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Rudiy27
3 years ago
15

A proton at location A makes an electricfield E-->1 atlocation B. A different proton, placed atlocation B, experiences a forc

e F--->1.
If |E-->|= 300 N/C, what is |F-->1|?
|F-->1|= N
Now the proton at B is removed and replaced by alithium nucleus, containing three protons and four neutrons. Theproton at location A remains in place.
What is the magnitude of the electric force on the lithium nucleus?
|F-->Li| = N
Now the Lithium nucleus is removed, and an electron is placed atlocation B. The proton atlocation A remains in place. What is the magnitude of the electric force on the electron?
|F-->1| = N
Physics
1 answer:
luda_lava [24]3 years ago
7 0

Answer:

Explanation:

 |F-->1| =  E-->1 x charge on proton

= 300 x 1.6 x 10⁻¹⁹

Magnitude of electric force on proton

|F-->1| = 480 x 10⁻¹⁹ N .

Inside lithium nucleus , there are three protons so charge on it

= 3 x 1.6 x 10⁻¹⁹ = 4.8 x 10⁻¹⁹ C .

Magnitude of electric force on lithium nucleus

|F-->Li|  =   E-->1 x charge on lithium nucleus

= 300 x  4.8 x 10⁻¹⁹

= 1440 x 10⁻¹⁹ N .

Electron has same charge as that possessed by proton so charge on electron = 1.6 x 10⁻¹⁹ C

Magnitude of force on electron

|F-->1| =  E-->1 x charge on electron

= 300 x 1.6 x 10⁻¹⁹

=  480 x 10⁻¹⁹ N .

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Answer:

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Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

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

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

<u>Wn = 9.14 x 10¹⁷ N</u>

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You are given a parallel plate capacitor that has plates of area 29 cm2 which are separated by 0.0100 mm of nylon (dielectric co
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Answer:

20.60 kV

Explanation:

Capacitance of parallel plates without dielectric between them is:

C=\frac{\varepsilon_{0}A}{d}

with d the distance between the plates, A the area of the plates and ε₀ the constant 8.85419\times10^{-12}\frac{C^{2}}{Nm^{2}}, so :

C_0=\frac{(8.85419\times10^{-12})(0.0029)}{0.0100\times10^{-3}}=2.57\times10^{-9} F

But the dielectric constant is defined as:

k=\frac{C}{C_{0}}

With C the effective capacitance (with the dielectric) and Co the original capacitance (without the dielectric). So, the new capacitance is:

C=kC_0

But capacitance is related with voltage by:

C=\frac{Q}{V}

with Q the charge and V the voltage, using the new capacitance and solving for V:

kC_0=\frac{Q}{V}

V=\frac{Q}{kC_0}=\frac{0.18\times10{-3}}{(3.4)(2.57\times10^{-9})}=20599.68V=20.60 kV

4 0
3 years ago
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The magnetic flux through each turn of a 110-turn coil is given by ΦB = 9.75 ✕ 10−3 sin(ωt), where ω is the angular speed of the
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Answer:

Explanation:

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N = 110 turns

And the flux is given as function of t

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ω = 8.70 ✕ 10² rev/min

Converting this to rad/sec

1 rev = 2πrad

Then,

ω = 8.7 × 10² × 2π / 60

ω = 91.11 rad/s

Now, we want to find the induced EMF as a function of time

EMF is given as

ε = —NdΦB/dt

ΦB = 9.75 ✕ 10^-3 sin(ωt),

dΦB/dt = 9.75 × 10^-3•ω Cos(ωt)

So,

ε = —NdΦB/dt

ε = —110 × 9.75 × 10^-3•ω Cos(ωt)

Since ω = 91.11 rad/s

ε = —110 × 9.75 × 10^-3 ×91.11 Cos(91.11t)

ε = —97.71 Cos(91.11t)

The EMF as a function of time is

ε = —97.71 Cos(91.11t)

Extra

The maximum EMF will be when Cos(91.11t) = -1

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Answer:

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Explanation:

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