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3 years ago

Which of the following statements is true?

Physics

1 answer:

Lilit [14]3 years ago

8 0

A. All natural radiation is at a level low enough to be safe

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Q.1. What determines the rate at which energy is delivered by a current ?

aksik [14]

Explanation:

The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.

8 0

3 years ago

A charge q1= 3nC and a charge q2 = 4nC are located 2m apart. Where on the line passing through these charges is the total electr

Ipatiy [6.2K]

Answer:

Explanation:

Electric field due to a charge Q at a point d distance away is given by the expression

E = k Q / d , k is a constant equal to 9 x 10⁹

Field due to charge = 3 X 10⁻⁹ C

E = E = $\frac{9\times 10^9\times3\times10^{-9}}{d^2}$

Field due to charge = 4 X 10⁻⁹ C

$E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}$

These two fields will be equal and opposite to make net field zero

$\frac{9\times 10^9\times3\times10^{-9}}{d^2}$ = $[tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}$ [/tex]

$\frac{3}{d^2} =\frac{4}{(2-d)^2}$

$\frac{2-d}{d} =\frac{2}{1.732}$

d = 0.928

5 0

3 years ago

Does the efficiency of a motor depend on mass?

Valentin [98]

Practically yes

Efficiency=Output/input

If mass is more output may come less so it affects the efficiency practically

But thepritically it doesn't

5 0

2 years ago

Read 2 more answers

What are ways to improve the design of this experiment? Check all that apply

Serhud [2]

Answer:

B,D,E

Explanation:

I got you

B. Experiment with a wider range of materials.

D.Use a laboratory galvanometer to make precise measurements.

E. Test the strength of the electromagnet by varying the number of wire coils.

3 0

2 years ago

A converging lens has a focal length of 20 cm. An object 1 cm tall is placed 10 cm from the center of the lens. What is the heig

SCORPION-xisa [38]

Answer: 2 cm

Explanation:

Given , for a converging lens

Focal length : $f=20\ cm$

Height of object : $h=1\ cm$

Object distabce from lens : $u=-10\ cm$

Using lens formula: $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ , we get

$\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{10}$ , where v = image distance from the lens.

On solving aboive equation , we get

$\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm$

Formula of Magnification : $m=\dfrac{v}{u}=\dfrac{h'}{h}$ , where h' is the height of image.

Put value of u, v and h in it , we get

$\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm$

Hence, the height of the image is 2 cm.

3 0

2 years ago