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OLga [1]
2 years ago
12

​Hooke's Law. The distance d when a spring is stretched by a hanging object varies directly as the weight w of the object. If th

e distance is 10 cm when the weight is 3 ​kg, what is the distance when the weight is 8 ​kg?
Physics
1 answer:
nadezda [96]2 years ago
7 0

Answer:

distance when the weight is 8 ​kg is 26.66 cm

Explanation:

given data

distance d2 = 10 cm

weight w2 = 3 ​kg

weight w1 = 8 kg

to find out

distance when the weight is 8 ​kg

solution

we consider here distance d1 when weight is 8 kg

so equation will be

d1/d2 = w1/w2

d/10 = 8/ 3

so d = 8/3 × 10

so d = 26.66

distance when the weight is 8 ​kg is 26.66 cm

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A 2.5g copper penny is given a charge of -4.0*10^-9c. how mny excess electrons are on the penny?
Lyrx [107]

Answer : The excess of electrons on the penny are, 2.5\times 10^{10} electrons

Solution : Given,

Total charge = -4.0\times 10^{-9}C

Charge on electron = -1.6\times 10^{-19}C

Formula used :

\text{Excess of electrons}=\frac{\text{Total charge}}{\text{Charge of electron}}

Now put all the given values in this formula, we get the excess of electrons present on the penny.

\text{Excess of electrons}=\frac{\text{Total charge}}{\text{Charge of electron}}=\frac{-4.0\times 10^{-9}C}{-1.6\times 10^{-19}C/e^-}=2.5\times 10^{10}e^-

Therefore, the excess of electrons on the penny are, 2.5\times 10^{10} electrons


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2 years ago
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LUCKY_DIMON [66]

Answer:

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Explanation:

6 0
2 years ago
A parallel-plate capacitor in air has circular plates of radius 3.0 cm separated by 1.1 mm. Charge is flowing onto the upper pla
Yuki888 [10]

Answer:

2 x 10^14 N/Cs

Explanation:

radius, r = 3 cm

Area , A = 3.14 x 3 x 3 = 28.26 cm^2 = 28.26 x 10^-4 m^2

d = 1.1 mm = 1.1 x 10^-3 m

i = 5 A

Let time be t and electric field strength is E.

Charge, q = i x t = 5 t

q = C V

q = C x E x d

5 t = \frac{\varepsilon _{0}A}{d}\times E\times d

E/t = \frac{5}{\varepsilon _{0}A}

E/t = \frac{5}{8.854\times 10^{-12}\times 28.26\times 10^{-4}}

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5 0
3 years ago
The power rating of a 400-ΩΩ resistor is 0.800 W.
sashaice [31]

Answer:

voltage= 17.88volts

current= 0.04 amps

Explanation:

Step one:

given data

resistance R=400 ohms

Power P= 0.8W

a.  What is the maximum voltage that can be applied across this resistor without damaging it?

the expression relating power and voltage is

P=V^2/R

substituting we have

0.8=V^2/400

V^2=0.8*400

V^2=320

V=√320

V=17.88 volts

the maximum voltage is 17.88volts

b.What is the maximum current it can draw?

we know that from Ohm' law

V=IR

17.88=I*400

I=17.88/400

I=0.04amps

3 0
2 years ago
If we add the number in that constant term is there will also be independence between them?(case of refractI've index of medium)
serious [3.7K]

Answer:

Yes

Explanation:

n1 sinθ1 = n2 sinθ2, where θ1 and θ2 are the angles of incidence and refraction.

n=\frac{c}{v}

changing D between them will not affect dependence

8 0
3 years ago
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