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scoundrel [369]
3 years ago
12

You construct a circuit containing some component C, along with other circuit elements. You want to simultaneously measure the c

urrent through C and the voltage across C while it remains connected to the rest of the circuit. Indicate where you would connect an ammeter (A) and a voltmeter (V) to perform these measurements. Fill any remaining gaps with wires as needed to complete the connection with the rest of the circuit.

Physics
1 answer:
astra-53 [7]3 years ago
6 0

Answer:

The position of the ammeter is in series down the component. While the voltmeter must be connected in parallel. Its position is the gap through the component C.

The positions of one or the other are indicated in the image.

Explanation:

An ammeter is defined as a device used to measure current. Its unit is the ampere. While a voltmeter is used in the measurement of the potential difference between two points. Its unit is the volt. The ammeter must be connected in series with the point at which the current is to be measured, while the voltmeter must be connected in parallel.

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Answer:

52 mm/s (approximately)

Explanation:

Given:

Initial speed of the projectile is, u=60.0\ mm/s

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In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.

So, the velocity in the horizontal direction always remains the same.

The horizontal component of initial velocity is given as:

u_x=u\cos\theta\\u_x=60\times \cos(30)\\u_x=30\sqrt3\approx52\ mm/s

Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.

Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.

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Answer:

Two point charges are separated by 6.4 cm . The attractive force between them is 10 N .

units.

Explanation:

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A point charge +2Q is at the origin and a point charge −Q is located along the x axis at x = d as in the figure below. Find a sy
Akimi4 [234]

Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis  

F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}

Explanation:

Let the force on +Q charge y-axis due to +2Q charge be F_1 and force on +Q charge y axis due to -Q charge on x-axis be F_2.

Distance between the +2Q charge and +Q charge = d units

Distance between the -Q charge and +Q charge = \sqrt{2}d units

k_e= Coulomb constant

F_1=k_e\frac{(+2Q)(+Q)}{d^2}=k_e\frac{+2Q^2}{d^2} N

F_2=k_e\frac{(-Q)(+Q)}{(\sqrt{2}d)^2}=k_e\frac{-Q^2}{2d^2} N

Net force on +Q charge on y-axis is:

F_x=F_2sin 45^o=k_e\frac{-Q^2}{2d^2}\times \frac{1}{\sqrt{2}} N

F_y=F_1-F_2cos45^o

F_y=(F_1-F_2cos45^o)=(k_e\frac{+2Q^2}{d^2})-(k_e\frac{-Q^2}{2d^2}\frac{1}{\sqrt{2}})

F_N=\sqrt{F_x^2+F_y^2}

|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|

The net froce on the +Q charge on y-axis is

F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}

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3 years ago
Read 2 more answers
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