Answer:
Explanation:
HELP ME WRITE A ONE PAGE ESSAY TO EXPLAIN THE AUTHOR'S PURPOSE IN WRITING HOM SMART ARE ANIMAL
Answer:
3 820 885 N
Explanation:
Gravitational equation
F = G m1 m2 / r^2
G = gravitational constant = 6.6713 x 10^-11 m^3/kg-s^2
F = 6.6713 x 10^-11 * 4.41 x 10^5 * 5.97 x 10^24 / ( 6.78x 10^6)^2
= 3820885 .3 N
Hey there!
Seems like you're looking for the size and direction to the final velocity of the two cars. To find it, you must solve it like this.
0.4 kg(3 m/s) + 0.8kg(–2 m/s) = 1.2 kg m/s -1.6 kg m/s = –0.4 kg m/s
–0.4 kg m/s = 1.2 kg(v) = (–0.4 kg m/s)/(1.2 kg) = v = –0.33 m/s
So, the cars are traveling at -0.33 m/s in the direction of the second car.
Hope this helps
<em>Tobey</em>
A decrease in velocity is referred to as deceleration. If car is moving at 30 m/s and stop in 50 m .The value of deceleration is 11.56 ms−2.
<h3>How to calculate deceleration ?</h3>
While acceleration is motion in which an object's speed varies every second, deceleration is motion that causes an object to slow down.
We are aware that acceleration refers to an object's rate of increase in speed, and deceleration refers to an object's rate of decrease in speed. For instance, when we apply the brakes while driving, we benefit from the vehicle's ability to decelerate and slow down.
The Deceleration Formula is the final velocity minus the initial velocity, with a negative sign in the result because the velocity is decreasing, if starting velocity, final velocity, and time taken are given.
velocity of car = 30 m/s
car need to stop in 50m
Deceleration a = v^2 – u^2 / 2s
= 0^2 - 50^2 / 2*30
= 11.56
Deceleration of the care = 11.56 ms−2
To learn more about deceleration refer :
brainly.com/question/75351
#SPJ4
Question:
A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.
a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.
Express your answer using two significant figures.
Answer:
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 
Explanation:
A point charge ,q = 
is located in the center of a spherical cavity of radius , 
m inside an insulating spherical charged solid.
The charge density in the solid , d = 
Distance from the center of the cavity,R =
Volume of shell of charge= V =![(\frac{4\pi}{3})[ R^3 - r^3 ]](https://tex.z-dn.net/?f=%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D)
Charge on the shell ,Q = 
![Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d](https://tex.z-dn.net/?f=Q%20%3D%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D%20%5Ctimes%20d)
![Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%2A%5Ctimes%2010%5E%7B-4%20%7D%5B8.57375%20-%202.81011%20%5D%5Ctimes%207.35%5Ctimes%2010%5E%7B-4%7D)
![Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%5Ctimes%2010%5E%7B-4%7D%20%5B5.76364%20%5D%20%5Ctimes%207.35%20%5Ctimes%2010%5E%7B-4%7D)
Electric field at 
m due to shell
E1 = 
Electric field at due to 'q' at center 
E2 =
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity
= E2- E1
![=[ 2.134 - 1.769 ]\times 10^6](https://tex.z-dn.net/?f=%3D%5B%20%202.134%20%20-%201.769%20%5D%5Ctimes%2010%5E6)
