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worty [1.4K]
2 years ago
15

5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be

changed to raise the wave speed to 180 m/s?
Physics
1 answer:
ANTONII [103]2 years ago
3 0

Answer:

The tension on string when the speed was raised is 134.53 N

Explanation:

Given;

Tension on the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The speed of the wave is given as;

v = \sqrt{\frac{T}{\mu} }

where;

μ is mass per unit length

v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2}  = \frac{T_2}{v_2^2}

The final tension T₂ will be calculated as;

T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N

Therefore, the tension on string when the speed was raised is 134.53 N

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Answer:

D. has no overall force acting on it.

Explanation:

Why?

Because in a straight line at the constant speed means the car moving in the same velocity, which is not acceleration neither deceleration, and it cannot be on a downhill slope. So the correct answer is

<h3>→ D. has no overall force acting on it.</h3>
8 0
3 years ago
What is a element that tends to be shiny, easy shaped, and a good conductor of heat?
ozzi
The answer to your question is Metal
8 0
2 years ago
How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example
MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the (velocity)^2 and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

\text { Kinetic Energy }=\frac{1}{2} m v^{2}

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}

\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

4 0
2 years ago
The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,
kondaur [170]

Answer:

a= -2\ m/s^2

a=-12.5\ m/s^2

x=2.17 m

x=8.4 m

Explanation:

Given that

v=A+Bt^{-1}

v=2+0.25t^{-1}

To find acceleration :

we know that

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=0-0.5t^{-2}

a=-0.5t^{-2}

Acceleration at t= 2 s

a=-0.5\times 2^{-2}

a= -2\ m/s^2

Acceleration at t= 5 s

a=-0.5\times 5^{-2}

a=-12.5\ m/s^2

We know that

v=\dfrac{dx}{dt}

dx=\left(2+\dfrac{1}{4t}\right)dt

Position at t= 2 s:

\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{2}

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{5}

x=8+0.25 ln5

x=8.4 m

4 0
2 years ago
Leah is making a chart to compare and contrast renewable and non-renewable resources. Which of the following would best complete
Sedbober [7]

Answer:

d. Can be replaced at the same rate or faster than they are used

Explanation:

I just took the test, and its the right answer. Your welcome ✌

6 0
2 years ago
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