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madam [21]
3 years ago
5

A 0.19 F capacitor is desired. What area must the plates have if they are to be separated by a 2.0 mm air gap? a= ?? m^2

Physics
1 answer:
Aliun [14]3 years ago
4 0
.125 
Hope this answer helps you

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PLEASE HELP!!
IrinaK [193]

Answer:

A

Explanation

i think its a not 100% sure though.

7 0
3 years ago
Orbit around hub and arms of spirals ( with name )
const2013 [10]
 the answer is the Spiral Galaxy
6 0
3 years ago
When a force is applied to an object, no work is being preformed on the object unless the object____.
Temka [501]
No work is being performed on the object unless the object is moving, as work done = force x distance
8 0
3 years ago
g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses
statuscvo [17]

Answer:

v₁f = 0.5714 m/s   (→)

v₂f = 2.5714 m/s   (→)

e = 1  

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i +  ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) +  ((2*6m) / (m + 6m)) * (2)  

v₁f = 0.5714 m/s   (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i +  ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s   (→)

e = - (v₁f - v₂f) / (v₁i - v₂i)   ⇒   e = - (0.5714 - 2.5714) / (4 - 2) = 1  

It was a perfectly elastic collision.

8 0
3 years ago
The charge of an electron is -1.60x10-19 C. A current of 1 A flows in a wire carried by electrons. How many electrons pass throu
faltersainse [42]

Answer: 6.241\times 10^{18} electrons pass through a cross section of the wire each second.

Explanation:

According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

Given : Charge on 1 electron = 1.6\times 10^{-19}C

Charge on 1 mole of electrons = 1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C

To calculate the charge passed we use the equation:

I=\frac{q}{t}

where,

I = current passed = 1 A

q = total charge = ?

t = time required = 1 sec

Putting values in above equation, we get:

1A=\frac{q}{1s}\\\\q=1A\times 1s=1C

When 96500C of electricity is passed , the electrons passed = 6.022\times 10^{23}

1 C of electricity is passed , the electrons passed = \frac{6.022\times 10^{23}}{96500}\times 1C=6.241\times 10^{18}

Hence, 6.241\times 10^{18} electrons pass through a cross section of the wire each second.

4 0
3 years ago
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