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2 years ago

How is the density of water different from the densities of most other substances?

Physics

1 answer:

Ket [755]2 years ago

4 0

Answer:

In practical terms, density is the weight of a substance for a specific volume. The density of water is roughly 1 gram per milliliter but, this changes with temperature of if there are substances dissolved in it. Ice is less dense than liquid water which is why your ice cubes float in your glass.

Explanation:

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3 years ago

It usually takes more force to start an object sliding than it does to keep an object sliding because static friction is usually

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3 years ago

A football is kicked into the air from an initial height of 4 feet. The height, in feet, of the football above the ground is giv

kakasveta [241]

Answer: 0.5 seconds or 2.625 seconds

Explanation:

At t = 0, The ball is 4 ft above the ground.

The height of the football varies with time in the following way:

s(t) = -16 t² + 50 t + 4

we need to find the time in which the height would of the football would be 25 ft:

⇒25 = -16 t² + 50 t + 4

we need to solve the quadratic equation:

⇒ 16 t² - 50 t + 21 = 0

$t = \frac{50 \pm \sqrt{50^2-4\times 16\times 21}}{2\times16}$

⇒ t = 0.5 s or 2.625 s

Therefore, at t = 0.5 s or 2.625 s, the football would be 25 ft above the ground.

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3 years ago

Read 2 more answers

If a body accelerates from a state of relative rest in a gravity field, where does the energy come from to transfer as kinetic e

Delvig [45]

The energy that transforms into kinetic energy is the Potential Energy. It happens that objects can store energy as a result of its position. Image for example a slingshot. When you stretch the slingshot, it stores energy, this energy would be the energy you used to stretch the slingshot, the material aborbs it and then release to throw the projectile.

Now, on earth and everywhere in the universe where you are close to an object with mass, it exists a force called gravity that attracts you towards that object. Every object that has mass exercises gravitational attration towards the other objects. It just happens that Earth is has so much mass that its gravitational pull is way stronger that the gravitational pull of another object on its surface. This means things will tend to be as close as earth as possible, and in order to move something away from earth, you will have to perform a force in the opposite direction to Earth and, therefore, consume energy. This energy will be store as potential energy, and when you drop the object, the potential energy will be the energy that will transform to kinetic energy.

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2 years ago

A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is<u> positive.</u>

<u></u>

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

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The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is <u>positive.</u>

<u></u>

<u></u>

<u>Given:</u>

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

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2 years ago