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lbvjy [14]
2 years ago
7

How do simple machines make it easier to move an object?

Physics
1 answer:
natima [27]2 years ago
5 0

Answer:

by Multiply Force,change direction and speed multiply

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In a circuit of 2 lamps in parallel if the current in one lamp is 2a the current in the other lamp is?
ioda

In a circuit having 2 lamps are connected in parallel to a battery

then the two lamps will be having the same potential as the battery

i.e

V_{1} = V_{2} = V_{battery}

As per Ohm's law,

I_{1} = \frac{V_{1}}{R_{1}} and I_{2} = \frac{V}{R_{2} }

In other words, each lamp's current is inversely related to its individual resistance. We only know the current in one of the bulbs in this specific instance. We would therefore need further information in order to calculate the current in the other light. Therefore, there isn't enough data to make a statement.

Under the assumption that all physical parameters, including temperature, remain constant, Ohm's law asserts that "the voltage across a conductor is directly proportional to the current flowing through it".

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All you have to do is find the description for a b c and d words and find the meaning 8 9 10 11 and 12
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A steady electric current flows through a wire. If 9.0 C of charge passes a particular spot in the wire in a time period of 2.0
defon

1) Current: 4.5 A

2) Time taken: 4.7 s

Explanation:

1)

The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:

I=\frac{q}{t}

where

I is the current

q is the amount of charge passing a given point in a time t

For the wire in this problem, we have

q = 9.0 C is the amount of charge

t = 2.0 s is the time interval

Solving for I, we find the current:

I=\frac{9.0}{2.0}=4.5 A

2)

To solve this problem, we can use again the same formula

I=\frac{q}{t}

where

I is the current

q is the amount of charge passing a given point in a time t

In this problem, we have:

I = 3.0 A (current)

q = 14.0 C (charge)

Therefore, the time taken for the charge to move past a particular spot in the wire is

t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s

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