A 2kg object initially going 4m/s to the right is later going 8m/s. whats the change in velocity?

How is artificial gravity created in spaceships and sattelites ?????? U need some IQ for this

goldfiish [28.3K]

Explanation:

Artificial gravity can be created using a centripetal force. A centripetal force directed towards the center of the turn is required for any object to move in a circular path. In the context of a rotating space station it is the normal force provided by the spacecraft's hull that acts as centripetal force.

Hope it helps.

4 0

3 years ago

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Elements are arranged in groups by similar atomic structure on the periodic table. This allows for an elements properties to be

DiKsa [7]

C

Atomic radius is the distance between the center of the nucleus to the outermost orbital shell of the atom. Assume the atom is like a football stadium and the nucleus of the atom is a ball placed at the center of the pitch. The atomic radius is from the center of the ball to the edge of the football stadium.

Explanation:

This atomic radius decreases from left to right of a periodic table because of increases in protons in the nucleus along the periodic table. This increased proton count has a higher attractive force on the electron orbitals of the atom. This decreases the atomic radius

The atomic radius of atoms down a column of the periodic table increase because an extra orbital shell is added to the atoms with every period down the column.

Learn More:

brainly.com/question/1604565

brainly.com/question/13126562

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5 0

4 years ago

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A train moving at a constant speed is passing a stationary observer on a platform. On one of the train cars, a flute player is c

Ilya [14]

Answer:

Explanation:

We shall apply the formula of Doppler effect here

F( APPARENT) = F( REAL ) X V/(V + Vs) [ v is velocity of sound and Vs is velocity of source.

415 = 440 X 343/343+Vs

142345 + 415Vs = 150920

415 V₀ = 8575

V₀ = 20.66 m/s.

8 0

3 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

3 years ago

Katie and Mark sit next to one another in class. She has a mass of 40 kg and his mass is 65 kg.

OlgaM077 [116]

A :-) for this question , we should apply
F = GMm by d^2
( For making the calculation easy , first remove the decimals )
Given : G = 6.7 x 10^-11 Nm^2 / kg^2
= 67 x 10^-12 Nm^2 / Kg^2
M = 65 kg
m = 40 kg
d = 0.5 m
Solution -
F = GMm by d^2
F = 67 x 10^-12 x 65 x 40 by 0.5 x 0.5
F = 4355 x 40 x 10^-12 by 0.25
F = 174200 x 10^-12 by 0.25
F = 696800 x 10^-12

.:. The Gravitational force between mark and Katie is 696800 x 10^-12

6 0

3 years ago