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Genrish500 [490]
3 years ago
14

When a voltage difference is applied to a piece of metal wire, a 10.0 mA current flows through it. If this metal wire is now rep

laced with a silver wire having twice the diameter of the original wire, how much current will flow through the silver wire?
Physics
1 answer:
Mrrafil [7]3 years ago
6 0

Answer:

10 mA

Explanation:

We have given when a potential difference is applied then a current of 10 mA is flows so current i = 10 mA

According to ohms law voltage V =i R

And resistance is given as R= \frac{\rho l}{A}

From the relation it is clear that resistance is inversely proportional to the area

As in new case the wire have twice diameter means 4 times the area so the resistance will decrease by 4 times from the previous resistance

As current i=\frac{V}{R} current is inversely proportional to resistance so current will become 4 times of previous value

So current = 10×4=40 mA

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Gravitational force depends on the mass of an object.
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Answer:

sun, earth, capitol building, human, atom

Explanation:

Gravity is directly proportional to the size of an object. Therefore, the object with the most mass will have the greatest force of gravity. We know this from the equation. F = G\frac{m_1m_2}{r^2}

4 0
3 years ago
It took a student 30 minutes to drive from his home to campus on
Gennadij [26K]

Answer:

48 i believe

Explanation:

3 0
4 years ago
The gage pressure in a liquid at a depth of 3 m is read to be 39 kPa. Determine the gage pressure in the same liquid at a depth
ioda

Answer: 117 kPa

Explanation:

For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa

For the liquid at depth 9m, the gauge pressure is equal to= P₂

Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.

So, For finding gauge pressure we have formula P= ρ * g * h

Also gravity also remains same for both liquids

So taking ratio of their respective pressures we have

\frac{P_{1} }{\\P_2}= \frac{density * g * h_1}{density * g * h_2}

So \frac{39}{P_2}= \frac{3}{9}

Or P₂= 39 * 3 = 117 kPa

5 0
3 years ago
Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t
stepan [7]
We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
F=ma
where F=-50 N is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
a= \frac{v_f^2-v_i^2}{2d}
where
v_f=0 is the final speed of the sled
v_i=5 m/s is the initial speed
d is the distance covered

By rearranging the equation, we find d:
d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m
3 0
3 years ago
List five examples from daily life in which you see periodic motion caused by a spring or cord.
dmitriy555 [2]

car suspension

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clips for hanging clothes

keyboards buttons

elevators

4 0
3 years ago
Read 2 more answers
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