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PtichkaEL [24]
3 years ago
6

A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle θ = 30°. The sph

ere has mass M = 8 kg and radius R = 0.19 m . The coefficient of static friction between the sphere and the plane is μ = 0.64. What is the magnitude of the frictional force on the sphere?
Physics
1 answer:
ale4655 [162]3 years ago
7 0

Answer:

40N

Explanation:

frictional force = mgsin©

frictional force = 8*10*sin30

frictional force = 8*10*0.5= 40N//

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Answer: anlien, enemy gnome, spaceship

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You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resul
photoshop1234 [79]
1) First of all, let's find the resistance of the wire by using Ohm's law:
V=IR
where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have:
R= \frac{V}{I}= \frac{5.70 V}{17.6 A}=0.32 \Omega

2) Now that we have the value of the resistance, we can find the resistivity of the wire \rho by using the following relationship:
\rho =  \frac{RA}{L}
Where A is the cross-sectional area of the wire and L its length.
We already have its length L=2.90 m, while we need to calculate the area A starting from the radius:
A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2

And now we can find the resistivity:
\rho =  \frac{RA}{L}= \frac{(0.32 \Omega)(1.34 \cdot 10^{-6}m^2)}{2.90m}=  1.48 \cdot 10^{-7}\Omega \cdot m
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Is ice a fluid at room temperature?
Margarita [4]

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Explanation:

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Four velcro-lined air-hockey disks collide with each other in a perfectly
Reil [10]

Answer:

The magnitude of the final velocity is approximately 0.526 m/s in approximately the direction of 8.746° East of South

Explanation:

The given collision parameters are;

The kind of collision experienced by the four velcro-lined air-hockey disk = Inelastic collision

The mass of the first disk, m₁ = 50.0 g

The velocity of the first disk, v₁ = 0.80 m/s West = -0.8·i

The mass of the second disk, m₂ = 60.0 g

The velocity of the second disk, v₂ = 2.50 m/s North = 2.5·j

The mass of the third disk, m₃ = 100.0 g

The velocity of the third disk, v₃ = 0.20 m/s East = 0.20·i

The mass of the fourth disk, m₄ = 40.0 g

The velocity of the fourth disk, v₄ = 0.50 m/s South = -0.50·j

Therefore, the total initial momentum of the four velcro-lined air-hockey disk, \Sigma P_{initial} is given as follows;

\Sigma P_{initial} = m₁·v₁ + m₂·v₂ + m₃·v₃ + m₄·v₄ = 50.0×(-0.80·i) + 60.0×(2.50·j) + 100 × (0.20·i) + 40.0 × (-0.50·j)

∴ \Sigma P_{initial} = -40·i + 150·j + 20·i - 20·j = -20·i + 130·j

∴ \Sigma P_{initial} = -20·i + 130·j

By the law of conservation of linear momentum, we have;

\Sigma P_{initial} = \Sigma P _{final} = -20·i + 130·j

Therefore, given that the collision is perfectly inelastic, the disks move as one after the collision and the four masses are added to form one mass, "m", m = m₁ + m₂ + m₃ + m₄ = 50.0 + 60.0 + 100.0 + 40.0 = 250.0

∴ m = 250.0 g

Let, "v" represent the final velocity of the four disks moving as one after the collision

We have;

\Sigma P _{final} = m × v = 250.0 × v = -20·i + 130·j

∴ v = -20·i/250 + 130·j/250 = -0.08·i + 0.52·j

The final velocity of the four disks after collision, v = -0.08·i + 0.52·j

The magnitude of the final velocity, \left | v \right | = √((-0.08)² + (0.52)²) ≈ 0.526

\left | v \right | ≈ 0.526 m/s

The direction of the final velocity, θ = arctan(0.52/(-0.08)) ≈ -81.254°

The direction of the final velocity, θ ≈ -81.254° which is 8.746° East of South

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