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iragen [17]
3 years ago
12

The pointer of an analog meter is connected to a

Physics
2 answers:
dangina [55]3 years ago
8 0
C. coil suspended by bearings. 
<span>but im not 100% sure</span>
Rom4ik [11]3 years ago
5 0

Answer: the correct answer is c coil suspended by bearings.

Explanation: this is so that whenever the coil detect a parameter to be measured say current voltage or resistance it energizes and the bearings moves accordingly to indicate the corresponding reading

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Calculate the acceleration of the car for each set of conditions using the formula a = (v2 – v1) / (t2 – t1) where v2 and v1 are
Arisa [49]

Answer:  1.  0.19

               2.  0.33

               3.  0.47

               4.  0.62

Explanation:  just did it & these are the correct answers

4 0
3 years ago
Read 2 more answers
Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with h
Sophie [7]
I disagree.
Many substances show fluorescence under ULTRAVIOLET light, NOT microwaves. :)
5 0
3 years ago
Read 2 more answers
GIVING BRAINLIEST PLEASE HELP!!
erik [133]

Explanation:

the other 40% is used to power the 60% making it only capable of 60% efficiency

4 0
2 years ago
A small bolt with a mass of 33.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical dire
densk [106]

Answer:

0.027m

Explanation:

the bolt loses contact with the piston only when acceleration due to gravity equals acceleration of piston

ω² * A = g where ω is angular velocity, A amplitude, g acceleration due to gravity

ω is given by 2πf, ω² is 4π²f²

A= g/4π²f² depending on the value of g used either 10m/s² or 9.8m/s²,

i used 10m/s² in this answer

5 0
3 years ago
A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
3 years ago
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