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irga5000 [103]
3 years ago
10

ransverse waves are sent along a 5.00-m-long string with a speed of 30.00 m/s. The string is under a tension of 10.00 N. What is

the mass of the string
Physics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

0.055 kg

Explanation:

According to the given situation the solution of the mass of the string is shown below:-

Speed of the wave is

v = \sqrt{\frac{F_T\times Length\ of\ string}{Mass\ of\ string}}

30.0 m/s = \sqrt{\frac{10 kg m/s^2\times 5.00 m}{Mass\ of\ string}

Mass of string is

= \sqrt{\frac{10 kg m^2/s^2\times 5.00 m}{900 m^2 s^2}

After solving the above equation we will get the result that is

= 0.055 kg

Therefore for calculating the mass of the string we simply applied the above formula.

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D is the answer..........

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Help wit these questions someone.
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In series circuit, Req = R₁ + R₂ + R₃ + ···

In parallel circuit, \frac{1}{Req}  = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} +...

<h3>Q7.</h3>

total resistance in the upper branch = R₂ + R₃ = R₂ + 2

\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}

\frac{1}{4} = \frac{1}{R2+2} +\frac{1}{6}

R₂ + 2 = 12

R₂ = 10Ω

<h3>Q8.</h3>

\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}

\frac{1}{Req} = \frac{1}{2+1} + \frac{1}{4}

Req = 1.7Ω

7 0
3 years ago
A woman climbs up a ladder in 1.37 s at 2.20 m/s. How tall is the ladder?
ArbitrLikvidat [17]

Answer:

The ladder is 3.014 m tall.

Explanation:

To solve this problem, we must use the following formula:

v = x/t

where v represents the woman’s velocity, x represents the distance she climbed (the height of the ladder), and t represents the time it took her to move this distance

If we plug in the values we are given for the problem, we get:

v = x/t

2.20 = x/1.37

To solve this equation for x (the height of the ladder), we must multiply both sides by 1.37. If we do this, we get:

x = (2.20 * 1.37)

x = 3.014 m

Therefore, the ladder is 3.014 m tall.

Hope this helps!

6 0
4 years ago
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A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm
iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

m = m_{o}* m_{e}

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

m_{o} = \frac{v_{o}}{-u _{0}}

and me as

m_{e} = 1+\frac{D}{f_{e}}

d is distant of distinct vision = 25.0 cm for normal eye

fe =  focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}

\frac{1}{v_{o}} = 2.1 cm

m_{o} = \frac{2.1}{0.150} = 14

m_{e} = 1+\frac{25}{1}

m_{e} =26

m = m_{o}* m_{e}

m = 14*26 = 364

4 0
3 years ago
What is the most potentially destructive way to deal with stress while at college? a. texting friends back home b. watching TV c
diamong [38]

Answer:

Option (C)

Explanation:

Recreational drugs are usually referred to those drugs or medicines that are legally or illegally used without any prescription given by the doctors. These drugs include marijuana, cocaine, alcohol, nicotine, heroin, and many others. These drugs provide relaxation for a shorter period of time, but using of these substances, it makes an individual addicted to drugs.

So, recreational drugs are one of the most potentially destructive ways to handle stress when a student is in college.

Thus, the correct answer is option (C).

7 0
3 years ago
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