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2 years ago

SUPER EASY POINTS!!!!!!!!!!

Physics

2 answers:

ivolga24 [154]2 years ago

7 0

Answer:

from all its mass

Explanation:

write by doing arrangement

saw5 [17]2 years ago

5 0

1.The gravity of Earth, denoted by g, is the net acceleration that is imparted to objects due to the combined effect of gravitation (from mass distribution within Earth) and the centrifugal force (from the Earth's rotation).

2.How does gravity work in space? Gravity is a very important force. Every object in space exerts a gravitational pull on every other, and so gravity influences the paths taken by everything traveling through space. It is the glue that holds together entire galaxies. It keeps planets in orbit.

3.Our weight on moon is less than it would be on Earth due to a difference of the strength of gravity on the moon. The moon's gravitation force is determined by the mass and the size of the moon. Since the moon has significantly less mass than the Earth, it will not pull objects toward itself at the strength that Earth will.

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At a playground, a child slides down a slide that makes a 42° angle with the horizontal direction. The coefficient of kinetic fr

kumpel [21]

Answer:

$a = 5.1\ m/s^2$

Explanation:

Given,

The angle of the slide= $42^\circ$

The mass of the child is= m

coefficient of friction = 0.20

when she slides down now apply Newton's law

$ma =mg \sin\theta - f$

$ma = mg\sin \theta -\mu mg \cos \theta$

therefore the acceleration

$a=g \sin\theta -\mu gcos\theta$

$a=g[\sin \theta -\mu \cos \theta]$

$a=9.8\times [\sin 42^\circ -0.2\times \cos 42^\circ]$

$a = 5.1\ m/s^2$

hence, the magnitude of acceleration during her sliding is equal to $a = 5.1\ m/s^2$

4 0

3 years ago

you slide abox of books at constant speed up a30 degree ramp, applying a force of 200 newtons directed up the slope. The coeffic

Tomtit [17]

The work done is 400 J

Explanation:

The work done by you in pushing the box along the slope is given by

$W=Fd$

where

F is the magnitude of the force applied

d is the distance covered by the box along the slope

Here we have the following:

F = 200 N is the magnitude of the force applied

$d=\frac{h}{sin \theta}$ is the distance covered, where

h = 1 m is the vertical rise

$\theta=30^{\circ}$ is the slope of the plane

Substituting and solving, we find

$W=F\frac{h}{sin \theta}=\frac{(200)(1)}{sin 30^{\circ}}=400 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

5 0

2 years ago

In what factors sound can be heard

MariettaO [177]

Explanation:

Loudness of sound is a measure of response of sound to our ear. Loudness of sound is not simply the energy reaching the human ear, but it also tells about the sensitivity of human ear detecting this energy. Loudness of sound is measured in decibel (dB). As energy reaching the ear depends on square of amplitude, loudness of sound depends on various factors namely,

(i) Amplitude of sound waves

(ii) Sensitivity of ear

(iii) Distance from the source of the sound and the listener.

6 0

2 years ago

Read 2 more answers

A boat is traveling with a velocity of 18 meters/second relative to water and the river is flowing at a velocity of 2.5 meters/s

ddd [48]

Vt = Vboat - Vriver
Vt = 18 - 2.5 = 15.5 m/s

If the boat's direction is the same as the water, you sum the velocities of the river and the boat .

8 0

3 years ago

Read 2 more answers

A balloon contains 2.3 mol of helium at 1.0 atm , initially at 240 ∘C. What's the initial volume? What's the volume after the ga

pashok25 [27]

A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
$p_i V_i = nRT_i$
where
$p_i=1.0 atm=1.01 \cdot 10^5 Pa$ is the initial pressure of the gas
$V_i$ is the initial volume of the gas
$n=2.3 mol$ is the number of moles
$R=8.31 J/K mol$ is the gas constant
$T_i=240^{\circ}C=513 K$ is the initial temperature of the gas

By re-arranging this equation, we can find $V_i$ :
$V_i = \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3$

2) Now the gas cools down to a temperature of
$T_f = 14^{\circ}C=287 K$
while the pressure is kept constant: $p_f = p_i = 1.01 \cdot 10^5 Pa$ , so we can use again the ideal gas law to find the new volume of the gas
$V_f = \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3$

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
$W=p \Delta V= p(V_f -V_i)$
by using the data we found at point 1) and 2), we find
$W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J$
where the negative sign means the work is done by the surrounding on the gas.

5 0

3 years ago