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ddd [48]
3 years ago
7

A string of mass 60.0 g and length 2.0 m is fixed at both ends and with 500 N in tension. a. If a wave is sent along this string

, what will be the wave's speed? A second wave is sent in the string, what is the new speed of each of the two waves?
Physics
1 answer:
Darya [45]3 years ago
8 0

Answer:

a

The  speed of  wave is   v_1  = 129.1 \ m/s

b

The new speed of the two waves is v =  129.1 \ m/s

Explanation:

From the question we are told that

    The mass of the string is  m  =  60 \ g  =  60 *10^{-3} \ kg

    The length is  l  =  2.0 \ m

    The tension is  T  = 500 \ N

Now the velocity of the first wave is mathematically represented as

     v_1  = \sqrt{ \frac{T}{\mu} }

Where  \mu is the linear density which is mathematically represented as

      \mu  =  \frac{m}{l}

substituting values    

     \mu  =  \frac{ 60 *10^{-3}}{2.0 }

     \mu  =  0.03\ kg/m

So

   v_1  = \sqrt{ \frac{500}{0.03} }

   v_1  = 129.1 \ m/s

Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )

     

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2) On Mars, the value of the gravitational acceleration is different:g=3.7 m/s^2. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus: 
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<span>6) On Earth, the gravity acceleration is </span>g=9.81 m/s^2<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is 
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5 0
2 years ago
4025.05m +20.0m +0.050004m
PIT_PIT [208]
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Determine the thrust that a boat with a volume of 1.2m³ receives when it is stranded at sea. The density of seawater is 1020kg /
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Answer:

The maximum possible up-thrust on the boat is 11,995.2 N

Explanation:

According to Archimedes' principle, the thrust received by an object immersed a fluid is equal to the weight of the fluid displaced;

The given parameter of the boat in sea water are;

The volume of the boat = 1.2 m³

The density of seawater = 1020 kg/m³

Density = Mass/Volume

Therefore, Mass = Density × Volume

The maximum volume of water that the boat displaces = 1.2 m³

The mass of the water displaced by the boat = (Density of seawater) × (Volume of seawater displaced)

∴ The maximum possible mass of the water displaced by the boat = 1.2 m³ × 1020 kg/m³ = 1224 kg

The maximum possible mass of the water displaced by the boat, m = 1224 kg

Weight = Mass, m × g

Where;

g = The acceleration due to gravity = 9.8 m/s²

The up-thrust on the boat = The weight of the seawater displaced

∴ The maximum possible up-thrust on the boat = m × g = 1224 kg × 9.8 m/s² = 11,995.2 N

The maximum possible up-thrust on the boat = 11,995.2 N.

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Answer:

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