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daser333 [38]
3 years ago
8

A circus performer stretches a tightrope between two towers. He strikes one end of the rope and sends a wave along it toward the

other tower. He notes that it takes the wave 0.775 s to reach the opposite tower, 20.0 m away. If a 1 meter length of the rope has a mass of 0.300 kg, find the tension in the tightrope. N
Physics
1 answer:
IgorC [24]3 years ago
5 0

Explanation:

The given data is as follows.

      Distance (s) = 20 m

       time (t) = 0.775 s

Also, it is given that mass per 1 meter length (m) = 0.300 kg

Formula to calculate the velocity is as follows.

        Velocity (v) = \frac{s}{t}

Putting the given values into the above formula as follows.

              v = \frac{s}{t}

                 = \frac{20 m}{0.775 s}

                 = 25.80 m/s

We know that,

              v = \sqrt{\frac{T}{m}}

Taking square on both the sides, the formula will become as follows.

                      T = mv^{2}

                         = 0.3 kg \times 25.80          

                         = 199.692 N

Therefore, we can conclude that tension in the given tightrope is equal to 199.692 N .

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Evgen [1.6K]

Answer:

Circuit one will have more current than circuit two

Explanation:

I am assuming that you have to see which circuit has the greater current in this case. Well, this is the perfect example of Ohm's Law, which states the following -

V = IR,

where V = voltage / potential difference, I = current, and R = resistance

If one circuit has twice the voltage and half the resistance of the second circuit, as voltage is directly proportional to the resistance -

2V = I( 1 / 2R ),

4V = IR,

I = 4V / R

Whereas in the second circuit -

V = IR,

I = V / R

As you can note, voltage is directly proportional to the current ( I ) as well as the resistance. The only difference between the two formulas I = 4V / R, and I = V / R is the difference in the voltage. With the voltage being 4 times greater in the first circuit, and current is 4 times greater in the first circuit as well.

<u><em>Hence, circuit one will have more current than circuit two</em></u>

5 0
3 years ago
Read 2 more answers
Marks: 1
____ [38]

Answer:

W = 5W

Explanation:

1W = 1J/s

W = Watts

J = Joules

s = seconds

W = J/s

W = 300J/60s

W = 5W

6 0
2 years ago
A simple pendulum 2 m long swings through a maximum angle of 30 ? with the vertical.
Alexus [3.1K]
The complete question was calculate the period T assuming the smallest amplitude.
Using the equation;
T = 2 π√(L/g)
Where T is the period in seconds, L is the length of the rod or wire in meters and g is the acceleration due to gravity. 
Hence; T = 2×3.14 × √(2/9.81)
                = 6.28 × 0.4515
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6 0
3 years ago
Please help me with this question​
Alexandra [31]

Answer:

(4) A = 3 A, A₂ = 11 A

(5) 7 A

Explanation:

(4)

From the diagram,

A = 3+6+2

A = 11 A

V = A₂R

A₂ = V/R₂............ Equation 1

Given: V = 12 V,  R₂ = 4 Ω

Substitute these values into equation 1

A₂ = 12/4

A₂ = 3 A

(5) Applying,

V = IR'

I = V/R'............ Equation 1

Where V = Voltage, I = cuurent, R' = total resistance.

But,

1/R' = (1/3)+(1/4)

1/R' = (3+4)/12

1/R' = 7/12

R' = 12/7 Ω

Given: V = 12 V

Substitute these values into equation 1

I = 12/(12/7)

I = 7 A

Therefore

A = 7 A

5 0
2 years ago
A positron is an elementary particle identical to an electron except that its charge is . An electron and a positron can rotate
denis23 [38]

Explanation:

According to the energy conservation,

          F_{centripetal} = F_{electric}

            \frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}

           v^{2} = \frac{kq^{2}r}{d^{2}m}

                 = \frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}

                = 8.430 \times 10^{10} m^{2}/s^{2}

             v = \sqrt{8.430 \times 10^{10} m^{2}/s^{2}}

                = 2.903 \times 10^{5} m/s

Formula for distance from the orbit is as follows.

               S = 2 \pi r

                  = 2 \times 3.14 \times 0.75 \times 10^{-9} m

                  = 4.71 \times 10^{-9} m

Now, relation between time and distance is as follows.

                T = \frac{S}{v}

       \frac{1}{f} = \frac{S}{v}

or,           f = \frac{v}{S}          

                = \frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}      

                = 6.164 \times 10^{13} Hz

Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is 6.164 \times 10^{13} Hz.

7 0
3 years ago
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