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2 years ago

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Physics

1 answer:

tensa zangetsu [6.8K]2 years ago

3 0

Answer:

1) Determine the domain of the following functions: d ... 3) If g(x) = x + 3 and f(x)= x² – 2x, find the value of f(g(a)). ... 6) Given the graph of f(x) to the right, determine: ... 8) Given f(x)= x? and g(x)= 2* The inverse of g is a function, but the inverse off is ... -3(x-1)= -5 4 (-3). -3x+ 3 = y. 10) The graph of a function f (x) is given at the.

Explanation:

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Two long, straight, parallel wires 7.2 cm apart carry currents of equal magnitude I. They repel each other with a force per unit

Bogdan [553]

Answer:

current I = 38 mA

Explanation:

given data

distance r = 7.2 cm

repel each other force per unit length \frac{F}{l} = 4.2 nN/m

solution

we know 2 wire is parallel and when current flow through these wire they exert force each other due to magnetic field

and current I(1) = I(2) ................1

$\frac{F}{l} = \frac{\mu _o}{2\pi } \times \frac{I(1) \times I(2)}{r}$ ..................2

put here value

4.2 × $10^{-9}$ = $\frac{4\pi \times 10^{-7}}{2\pi } \times \frac{I^2}{7.2\times 10^{-2}}$

solve it we get

I = 0.038884 A

current I = 38 mA

5 0

2 years ago

What element has 81 protons in the nuclei of atoms

Margarita [4]

Thallium. 81 Protons, post-transitional metal that is not found in nature.

3 0

3 years ago

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

Anarel [89]

Answer:

$R = 24.3 m$

Explanation:

As we know that the orbital speed is given as

$v = \sqrt{\frac{GM}{R + h}}$

here we know that

v = 5500 m/s

$R = 4.48 \times 10^6 m$

$h = 630 km$

now we have

$5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}$

$M = 2.34 \times 10^24 kg$

now acceleration due to gravity of planet is given as

$a = \frac{GM}{R^2}$

$a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}$

$a = 7.7 m/s^2$

now range of the projectile on the surface of planet is given as

$R = \frac{v^2 sin2\theta}{g}$

$R = \frac{14.6^2 sin(2\times 30.8)}{7.7}$

$R = 24.3 m$

3 0

3 years ago

Which one of the following statements is true concerning the magnitude of the electric field at a point in space? It is a measur

IceJOKER [234]

Answer:

It is a measure of the electric force per unit charge on a test charge.

Explanation:

The magnitude of the electric field is defined as the force per charge on the test charge.

Since we define electric field as the force per charge, it will have the units of force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).

5 0

2 years ago

an electric current of 285.0 ma flows 674. milliseconds. Calculate the amount of electric charge transported. Be sure your answe

yulyashka [42]

Answer:

The amount of electric charge transported = 0.192 C

Explanation:

Electric Charge: This is defined as the product of electric current and time in an electric circuit, The S.I unit of electric charge is Coulombs (C)

Q = It..................... Equation 1

Where Q = Electric charge, I = electric current, t = time.

Given: I = 285 mA, t = 674 milliseconds.

Conversion: (i) Convert from 285 mA to A = (285/1000) A = 0.285 A

(ii) convert from 674 milliseconds to seconds = (674/1000) s = 0.674 s

Substituting these values into equation 1

Q = 0.285 × 0.674

Q = 0.192 C

Therefore the amount of electric charge transported = 0.192 C

5 0

2 years ago