Pleaaaase help meeee

Which of the following would create voltage in a coil of wire?

zubka84 [21]

Its d all the above your welcome

8 0

3 years ago

I have four questions.

Inga [223]

Answer:

The potential energy of the ball is 784 joules. And the kinetic energy of it is 392 while falling halfway down.

Explanation:

PE = mass (2kg) * Gravitational acceleration (9.8 m/s^2)* height (40 meters)

KE = 1/2 mass (1 kg) * velocity^2 (19.8)

7 0

3 years ago

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Please help me with this...<br>And Show all steps correctly

ipn [44]

Answer:

4 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 10 m/s

Acceleration (a) = –8m/s²

Time (t) = 2 s

Displacement (s) =?

The displacement of the object can be obtained as follow:

s = ut + ½at²

s = (10 × 2) + (½ × –8 × 2²)

s = 20 + (–4 × 4)

s = 20 + (–16)

s = 20 – 16

s = 4 m

Thus, the displacement of the object is 4 m.

4 0

3 years ago

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.

tatiyna

(a) $2.7\cdot 10^{25} kg$

The acceleration due to gravity on the surface of the planet is given by

$g=\frac{GM}{R^2}$ (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

Here we know:

$g=22.4 m/s^2$

$d=1.8\cdot 10^7 m$ is the diameter, so the radius is

$R=\frac{d}{2}=\frac{1.8\cdot 10^7 m}{2}=9\cdot 10^6 m$

So we can re-arrange eq.(1) to find M, the mass of the planet:

$M=\frac{gR^2}{G}=\frac{(22.4 m/s^2)(9\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=2.7\cdot 10^{25} kg$

(b) $4.8\cdot 10^{31}kg$

The planet is orbiting the star, so the centripetal force is equal to the gravitational attraction between the planet and the star:

$m\frac{v^2}{r}=\frac{GMm}{r^2}$ (1)

where

m is the mass of the planet

M is the mass of the star

v is the orbital speed of the planet

r is the radius of the orbit

The orbital speed is equal to the ratio between the circumference of the orbit and the period, T:

$v=\frac{2\pi r}{T}$

where

$T=402 days = 3.47\cdot 10^7 s$

Substituting into (1) and re-arranging the equation

$m\frac{4\pi r^2}{rT^2}=\frac{GMm}{r^2}\\\frac{4\pi r}{T^2}=\frac{GM}{r^2}\\M=\frac{4\pi r^3}{GMT^2}$

And substituting the numbers, we find the mass of the star:

$M=\frac{4\pi^2 (4.6\cdot 10^{11} m)^3}{(6.67\cdot 10^{-11})(3.47\cdot 10^7 s)^2}=4.8\cdot 10^{31}kg$

4 0

3 years ago

If a friend is making lemonade from an instant mix, which set of conditions would lead to a faster rate of dissolving the mix in

Wittaler [7]

Answer:

C: Warm water and Powdered Lemonade.

Explanation:

I got it right first try trust me, It is the right answer.

3 0

3 years ago

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