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lisov135 [29]
3 years ago
7

A small sphere with mass 5.00×10−7 kg and charge +8.00 μC is released from rest a distance of 0.500 m above a large horizontal i

nsulating sheet of charge that has uniform surface charge density σ = +8.00 pC/m2.
Using energy methods, calculate the speed of the sphere when it is 0.100 m above the sheet.
Physics
1 answer:
Damm [24]3 years ago
6 0
<span>Work done on charge is W = Eqd = σ/(2ε₀) x q x d = {(8.00 x 10⁻¹²)/(2 x 8.854187 x 10⁻¹²)} x 3.00 x 10⁻⁶ x (0.650 - 0.250) = 5.42116402J. KE of sphere = 0.5mv² = 0.5 x 5.00 x 10⁻⁷v² = work done by E-field on charge during its fall = 5.42116402→ v = 4657 m/s.</span>
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A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It ta
Tomtit [17]

Answer:

(a) 185 N/m

(b) 3.083 kg

Explanation:

(a)

Using,

E = 1/2ke²....................... Equation 1

Where E = work done to compress the spring, k = spring constant of the spring, e = compression of the spring.

make k the subject of the equation

k = 2E/e²............... Equation 2

Given: E = 3.7 J, e = 0.20 m

Substitute into equation 2

k = 2(3.7)/0.2²

k = 185 N/m.

(b)

Using,

F = ma.............. Equation 2

Where F = force applied to the spring, m = mass attached to the spring, a = acceleration of the spring.

But from hook's law,

F = ke................. Equation 3

substitute equation 3 into equation 2

ke = ma

make m the subject of the equation

m = ke/a................ Equation 4

Given: k = 185 N/m, e = 0.2 m, a = 12 m/s²

Substitute into equation 4

m = 185(0.2)/12

m = 3.083 kg

3 0
3 years ago
Read 2 more answers
an object tied to the end of the string moves in a circle . the force exerted by the string depends on the mass of the object it
otez555 [7]

Answer:Fc=mv^2/r

Explanation:

6 0
3 years ago
What are noble gases?
Bas_tet [7]

Answer:

noble gases are basically a group of gases that are similar in their chemical compounds, theres six of them : helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and the radioactive radon (Rn).

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6 0
3 years ago
A soccer player pumps air into a soccer ball until no more air can be pushed inside. Describe the air inside the soccer ball com
inessss [21]

Answer:

the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.

Explanation:

This exercise asks to describe the inflation situation of a spherical fultball.

Initially the balloon is deflated, therefore the internal pressure is equal to the pressure of the air outside, atmospheric pressure, when it begins to inflate the balloon with a pump this creates a pressure in the inlet valve and as it is greater than the pressure inside, the air enters it, this is repeated in each filling cycle, manual pump.

When the ball is full we have two forces, the one created by the external walls and the one aired by the pressure of the pump, these forces are directed towards the inside, but the air molecules exert a pressure towards the outside, which translates into a force. When these two forces are equal, the pump is no longer able to continue introducing air into the balloon.

Consequently the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.

4 0
3 years ago
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
schepotkina [342]

Answer:

6.77 minutes

Explanation:

172 degree - 78 degree = (185 degree - 78 degree)e−2 k

=> 94 = 107

e−2 k => 94 ÷ 107

k => ln (94÷107) / 2

147 - 78 = (185 - 78)e ^[ln (94÷107) / 2]

=> 69 = 107 e^ [ln (94÷107) / 2]

e^[ln (94÷107) / 2] =69 ÷ 107

=> t = [ln (69 ÷ 107)] ÷ [ln (94÷107) / 2]

t=> -0.4387 ÷ -0.0648

t => 6.77 minutes.

Therefore, the final answer to the question is 6.77 minutes.

4 0
3 years ago
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