PLEASE ASAP<br><br> Giving a brainlist

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

Mice21 [21]

Answer:

r = 5.08 m

Explanation:

The electric force of attraction or repulsion is given by :

$F=\dfrac{kq_1q_2}{r^2}$

We need to find how far above the electron would the proton have to be if you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it.

So, the force from the proton is balanced by the mass of the electron.

$\dfrac{kq_pq_e}{r^2}=mg$

r is distance

$r=\sqrt{\dfrac{kq_pq_e}{mg}} \\\\r=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{9.11\times 10^{-31}\times 9.8}} \\\\r=5.08\ m$

So, proton have to be at a distance of 5.08 meters above the electron.

3 0

3 years ago

A 160 g basketball has a 32.7 cm diameter and may be approximated as a thin spherical shell. Starting from rest, how long will i

mafiozo [28]

Answer:

t = 0.24 s

Explanation:

As seen in the attached diagram, we are going to use dynamics to resolve the problem, so we will be using the equations for the translation and the rotation dyamics:

Translation: ΣF = ma

Rotation: ΣM = Iα ; where α = angular acceleration

Because the angular acceleration is equal to the linear acceleration divided by the radius, the rotation equation also can be represented like:

ΣM = I(a/R)

Now we are going to resolve and combine these equations.

For translation: Fx - Ffr = ma

We know that Fx = mgSin27°, so we substitute:

(1) mgSin27° - Ffr = ma

For rotation: (Ffr)(R) = (2/3mR²)(a/R)

The radius cancel each other:

(2) Ffr = 2/3 ma

We substitute equation (2) in equation (1):

mgSin27° - 2/3 ma = ma

mgSin27° = ma + 2/3 ma

The mass gets cancelled:

gSin27° = 5/3 a

a = (3/5)(gSin27°)

a = (3/5)(9.8 m/s²(Sin27°))

a = 2.67 m/s²

If we assume that the acceleration is a constant we can use the next equation to find the velocity:

V = √2ad; where d = 0.327m

V = √2(2.67 m/s²)(0.327m)

V = 1.32 m/s

Because V = d/t

t = d/V

t = 0.327m/1.32 m/s

t = 0.24 s

7 0

3 years ago

A cube of side 6.50 cm is placed in a uniform field E = 7.50 × 10^3 N/C with edges parallel to the field lines (field enters the

Svetllana [295]

Answer:

a) $\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}$ , b) $\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}$ , c) $\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}$

Explanation:

a) The net flux through the cube is:

$\Phi_{net}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}+(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}$

b) The flux through the right face is:

$\Phi_{right}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}$

c) The flux through the left face is:

$\Phi_{left}=(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}$

5 0

3 years ago

Read 2 more answers

Kinetic energy is the energy of ______. It increases with mass ______ and _______ .Potential energy is the energy of _______or__

Alona [7]

Answer:

velocity. height. weight. possition. place. energy. force.

Explanation: 50/50 % chance they are wrong and write.

5 0

3 years ago

4) You have a 10 ohm resistor connected in series with an ammeter. The voltage applied to the whole circuit is 1.2 volts. At the

GuDViN [60]

Answer:

0.6 Ω

Explanation:

As shown in the diagram below,

Since the resistance and the ammeter are connected in series,

(i) The same amount of current flows through them.

(ii) The sum of their individual individual voltage is equal to the total voltage of the circuit.

Applying ohm's law,

V = IR................ Equation 1

Where V = Voltage across the ammeter, I = current flowing through the ammeter, R = resistance of the ammeter.

make R the subject of the equation

R = V/I............... Equation 2

Given: V = 1.2-0.9 = 0.3 V, I = 0.5 A.

Substitute into equation 2

R = 0.3/0.5

R = 0.6 Ω

3 0

3 years ago

Read 2 more answers

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