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3 years ago

I Do you guys think people used energy before the modern times? Why or why not?

Physics

1 answer:

zepelin [54]3 years ago

3 0

Answer:

do you mean before we used fossil fuels? steam.before steam? fire was used for like ...125,000 years ago. we were not smart back then but we were smart enough.

Explanation:

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What is the thermal energy of an object?

Vedmedyk [2.9K]

The total kinetic and potential energies

3 0

4 years ago

The skateboarder has a mass of 100 kg. When traveling downward from E to D, he reaches a velocity of 11 m/s. Calculate his kinet

patriot [66]

6050 J is the kinetic energy at D

<u>Explanation:</u>

In physics, the object's kinetic energy (K.E) defined as the energy it possesses during movement. It can be defined as the required work to accelerate a certain body weight in order to rest at a certain speed. When the body receives this energy as it speeds up (accelerates), it retains this energy unless speed varies. The equation is given as,

$K . E=\frac{1}{2} \times m \times v^{2}$

Where,

m - mass of an object

v - velocity of the object

Here,

Given data:

m = 100 kg

v = 11 m/s

By substituting the given values in the above equation, we get

$K . E=\frac{1}{2} \times 100 \times(11)^{2}=\frac{1}{2} \times 100 \times 121=\frac{12100}{2}=6050\ \mathrm{J}$

6 0

4 years ago

Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5

pishuonlain [190]

Answer

given,

force per unit length = 350 µN/m

current, I = 22.5 A

y = y = 0.420 m

$\dfrac{F}{L}= \dfrac{KI_1I_2}{d}$

$I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}$

$I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}$

I₂ = 32.67 A

distance where the magnetic field is zero

$\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}$

$y_1 = 0.248\ m$

there the distance at which the magnetic field is zero in the two wire is at 0.248 m.

3 0

3 years ago

Question 3 Please, i need help! Thank you

Tom [10]

I don't know this 1 I'm sorry I can't help you

4 0

4 years ago

Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

$T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}}$ (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

$\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}$

$\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2$ (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio $\frac{T^{2}}{a^{3}}$ is constant for all the planets orbiting the same sun, so we can said that:

$\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}$

$\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83$

6 0

4 years ago

Read 2 more answers