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mote1985 [20]
2 years ago
9

A rare and valuable antique chest is being moved into a truck using a 4.00 m long ramp. the kj weight of the chest plus packing

material is 1,500 n. if the truck bed is 1.00 m above the ground, find the work done by the movers as they slide the chest up the ramp if the coefficient of friction between the chest and the ramp is 0.200.
Physics
1 answer:
Lisa [10]2 years ago
7 0

First let us calculate for the angle of inclination using the sin function,

sin θ = 1 m / 4 m

θ = 14.48°

 

Then we calculate the work done by the movers using the formula:

W = Fnet * d

 

So we must calculate for the value of Fnet first. Fnet is force due to weight minus the frictional force.

Fnet = m g sinθ – μ m g cosθ

Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48

Fnet = 84.526 N

 

So the work exerted is equal to:

W = 84.526 N * 4 m

<span>W = 338.10 J</span>

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Answer:

Length of copper wire, l = 785 meters

Explanation:

Given that,

Resistance of the copper wire, R = 0.172 ohms

Area of cross section, A=7.85\times 10^{-5}\ m^2

Resistivity of copper, \rho=1.72\times 10^{-8}\ \Omega-m

The resistance of a wire is given by :

R=\rho\dfrac{l}{A}

l=\dfrac{RA}{\rho}

l=\dfrac{0.172\ \Omega\times 7.85\times 10^{-5}\ m^2}{1.72\times 10^{-8}\ \Omega-m}

l = 785 meters

So, the length of the copper wire is 785 meters. Hence, this is the required solution.

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answers Collision derivation problem. If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the in
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Answer:

4.8967m

Explanation:

Given the following data;

M = 0.2kg

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Firstly, we use conservation of momentum to find the velocity

Therefore, ∆p = MV

0.58kgm/s = 0.2V

V = 0.58/2

V = 2.9m/s

Then, we can use the conservation of energy to solve for maximum height the car can go

E(i) = E(f)

1/2mV² = mgh

Mass cancels out

1/2V² = gh

h = 1/2V²/g = V²/2g

h = (2.9)²/2(9.8)

h = 8.41/19.6 = 0.429m

Since we have gotten the heigh, the next thing is to solve for actual slant of the ramp and initial displacement using similar triangles.

h/w = 0.429/x

X = 0.429×75/12

X = 2.6815

Therefore, by Pythagoreans rule

S(ramp) = √2.68125²+0.429²

S(ramp) = 2.64671

Finally, S(t) = S(ramp) + S(i)

= 2.64671+2.25

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A 92-kg skier is sliding down a ski slope that makes an angle of 30 degrees above the horizontal direction. The coefficient of k
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Answer:

a = 4.05 m/s²

Explanation:

Known data

m= 92 kg  : mass of the  skier

θ =30°  :angle θ of the ski slope  with respect to the horizontal direction

μk= 0.10 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the skier

W: Weight of the skier : In vertical direction

N : Normal force : perpendicular to the ski slope

f : Friction force: parallel to the ski slope

Calculated of the W

W= m*g

W=  92kg* 9.8 m/s² = 901,6 N

x-y weight components

Wx= Wsin θ= 901,6 N *sin 30° = 450.8 N

Wy= Wcos θ = 901,6 N *cos 30° =780.8 N

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We apply the formula (1)

∑Fy = m*ay    ay = 0

N - Wy = 0

N = Wy

N = 780.8 N

Calculated of the f

f = μk* N=  0.10*780.8 N  

f = 78.08 N

We apply the formula (1) to calculated acceleration of the skier:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx - f = m*a

450.8- 78.08 = ( 92)*a

372.72 =  (92)*a

a = (372.72)/ (92)

a = 4.05 m/s²

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