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2 years ago

A student is conducting a physics experiment and rolls four different-

Physics

1 answer:

joja [24]2 years ago

4 0

Answer:

I think the answer is C

Explanation:

Its heavy but not too heavy

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three condensers are connected in series across a 150 volt supply. The voltages across them are 40,50 and 60 volts respectively,

ioda

Explanation:

Given that,

The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.

(a) Capacitance of capacitor 1,

$C_1=\dfrac{Q}{V_1}\\\\C_1=\dfrac{6\times 10^{-8}}{40}\\\\C_1=1.5\times 10^{-9}\ F\\\\C_1=1.5\ nF$

Capacitance of capacitor 2,

$C_2=\dfrac{Q}{V_2}\\\\C_2=\dfrac{6\times 10^{-8}}{50}\\\\C_2=1.2\times 10^{-9}\ F\\\\C_2=1.2\ nF$

Capacitance of capacitor 3,

$C_3=\dfrac{Q}{V_3}\\\\C_3=\dfrac{6\times 10^{-8}}{60}\\\\C_3=1\times 10^{-9}\ F\\\\C_3=1\ nF$

(b) The equivalent capacitance in series combination is :

$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\\\\\dfrac{1}{C}=\dfrac{1}{1.5}+\dfrac{1}{1.2}+\dfrac{1}{1}\\\\C=0.4\ nF$

Hence, this is the required solution.

5 0

3 years ago

Find the speed of light in each of the following materials. (a) gallium phosphide m/s (b) carbon disulfide m/s (c) benzene

Oksanka [162]

Explanation:

We need to calculate the speed of light in each materials

(I). Gallium phosphide,

The index of refraction of Gallium phosphide is 3.50

Using formula of speed of light

$v=\dfrac{c}{\mu}$ ....(I)

Where, $\mu$ = index of refraction

c = speed of light

Put the value into the formula

$v=\dfrac{3\times10^{8}}{3.50}$

$v=8.6\times10^{7}\ m/s$

(II) Carbon disulfide,

The index of refraction of Gallium phosphide is 1.63

Put the value in the equation (I)

$v=\dfrac{3\times10^{8}}{1.63}$

$v=1.8\times10^{8}\ m/s$

(III). Benzene,

The index of refraction of Gallium phosphide is 1.50

Put the value in the equation (I)

$v=\dfrac{3\times10^{8}}{1.50}$

$v=2\times10^{8}\ m/s$

Hence, This is the required solution.

7 0

3 years ago

To store stacks of clean plates, a cafeteria uses a closed cart with a spring-loaded shelf inside. Customers can take plates off

ruslelena [56]

The answer for the following problem is mentioned below.

The option for the question is "A" approximately.

Therefore the elastic potential energy of the string is 20 J.

Explanation:

Given:

Spring constant (k) = 240 N/m

amount of the compression (x) = 0.40 m

To calculate:

Elastic potential energy (E)

We know;

According to the formula;

E = $\frac{1}{2}$ × k × x × x

E = $\frac{1}{2}$ × k ×(x)²

where;

E represents the elastic potential energy

K represents the spring constant

x represents amount of the compression in the string

So therefore,

Substituting the values in the above formula;

E = $\frac{1}{2}$ × 240 × (0.40)²

E = $\frac{1}{2}$ × 240 × 0.16

E = $\frac{1}{2}$ × 38.4

E = 19.2 J or approximately 20 J

Therefore the elastic potential energy of the string is 20 J.

5 0

3 years ago

2. A construction consultant may be responsible for

Brrunno [24]

This is a tricky one but on my part I'd have to say depending on the contract A,B,C.

3 0

3 years ago

Which two options are examples of waves reflecting?

Zinaida [17]

I think the answer is b

3 0

2 years ago

Read 2 more answers