Please help me on this

natka813 [3]

1). The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.

2). The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.

3). The block is also affected by friction with the air (air resistance) as it
falls to the ground. Friction robs some kinetic energy.

8 0

3 years ago

Describe a situation when you might travel at a high velocity, but with low acceleration

Alisiya [41]

Reading a book in your warm, comfy seat ... in Row-27 of a
passenger airliner cruising at 450 miles per hour.

5 0

3 years ago

Explain how Pascal's principle can be used to design a fluid power system and describe how a fluid power system works.

zhenek [66]

Pascal's law of fluid transfer states that when there is an increase in fluid pressure, the rest of the extrinsic variables also increases. For example, in a flow of liquid in an orifice, there is a contraction of diameter in the orifice part. The fluid that will go in there increases in pressure and thereby an increase in velocity as well.

4 0

3 years ago

What is kepler's law??

Elan Coil [88]

<h2>QUESTION:- </h2>

➜what is kepler's law??

$\huge\red{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W}\orange{ER}}}}$

Kepler gave the three laws or theorems of motion of the orbitals bodies

${\huge {\bold{ \red{ \star}}}}{ \blue{ \bold{FIRST \: \: \: LAW}}}$

This law state that the celestial bodies revolves around the stars in elliptical orbit and star as a single focus.

Example :- Earth revolves around the Sun as assuming it as single focus

This also shows that earth revolves around the sun in elliptical orbit.

${\huge {\bold{ \blue{ \star}}}}{ \green{ \bold{SECOND \: \: \: LAW}}}$

Area covered by the planet is equal in equal duration of time irrespective of the position of the planet.

It also states that Angular momentum is constant

As Angular momentum is constant it means areal velocity is also constant.

$\frac{ \triangle \: A}{ \triangle \: T} = \frac{L}{2m}△T△A=2mL$

where:-

A is the area.

T is the time.

L is the angular momentum.

M is the mass of the body.

${\huge {\bold{ \green{ \star}}}}{ \purple{ \bold{THIRD \: \: \: LAW}}}$

square of the time of the revolution is directly proportional to the cube of the distance between the planet and star in Astronomical unit.

${T}^{2} = {a}^{3}T2=a3$

where:-

T = time of revolution

a is the distance between the planet and star.

$\purple\star \: {Thanks \: And \: Brainlist} \blue \star \\ {\orange{ \star}}{if \: U \: Like d \: My \: Ans} {\green{ \star }}$

8 0

3 years ago

Four particles are in a 2-d plane with masses, x- and y- positions, and x- and y- velocities as given in the table below: what i

Arte-miy333 [17]

I attached the picture of the missing table.
Center of mass is the point such that if you apply force to it, the system would move without rotating.
We can use following formula to calculate the center of mass:
$R=\frac{1}{M}\sum_{i=1}^{n=i}m_ir_i$
Where M is the sum of the masses of all particles.
Part 1
To calculate the x coordinate of the center of mass we will use this formula:
$R_x=\frac{1}{M}\sum_{i=1}^{n=i}m_ix_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_x=0.96m$
Part 2
To calculate the y coordinate of the center of mass we will use this formula:
$R_y=\frac{1}{M}\sum_{i=1}^{n=i}m_iy_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_y=-0.84m$
Part 3
We will calculate speed along x and y-axis separately and then will add them together.
$v_x=\frac{\sum_{i=1}^{n=i}m_iv_x_i}{M}$
$v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}$
Total velocity is:
$v=\sqrt{v_x^2+v_y^2}$
Once we calculate velocities we get:
$v_x=-1.08\frac{m}{s}\\ v_y=-0.03\frac{m}{s}\\ v=\sqrt{(-1.08)^2+(-0.03)^2}=1.08\frac{m}{s}$
Part 4
Because origin is left to our center of mass(please see the attached picture) placing fifth mass in the origin would move the center of mass to the left along the x-axis.
Part 5
If you place fifth mass in the center of the mass nothing would change. The center of mass would stay in the same place.
Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing

3 0

3 years ago