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miv72 [106K]
3 years ago
14

You want to push a 71-kg box up a 16° ramp. The coefficient of kinetic friction between the ramp and the box is 0.23. With what

magnitude force parallel to the ramp should you push on the box so that it moves up the ramp at a constant speed?
Physics
1 answer:
ycow [4]3 years ago
8 0

Answer:

<em>345.60N</em>

Explanation:

The magnitude of the force parallel to the ramp that will push the box up the ramp are the frictional force(Ff) and the moving force Fm

Total force = Ff + Fm

Ff = ηR

Ff = ηmgcosθ

Fm = Wsinθ

Fm = mgsinθ

Total force = ηmgcosθ + mgsinθ

Total force = mg(ηcosθ + sinθ)

Given

mass m = 71kg

acceleration due to gravity g = 9.8m/s²

angle of inclination θ = 16°

coefficient of friction η = 0.23

Substitute into the formula;

Total force = 71(9.8)(0.23cos16 + sin16)

Total force = 695.8(0.2211+0.2756)

Total force = 695.8(0.4967)

Total force = 345.60N

<em>Hence the magnitude force parallel to the ramp that you should push on the box so that it moves up the ramp at a constant speed is 345.60N</em>

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4 0
3 years ago
The potential difference between two parallel conducting plates in vacuum is 165 V. An alpha particle with mass of 6.50×10-27 kg
shepuryov [24]

Answer:

kinetic energy (K.E) = 5.28 ×10⁻¹⁷            

Explanation:

Given:

Mass of  α particle (m) = 6.50 × 10⁻²⁷ kg

Charge of  α particle (q) = 3.20 × 10⁻¹⁹ C

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kinetic energy (K.E)

Computation:

kinetic energy (K.E) = (ΔV)(q)

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4 0
3 years ago
A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back
Kobotan [32]

Answer:

a) 5 m/s

b) 17.8542 m/s

c) 24.7212 m/s

0.229

Explanation:

t = Time taken

u = Initial velocity = 5 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow 0=5-9.81\times t\\\Rightarrow \frac{-5}{-9.81}=t\\\Rightarrow t=0.51 \s

s=ut+\frac{1}{2}at^2\\\Rightarrow s=5\times 0.51+\frac{1}{2}\times -9.81\times 0.51^2\\\Rightarrow s=1.27\ m

So, the stone would travel 1.27 m up

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.27+0^2}\\\Rightarrow v=5\ m/s

Velocity as the rock passes through the original point is 5 m/s

s=ut+\frac{1}{2}at^2\\\Rightarrow 1.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.27\times 2}{9.81}}\\\Rightarrow t=0.51\ s

Time taken to reach the original point is 0.51+0.51 = 1.02 seconds

So, total height of the rock would fall is 30+1.27 = 31.27 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 16.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{16.27\times 2}{9.81}}\\\Rightarrow t=1.82\ s

Time taken by the stone to reach 15 m above the ground is 1.82+0.51 = 2.33 seconds

v=u+at\\\Rightarrow v=0+9.81\times 1.82\\\Rightarrow v=17.8542\ m/s

Speed of the ball at 15 m above the ground is 17.8542 m/s

s=ut+\frac{1}{2}at^2\\\Rightarrow 31.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{31.27\times 2}{9.81}}\\\Rightarrow t=2.52\ s

v=u+at\\\Rightarrow v=0+9.81\times 2.52\\\Rightarrow v=24.7212\ m/s

Speed of the stone just before it hits the street is 24.7212 m/s

F = Force

m = Mass = 100 kg

g = Acceleration due to gravity = 9.81 m/s²

s = Displacement = 4 km = 4000 m

P = Power = 1 hp = 745.7 Watt

t = Time taken = 20 minutes = 1200 seconds

μ = Coefficient of sliding friction

F = μ×m×g

⇒F = μ×100×9.81

W = Work done = F×s

P = Work done / Time

⇒P = F×s / t

745.7=\frac{\mu \times 981\times 4000}{1200}\\\Rightarrow \mu=\frac{747.5\times 1200}{981\times 4000}\\\Rightarrow \mu=0.229

Coefficient of sliding friction is 0.229

8 0
3 years ago
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