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3 years ago

An electron and a proton are separated by a distance of 1.6×10−10m

Physics

1 answer:

lesya [120]3 years ago

4 0

Answer:

Fe = 9 x 10⁻⁹ N

Fg = 3.97 x 10⁻⁴⁶ N

Fe = 2.26 x 10³⁷ Fg

Explanation:

First we find electric force by Coulomb's Law as follows:

Fe = kq₁q₂/r²

where,

Fe = electric force = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = q₂ = charges on electron and proton = 1.6 x 10⁻¹⁹ C

r = distance between electron and proton = 1.6 x 10⁻¹⁰ m

Therefore,

Fe = (9 x 10⁹ N.m²/C²)(1.6 x 10⁻¹⁹ C)(1.6 x 10⁻¹⁹ C)/(1.6 x 10⁻¹⁰ m)²

Fe = 9 x 10⁻⁹ N

Now we find gravitational force by Newton's Law of Gravitation as follows:

Fg = Gm₁m₂/r²

where,

Fg = gravitational force = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of electron = 9.11 x 10⁻³¹ kg

m₂ = mass of proton = 1.673 x 10⁻²⁷ kg

r = distance between electron and proton = 1.6 x 10⁻¹⁰ m

Therefore,

Fg = (6.67 x 10⁻¹¹ N.m²/kg²)(9.11 x 10⁻³¹ kg)(1.673 x 10⁻²⁷ kg)/(1.6 x 10⁻¹⁰ m)²

Fg = 3.97 x 10⁻⁴⁶ N

Dividing both forces:

Fe/Fg = (9 x 10⁻⁹ N)/(3.97 x 10⁻⁴⁶ N)

Fe = 2.26 x 10³⁷ Fg

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The pressure of a box pushes down on the floor is 50 Pa if the box weighs 400 N what is the area of the base of the box

Wewaii [24]

Answer:8m^2

Explanation:

Area=force÷pressure

Area=400÷50

A=8m^2

5 0

3 years ago

Consider a uniform solid sphere of radius R and mass M rolling without slipping. Which form of its kinetic energy is larger, tra

castortr0y [4]

Answer:

A. Its translational kinetic energy is larger than its rotational kinetic energy.

Explanation:

Given that

Radius = R

Mass = M

We know that mass moment of inertia for the solid sphere

$I=\dfrac{2}{5}MR^2$

Lets take angular speed =ω

Linear speed =V

Condition for pure rolling , V= ω R

Rotation energy ,RE

$RE=\dfrac{1}{2}I\omega^2$

$RE=\dfrac{1}{2}\times \dfrac{2}{5}MR^2\times \omega^2$

$RE=\dfrac{1}{5}\times MR^2\times \omega^2$

$RE=\dfrac{1}{5}\times MV^2$

RE= 0.2 MV²

The transnational kinetic energy TE

$TE=\dfrac{1}{2}MV^2$

TE= 0.5 MV²

From above we can say that transnational energy is more than rotational energy.

Therefore the answer is A.

3 0

3 years ago

A 50 kg bumper car with a 40 kg child and it is at rest when a 60 kg child in her own bumper car slams into it the collision las

IrinaK [193]

Answer:

F = 99 v₂₀

v₂₀ = 1 m / s, F = 99 N

Explanation:

In this exercise it is asked to find the force during the collision, for this we use the relationship between the momentum and the momentum of car 1

I = Δp

F t = p_f- p₀

F t = m (v_f -v₀) (1)

We must find the final speed of car 1, for this we define a system formed by the two cars, in this case the forces during the collision are internal and the moment is conserved

initial instant. Before the crash

p₀ = 0 + m₂ v₂₀

final instant. After the crash

p_f = m₁ v₁ + m₂ v_{2f}

the moment is preserved

p₀ = p_f

m₂ v₂₀ = m₁ v_{1f} + m₂ v_{2f} (2)

m₂ (v₂₀ - v_2f}) = m₁ v_{1f}

as the collision is elastic the kinetic energy is also conserved

K₀ = K_f

½ m₂ v₂₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

m₂ (v₂₀² -v_{2f}²) = m₁ v_{1f}²

let's write our system of equations, using

a² - b² = (a + b) (a-b)

m₂ (v₂₀ - v_{2f}) = m₁ v_{1f}

m₂ (v₂₀ -v_{2f}) (v₂₀ + v_{2f}) = m₁ v_{1f}²

to solve we divide the equations

v₂₀ + v_{2f} = v_{1f}

with this we substitute in equation 2 and find the speed of each car, in this case we need the speed of car 1

m₂ v₂₀ = m₁ v_{1f} + m₂ (v_{1f}-v₂₀)

2m₂ v₂₀ = (m₁ + m₂) v_{1f}

v_{1f} = $\frac{2m_2}{m_1+m_2} v_{2o}$

We substitute in the drive ratio of car 1

F t = m (v_f -v₀)

F = m₁ (\frac{2m_2}{m_1+m_2} v_{2o} - 0) / t

F = $\frac{2m_1 m_2 }{m_1+m_2} \ \frac{v_{2o}}{t}$

the mass of each car is the mass of the car plus the mass of the boy

m₁ = 50 +40 = 90 kg

m₂ = 50 +60 = 110 kg

time is t = 1

we substitute the values

F = $\frac{ 2\ 90 \ 110}{90+110} \ \frac{v_{2o}}{1}$ 2 90 100/90 + 110 vo2 / 1

F = 99 v₂₀

The value of the initial velocity of car 2 is not indicated in the problem, if this velocity is known it can be included and the force value is obtained, suppose that the initial velocity v₂₀ = 1 m / s

F = 99 N

4 0

3 years ago

A parallel combination of a 1.13-μF capacitor and a 2.85-μF one is connected in series to a 4.25-μF capacitor. This three-capaci

Nata [24]

Answer:

(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = $\frac{C_{3}C_{4} }{C_{3} + C_{4} }$

Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = $\frac{4.25\times3.98 }{4.25 + 3.98 }$

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3 = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = $\frac{Q}{C_{4} }$ = $\frac{35.46\times10^{-6} }{3.98\times10^{-6}}$ = 8.90 volts