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Ahat [919]
2 years ago
11

Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a p slope at constant speed, as sho

wn below. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction as the sled moves 30.0 m along the hill
Physics
1 answer:
kondor19780726 [428]2 years ago
5 0

The work done by friction to move the sled is  - 1,323 J.

<h3>What is Coefficient of friction?</h3>
  • The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them.
  • Typically, it is represented by the Greek letter µ. In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
  • The coefficient of friction has no dimensions because both F and N are measured in units of force (such as newtons or pounds). For both static and kinetic friction, the coefficient of friction has a range of values.
  • When an object experiences static friction, the frictional force resists any applied force, causing the object to stay at rest until the static frictional force is removed. The frictional force opposes an object's motion in kinetic friction.

Solution:

Given that

Coefficient of friction (µ) = 0.10

Mass (m) = 90kg

distance covered (d) = 30m

We use the formula:

friction work = -µmgdcos∅

friction work = -0.100 × 90 kg × 9.8 m/s² × 30 m × cos 60°

friction work = - 1,323 J

Know more about Coefficient of friction numerical brainly.com/question/19308401

#SPJ4

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yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

\Delta V_{max} = 240 V

frequency, f = 50Hz

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Solution:

To calculate the parameters of the given circuit series RLC circuit:

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b). The capacitive reactance,  X_{C} is given by:

\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega

X_{C} = 0.636\Omega

c). Impedance, Z = \frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega

Z = 2400\Omega

d). Resistance, R is given by:

Z = \sqrt {R^{2} + (X_{L} - X_{C})}

2400^{2} = R^{2} + (47.12 - 0.636)^{2}

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e). Phase angle between current and the generator voltage is given by:

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\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})

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5 0
4 years ago
A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

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KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = \frac{1}{2}mv^{2} + 0

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Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

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Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

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