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dybincka [34]
2 years ago
11

A 66-kg base runner begins his slide into second base when he is moving at a speed of 3.4 m/s. The coefficient of friction betwe

en his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base(a) How much mechanical energy is lost due to friction acting on the runner. (b) How far does he slide?
Physics
1 answer:
Helen [10]2 years ago
5 0

Answer:

a ) = 381.48 J

b )=  84.25 cm

Explanation:

Kinetic energy of the runner

= 1/2 m v²

= .5 x 66 x 3.4²

= 381.48 J

The final kinetic energy of the runner is zero .

Loss of mechanical energy

= 381.48 J

This  loss in  mechanical energy is due to action of frictional force .

b )

Let s be the distance of slide

deceleration due to frictional force

= μmg/m

.7 x 66 x 9.8 / 66

a = - 6.86 m s⁻¹

v² = u² - 2 a s

0 = 3.4² - 2x6.86 s

s = 3.4² / 2x6.86

= .8425 m

84.25 cm

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Answer:

a) The potential energy in the system is greatest at X.

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U_{grav, X} = U_{grav, Y} + K_{Y} + \Delta W_{X \rightarrow Y} = U_{grav,Z}+\Delta W_{X \rightarrow Z}

Conclusions are showed as follows:

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2 years ago
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Two long, parallel transmission lines, 40.0cm apart, carry 25.0-A and 73.0-A currents.A). Find all locations where the net magne
In-s [12.5K]

Answer:

a) If the currents are in the same direction, the magnetic field is zero at x = 0.298 m = 29.8 cm

That is, in between the wires, 29.8 cm from the 73.0 A wire and 10.2 cm from the 25.0 A wire.

b) If the currents are in opposite directions, the magnetic field is zero at x = 0.608 m = 60.8 cm

That is, along the positive x-axis, 60.8 cm from the 73.0 A wire and 20.8 cm from the 25.0 A wire.

Explanation:

The origin is at the 73.0 A wire and the 25.0 A wire is at x = 0.40 m

The magnetic field in a current carrying wire at a distance r from the wire is given by

B = (μ₀I/2πr)

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

a) If the currents are in the same direction, at what positions is the magnetic field equal to 0.

According to laws describing the direction.of magnetic fields, this position will be at some point between the two wires.

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire points into the plane of the book, moving in the negative x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (0.4 - x)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(0.4-x)]

(73/x) = [25/(0.4-x)]

73(0.4-x) = 25x

29.2 - 73x = 25x

73x + 25x = 29.2

98x = 29.2

x = (29.2/98) = 0.298 m

b) If the currents are in the opposite directions, at what positions is the magnetic field equal to 0?

According to laws describing the direction.of magnetic fields, this position will be at some point beyond the second wire (since we're initially concerned about the positive x-direction).

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire (whose direction is now in the opposite direction to the current in the first wire) is also along the positive x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (x - 0.4)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(x-0.4)]

(73/x) = [25/(x-0.4)]

73(x-0.4) = 25x

73x - 29.2 = 25x

73x - 25x = 29.2

48x = 29.2

x = (29.2/48) = 0.608 m

Hope this Helps!!!

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a. 5.3

should be your answer

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