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3 years ago

Attraction is the measure of romantic love between two individuals. Please select the best answer from the choices provided T F

Physics

1 answer:

goldfiish [28.3K]3 years ago

6 0

Answer: false.

Explanation:

The attraction is a force (electromagnetic force) that acts between oppositely charged particles and draws them together.

(for example when you have two magnets that are stick to each other, when you want to pull them apart you can feel some resistance, that resistance is the attraction force between the magnets)

So the statement "Attraction is the measure of romantic love between two individuals. " is false.

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People could not walk without friction. What friction are we talking about here, kinetic friction or static friction?

3241004551 [841]

Answer:

Obviously, it is static type of friction because it is between two static bodies i.e., feet and ground at a particular time instant dt.

1. No “Hold” is not a good synonym because it involves a static body. In “Push” both of the bodies are in some kind of motion a given action time.

2. Similarly, “Support” is also not a good synonym because it involves a static body as compulsory. By the way it can a synonym for “Hold”.

3. Propel, Impel, Thrust and Shove are some basic synonyms that can be applied.

4 0

4 years ago

A positive point charge (q = +5.45 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 1.49 m. A p

ruslelena [56]

Answer:

$r_{B} = 3.34\ m$

Solution:

As per the question:

Point charge, q = $5.45\times 10^{- 8}\ C$

Test charge, $q_{o} = 4.96\times 10^{- 11}\ C$

Work done by the electric force, $W_{AB} = - 9.05\times 10^{- 9}\ J$

Now,

We know that the electric potential at a point is given by:

$V = \frac{kqq'}{r}$

where

r = separation distance between the charges.

Also,

The work done by the electric force i moving a test charge from point A to B in an electric field:

$W_{AB} = kqq_{o}(\frac{1}{r_{B} - \frac{1}{r_{A}}}$

$- 9.05\times 10^{- 9} = 9\times 10^{9}\times 5.45\times 10^{- 8}\times 4.96\times 10^{- 11}(\frac{1}{r_{B}} - \frac{1}{1.49}}$

$- 0.3719 = (\frac{1}{r_{B}} - 0.67}$

$r_{B} = \frac{1}{0.299} = 3.34\ m$

5 0

4 years ago

An electron with a speed of 1.9 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts

Nimfa-mama [501]

Explanation:

It is given that,

Speed of the electron in horizontal region, $v=1.9\times 10^7\ m/s$

Vertical force, $F_y=4.9\times 10^{-16}\ N$

Vertical acceleration, $a_y=\dfrac{F_y}{m}$

$a_y=\dfrac{4.9\times 10^{-16}\ N}{9.11\times 10^{-31}\ kg}$

$a_y=5.37\times 10^{14}\ m/s^2$ ..........(1)

Let t is the time taken by the electron, such that,

$t=\dfrac{x}{v_x}$

$t=\dfrac{0.024\ m}{1.9\times 10^7\ m/s}$

$t=1.26\times 10^{-9}\ s$ ...........(2)

Let $d_y$ is the vertical distance deflected during this time. It can be calculated using second equation of motion:

$d_y=ut+\dfrac{1}{2}a_yt^2$

u = 0

$d_y=\dfrac{1}{2}\times 5.37\times 10^{14}\ m/s^2\times (1.26\times 10^{-9}\ s)^2$

$d_y=0.000426\ m$

$d_y=0.426\ mm$

So, the vertical distance the electron is deflected during the time is 0.426 mm. Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

When you connect an unknown resistor across the terminals of a 1.50 V D-cell battery having negligible internal resistance, you

g100num [7]

Answer:

Explanation:

Using ohm's law

a) V = IR where V is voltage in Volt, I is current in Ampere and R is resistance in ohms

R = V / I = 1.50 V/ ( 2.05 /1000) A = 731.71 ohms

b) Power = IV = $\frac{V}{R}$ × v = $\frac{V^{2} }{R}$ = $\frac{9^{2} }{731.71}$ = 0.1107 W

c) E = IR + Ir = ( 731.71 × 0.0036) + ( 35 × 0.0036) = 2.76 V

d) Power use by the resistor = I²R = 0.0036² × 731.71 = 0.00948 W = 0.00948 W = 0.000009483 kw × ( 18 / 60 ) H = 2.84 × 10⁻⁶ KW-H

5 0

3 years ago

A parallel-plate capacitor has plates of area 0.40 m2 and plate separation of 0.20 mm. The capacitor is connected to a 9.0 V bat

mafiozo [28]

Answer:

a) E = 4.5*10⁴ V/m

b) C= 17.7 nF

c) Q = 159. 3 nC

Explanation:

By definition, the electric field is the electrostatic force per unit charge, and since the potential difference between plates is just the work done by the field, divided by the charge, assuming a uniform electric field, if V is the potential difference between plates, and d is the separation between plates, the electric field can be expressed as follows:

$E = \frac{V}{d} = \frac{9.0V}{2*10-4m} =4.5 * 10e4 V/m (1)$

For a parallel-plate capacitor, applying the definition of capacitance as the quotient between the charge on one of the plates and the potential difference between them, and assuming a uniform surface charge density σ, we get:

$Q = \sigma* A (2)$

From (1), we know that V = E*d, but at the same time, applying Gauss'

Law at a closed surface half within the plate, half outside it , it can be

showed than E= σ/ε₀, so finally we get:

$C = \frac{Q}{V} =\frac{\sigma*A}{E*d} = \frac{\sigma*A}{\frac{\sigma}{\epsilon_{o} } d} =\frac{\epsilon_{0}*A}{d} = \frac{8.85e-12F/m*0.4m2}{2e-4m} = 17.7 nF (3)$

From (3) we can solve for Q as follows:

$Q = C* V = 17.7 nF * 9.0 V = 159.3 nC (4)$

6 0

3 years ago