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Trava [24]
2 years ago
11

A small object slides along the frictionless loop-the-loop with a diameter of 3 m. what minimum speed must it have at the top of

the loop in order to remain in contact with the loop
Physics
1 answer:
otez555 [7]2 years ago
4 0
<span>3.834 m/s. In this problem we need to have a centripetal force that is at least as great as the gravitational attraction the object has. The equation for centripetal force is F = mv^2/r and the equation for gravitational attraction is F = ma Since m is the same in both cases, we can cancel it out and then set the equations equal to each other, so a = v^2/r Substitute the known values (radius is diameter/2) and solve for v 9.8 m/s^2 <= v^2/1.5 m 14.7 m^2/s^2 <= v^2 3.834057903 m/s <= v So the minimum velocity needed is 3.834 m/s.</span>
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Hanging by a thread. Two metal spheres hang from nylon threads and attract each other when brought close together. (i) What can
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Answer:

Explained

Explanation:

i)Two spheres hanging from nylon threads attract each other because either the two spheres are charged with opposite sign or only one of the spheres is charge so the other would be charge by induction of the charged sphere and hence attract each other.

ii)However, when they are touched the charges will be rearranged among the two sphere such that the two sphere have exact same magnitude and sign of charge and now they will repel each other or the  magnitude of charges on the two spheres become zero and they neither attract or repel each other.

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2 years ago
When explaining chemical reactions to a friend, Brianna models a reaction by taking several colors of modeling clay and making a
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Answer: synthesis

Explanation:

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2 years ago
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Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1
7nadin3 [17]

Answer:

A). σ = 3.823 x 10^{-5} C^{2}/N-m^{2}

B). \sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C). U=10.322 J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

\sigma =k.\varepsilon _{0}.E

where, E is resultant electric field = 1.2 x 10^{6} V/m

           \varepsilon _{0} is permittivity of free space = 8.85 x 10^{-12} C^{2}/N-m^{2}

           k is dielectric constant = 3.6

∴\sigma =k.\varepsilon _{0}.E

                     = 3.6 x 8.85 x10^{-12} x 1.2 x 10^{6}

                    = 3.823 x 10^{-5} C^{2}/N-m^{2}

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )

\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )

\sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C).

Area of the plate, A = 2.5 cm^{2}

                                 = 2.5 x 10^{-4}m^{2}

diameter of the plate, d = 1.8 mm

                                        = 1800 m

∴ Total energy stored in the capacitor

U=\frac{1}{2}k\varepsilon _{0}E^{2}Ad

U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800

U=10.322 J

4 0
2 years ago
An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini
arsen [322]
Kinetic energy is calculated through the equation,

   KE = 0.5mv²

At initial conditions,

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Due to the momentum balance,

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Substituting the known values,

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   V = 0.2977 m/s

The kinetic energy is,
   KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
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The difference between the kinetic energies is 0.0473 J. 
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The correct option is D. 
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Copernicus model was simpler and some of his claims were correct.<span />
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