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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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Answer:

The speed of the resistive force is 42.426 m/s

Explanation:

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mass of skydiver, m = 75 kg

terminal velocity, V_T = 60 \ m/s

The resistive force on the skydiver is known as drag force.

Drag force is directly proportional to square of terminal velocity.

F_D = kV_T^2

Where;

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k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}

When the new drag force is half of the original drag force;

F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}=  \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}=  0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s

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8 0
3 years ago
What is the gravitational relationship between two objects?
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Rudik [331]

Answer:

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b) 0.51 m/s

Explanation:

(a)

M = total mass of wagon, rider and the rock = 93.1 kg

V = initial velocity of wagon = 0.456 m/s

m = mass of the rock = 0.292 kg

v = velocity of rock after throw = 15.4 m/s

V' = velocity of wagon after rock is thrown

Using conservation of momentum

M V = m v + (M - m) V'

(93.1) (0.456) = (0.292) (15.4) + (93.1 - 0.292) V'

V' = 0.41 m/s

b)

M = total mass of wagon, rider and the rock = 93.1 kg

V = initial velocity of wagon = 0.456 m/s

m = mass of the rock = 0.292 kg

v = velocity of rock after throw = - 15.4 m/s

V' = velocity of wagon after rock is thrown

Using conservation of momentum

M V = m v + (M - m) V'

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6 0
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as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.61-T magnetic field. The p
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According to the given condition:

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where,

E = Magnitude of Electric Field = ?

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θ = Angle between speed and magnetic field = 90°

Therefore,

E = (3)(230\ m/s)(0.61\ T)Sin90^o

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Answer:

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