electric charge transferred over time t
Time should be converted to seconds:
1h 15 min= 75min= 4500s
I= $\frac{90C}{4500s}=0,02A$ Result is in unit-Ampere