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sergeinik [125]
3 years ago
8

A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in t

he wires is the same, and they have the same mass. A transverse wave travels on the shorter wire with a speed of 249 m/s. What would be the speed of the wave on the longer wire?
Physics
1 answer:
Margarita [4]3 years ago
3 0

Answer: 996m/s

Explanation:

Formula for calculating velocity of wave in a stretched string is

V = √T/M where;

V is the velocity of wave

T is tension

M is the mass per unit length of the wire(m/L)

Since the second wire is twice as far apart as the first, it will be L2 = 2L1

Let V1 and V2 be the speed of the shorter and longer wire respectively

V1 = √T/M1... 1

V2 = √T/M2... 2

Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1

The equations will now become

249 = √T/(m/L1) ... 3

V2 = √T/(m/2L1)... 4

From 3,

249² = TL1/m...5

From 4,

V2²= 2TL1/m... 6

Dividing equation 5 by 6 we have;

249²/V2² = TL1/m×m/2TL1

{249/V2}² = 1/2

249/V2 = (1/2)²

249/V2 = 1/4

V2 = 249×4

V2 = 996m/s

Therefore the speed of the wave on the longer wire is 996m/s

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8 0
3 years ago
2013 Indianapolis 500 champion Tony Kanaan holds his hand out of his IndyCar while driving through still air with standard atmos
Citrus2011 [14]

Answer:

(a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

Explanation:

Given that,

Speed v₁= 60 mph

Speed v₂ = 225 mph

Speed v₃ = 235 mph

(a). We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{1}=P+\dfrac{1}{2}\rho v^2+\rho gh

there, no vertical movement

So, on neglect of height term

P_{1}=P+\dfrac{1}{2}\rho v_{1}^2

Where, P= atmospheric pressure

\rho = air density

v = speed

Put the value in the equation

P_{1}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(60\times1.4667)^2)

P_{1}=2126.0\ lb/ft^2

P_{1}=\dfrac{2126.0}{144}

P_{1}= 14.76\ psi

(b). Speed v₂ = 225 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{2}=P+\dfrac{1}{2}\rho v_{2}^2

Put the value in the equation

P_{2}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(225\times1.4667)^2)

P_{2}=2246.17\ lb/ft^2

P_{2}=\dfrac{2246.17}{144}

P_{2}= 15.59\ psi

(c).  Speed v₃ = 235 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{3}=P+\dfrac{1}{2}\rho v_{3}^2

Put the value in the equation

P_{3}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(235\times1.4667)^2)

P_{3}=2257.93\ lb/ft^2

P_{3}=\dfrac{2257.93}{144}

P_{3}= 15.68\ psi

According to bernoulli's equation,

If the car increases the velocity the the pressure on the surface of the driver's hand increases.

The pressure from P₁ to P₃ are all near the value of one atmosphere.

So, the pressure difference of one atmosphere is not enough to break the driver's hand.

Hence, (a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

5 0
3 years ago
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8 0
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F
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3 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
guajiro [1.7K]

Answer:

f_{e} = 1.7 cm

Explanation:

The magnification of the compound microscope is given by the product of the magnification of each lens

        M = M₀ m_{e}

        M = - L/f₀  25/f_{e}

Where f₀ and f_{e} are the focal lengths of the lens and eyepiece, respectively, all values ​​in centimeters

In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece (f_{e})

         f_{e} = - L / f₀ 25 / M

Let's calculate

        f_{e} = - 16 / 0.6 25 / (-400)

        f_{e} = 1.67 cm

The minus sign in the magnification is because the image is inverted.

          f_{e} = 1.7 cm

6 0
3 years ago
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