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3 years ago

according to the law of conservation of energy what ultimatly happens to the total amount of energy in the universe?

1 answer:

8 0

<span>In physics, the law of conservation of energy states that the total energy of an isolated system remains constant—it is said to be conserved over time. Energy can neither be created nor destroyed; rather, it transforms from one form to another. For instance, chemical energy can be converted to kinetic energy in the explosion of a stick of dynamite.</span>

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A block is sitting on a platform 20 m high. It weighs 50 kg. How much potential energy does it contain?

kifflom [539]

Answer:

9806.65 Joules

Explanation:

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3 years ago

What is the force that opposes the movement of an object through water??

anastassius [24]

In air, you'd call it air resistance, pilots call it drag. In water, some call it
water resistance, or just plain drag.

Whatever it's called, it's the friction between the object and the fluid, plus
the force needed to push the fluid out of the way to let the object get through.

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3 years ago

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In a series circuit, when batteries are added to the circuit the voltage will _____ and the current will _____.

Eva8 [605]

Answer: voltage, explode is the answers

Explanation:

Bc of the device for each server

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3 years ago

An object is released from height of 17m. <br> The object will hit the ground approximately in

zzz [600]

$\text{Given that,}\\\\\text{Height, h = 17 m}\\\\\\\text{We know that,}\\\\h = v_0t + \dfrac 12 gt^2\\\\\implies h = \dfrac 12 gt^2\\\\\implies t^2 = \dfrac{2h}g\\\\\implies t =\sqrt{\dfrac{2h}g} = \sqrt{\dfrac{2(17)}{9.81}} = 1.87 ~ \text{sec}$

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2 years ago

A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0

3 years ago