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fenix001 [56]
3 years ago
7

When it is brought near a neutral object, a charged object can _________ a charge in a neutral object.A) induce, B) cancel,C) ne

utralize, D) add or subtract
Physics
1 answer:
Bess [88]3 years ago
3 0
The answer is a induce.
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Under metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a
joja [24]

Answer:

In metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a centimeter.

Explanation:

Here we asked to divide 1 meter in to 100 equal parts.

Let us find out what is the length of piece when 1 m is divided in to 100 equal parts.

Length

l =  \frac{1}{100}  = 0.01m

That is length of 1 m divided into 100 equal parts is 0.01m.

We know that 0.01 m is 1 centimeter.

So, in metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a centimeter.

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3 years ago
Why was Mars orbiter sent to mars?​
Dafna11 [192]

Explanation:

to develop the technologies required for designing, planning, management and operations of an interplanetary mission. The secondary objective is to explore Mars' surface features, morphology, mineralogy and Martian atmosphere using indigenous scientific instruments

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3 years ago
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Sandra joined a new company in HR. She believes most of the employees are overburdened and stressed. What action should she unde
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Arrange for a time management training meeting
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3 years ago
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A grandfather clock is controlled by a swinging brass pendulum that is 1.2 m long at a temperature of 27°C. (a) What is the leng
stealth61 [152]

Answer:

L2 = 1.1994 m

the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m

Explanation:

Given;

Initial length L1 = 1.2m

Initial temperature T1 = 27°C

Final temperature T2 = 0.0°C

Linear expansion coefficient of brass x = 1.9 × 10^-5 /°C

The change i length ∆L;

∆L = L2 - L1

L2 = L1 + ∆L ...........1

∆L = xL1(∆T)

∆L = xL1(T2 - T1) ......2

Substituting the given values into equation 2;

∆L = 1.9 × 10^-5 /°C × 1.2m × (0 - 27)

∆L = 1.9 × 10^-5 /°C × 1.2m × (- 27)

∆L = -6.156 × 10^-4 m

From equation 1;

L2 = L1 + ∆L

Substituting the values;

L2 = 1.2 m + (- 6.156 × 10^-4 m)

L2 = 1.2 m - 6.156 × 10^-4 m

L2 = 1.1993844 m

L2 = 1.1994 m

the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m

3 0
3 years ago
A 1.40 mH inductor and a 1.00 µF capacitor are connected in series. The current in the circuit is described by I = 14.0 t, where
4vir4ik [10]

Answer:

Explanation:

Inductance L = 1.4 x 10⁻³ H

Capacitance C = 1 x 10⁻⁶ F

a )

current I = 14 .0 t

dI / dt  = 14

voltage across inductor

= L dI / dt

= 1.4 x 10⁻³ x 14

= 19.6 x 10⁻³ V

= 19.6 mV

It does not depend upon time because it is constant at 19.6 mV.

b )

Voltage across capacitor

V = ∫ dq / C

= 1 / C ∫ I dt  

= 1 / C ∫ 14 t dt

1 / C x 14 t² / 2

= 7 t² / C

= 7 t² / 1 x 10⁻⁶

c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance

energy stored in inductor

= 1/2 L I²

energy stored in capacitor

= 1/2 CV²

After time t

1/2 L I² = 1/2 CV²

L I² =  CV²

L x ( 14 t )² = C x  ( 7 t² / C )²

L x 196 t² = 49 t⁴ / C

t² = CL x 196 / 49

t = 74.8 μ s

After 74.8 μ s energy stored in capacitor exceeds that of inductor.

7 0
3 years ago
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