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Viefleur [7K]
3 years ago
9

The electric field of a charge is defined by the force on: An electron A probe charge A proton. A source charge.

Physics
1 answer:
Alina [70]3 years ago
7 0

Answer:

A probe charge

Explanation:

As we know that electric field intensity is the force experienced by the probe charge which is placed in the electric field region

Here we can say it as

E = \frac{F}{q}

so here that probe charge should be very small so that it will not disturb the electric field in the space.

If the probe charge is of large magnitude then the field will get disturbed and the intensity which is to be measured is different from its actual value.

Also the sign of the probe charge is taken to be positive.

so correct answer here will be

A probe charge

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Participating in team sports can have a positive effect on all aspects of life. Please select the best answer from the choices p
AveGali [126]

Answer: True

Explanation:

Participating in team sports can lead to the development of healthy fellowship and brotherhood behavior among the participants. It induces team spirit and allows to lead a healthy competition. It can be a good recreational activity which can be used for removing the emotional and physiological stress. Hence, can improve positivity in all aspects of life.

5 0
3 years ago
A space station in the shape of a 100 m-diameter (50m radius) wheel is spinning so as to impart a linear tangential speed of 22.
zalisa [80]

Answer:

correct option is b. 31.3 m/s

Explanation:

given data

artificial gravity a1 = 1 g

artificial gravity a2 = 2 g

diameter = 100 m

radius  r= 50 m

speed v1 = 22.1 m/s

solution

As acceleration is  ∝ v²

so we can say

\frac{a2}{a1} = \frac{v2}{v1}    .....................1

put here value

\frac{2}{1} = \frac{v2}{22.1}  

solve it

v2 = \sqrt{2 } × 22.1

v2 = 31.25 m/s

so correct option is b. 31.3 m/s

4 0
3 years ago
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

  = 4.59\times 10^{-19} \ J

or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

  = 2.87 \ eV

(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

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List and describe three biological applications of electricity
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