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Tju [1.3M]
3 years ago
5

A box is sitting on a 2 m long board at one end. A worker picks up the board at the end with the box so it makes an angle with t

he ground of 35o. The coefficient of static friction between the box and the board is 0.5 and the coefficient of kinetic friction is 0.3 . A) Will the box slide down the ramp when it is at 35o? Prove it! B) How long will it take the box to slide to the other end of the ramp? (if it slides)
Physics
1 answer:
Vlada [557]3 years ago
3 0

Answer:

Explanation:

When the box is on the ramp , component of its weight along the ramp

= mg sinθ

Friction force acting on it in upward direction

=μ mg cosθ

For sliding

μ mg cosθ < mg sinθ

μ cosθ < sinθ

.5 x cos35 < sin35

.41 < .57

So the box will slide

When sliding starts , kinetic friction acts

Net force in downward direction

mgsinθ - μ mg cosθ

acceleration

= gsinθ - μ g cosθ

= 5.62 - .3 x 9.8 x cos35

= 5.62 - 2.4

= 3.22 m /s²

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net force= the sum of all the forces acting on the object

the frictional force is opposite to the direction of motion of the object

If the object moves to the right (positive) then the friction force moves to the left (negative).

so the net force: 50 N - 5 N = 45 N (B)

4 0
3 years ago
An electric dipole having dipole moment of magnitude p is placed in a uniform electric field having magnitude E. What is the mag
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Answer:

Explanation:

The formula for the potential energy of a dipole placed in an electric field is given by

U = - pE Cos θ

where, θ is the angle between dipole moment and the electric field vector.

For θ = 0°,

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For θ = 180°,

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Change in potential energy

ΔU = Uf - Ui

ΔU = pE - (-pE)

ΔU = 2pE

5 0
4 years ago
At a pressure 43 kPa, the gas in a cylinder has a volume of 12 liters. Assuming temperature remains the same, if the volume of t
Allisa [31]
P = 43 * 10^3
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P' =?
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PV = P'V'
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I know you can do it from here, mate ;)
7 0
3 years ago
A block of mass m1 = 290 g is at rest on a plane that makes an angle θ = 30° above the horizontal. The coefficient of kinetic fr
GrogVix [38]

Answer:

The speed of the block when is fallen 30cm

v=0.726\frac{m}{s}

Explanation:

∑F= (m2)g - ƒ - (m1)g*sin(θ) = (m1)a

g = 9.81 m/s²

ƒ  = μN = μ(m1)g

(m2)*g - u*(m1)*g - (m1)*g*sin(\alpha ) = (m1)*a

(0.200)(9.81) - (0.1)(0.290)(9.81) - (0.290)(9.81)sin(30°) = (0.290)a

(0.200kg)(9.81\frac{m}{s^{2}}) - (0.1)(0.290kg)(9.81\frac{m}{s^{2}}) - (0.290kg)(9.81\frac{m}{s^{2}})sin(30°) = (0.290kg)*a

0.511101=0.29*a\\a=0.879\frac{m}{s^{2} }

v_{f}^{2}=v_{o}^{2}+2*a(x_{f}^{2}-v_{o}^{2})\\v_{o}=0\\v_{o}=0\\v_{f}^{2}=2*a(x_{f})\\v_{f}=\sqrt{2*a(x_{f})}\\v_{f}=\sqrt{2*0.879\frac{m}{s^{2}}*0.30m} \\v_{f}=0.726 \frac{m}{s}

5 0
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