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Tatiana [17]
3 years ago
12

The process of changing the energy of a system by means of force. Force * Diatance

Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
7 0
This is known as work
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Answer:

it becomes a gas

Explanation:

the matter expands, turning into steam, a gas.

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A helium nucleus is composed of two protons (positively charged) and two neutrons (charge neutral). The distance between the two
lora16 [44]

Answer:

Using the given values

F = K q^2 / r^2 = 9 * 10E9 * (1.6 * E-19)^2 / (5.18 * E-15)^2 N

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2 years ago
Sharks and related fish can sense the extremely weak electric fields emitted by their prey in the surrounding waters. These dete
Sonja [21]

Answer

2) 1.5×10-2 m

Explanation

The potential difference is related to the electric field by:

\Delta V=Ed (1)

where

\Delta V is the potential difference

E is the electric field

d is the distance

We want to know the distance the detectors have to be placed in order to achieve an electric field of

E=1 V/cm=100 V/m

when connected to a battery with potential difference

\Delta V=1.5 V

Solving the equation (1) for d, we find

d=\frac{\Delta V}{E}=\frac{1.5 V}{100 V/m}=0.015 m=1.5 \cdot 10^{-2} m

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3 years ago
What is the relationship between inertia and mass?
ivann1987 [24]

Answer:

D

Explanation:

The greater the mass, the greater the inertia, and vice versa.

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8 0
3 years ago
How do resistors in parallel affect the total resistance?
4vir4ik [10]

Answer:

They're going to increase the total resistance as R_{T} = \sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}

Explanation:

If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then

I = I_1 + I_2 + ... + I_N

where

I_i = \frac{V_i}{R_i}

but

V_i = V_j = V for i,j= 1, 2,..., N

so

I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}

where

R_T = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)^{-1} =\sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}

4 0
3 years ago
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