ok so i dont know srry5
As the distance between the sun and earth decreases, the speed of the planet
a
increases
b
decreases
c
stays the same
Answer:
Explanation:
Increases. The force of gravity is distance dependent. Therefore, a smaller 'r' value will result in a larger force. Net force is proportional to the acceleration, so the planet will increase its speed.
If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is
364N
185N
173N
73N
ANSWER AND I WILL GIVE YOU BRAINILIEST
Answer:
73N
Explanation:Just multiply 1.2^2 by 50
Earth has seasons because _____.
it rotates on its axis as it moves around the sun
the temperature of the sun changes
its axis is tilted
the distance between Earth and the sun changes
Answer:
c, its axis is tilted
maybe
As it works its way around the sun, its tilted axis exposes different parts of earth.
C would be it because the roation of Earth on its axis doesn't have anything to do with the exposer of the revolution on its axis
HELP ! ILL MARK BRAINLIEST HELP ASAP
Answer:
A
Explanation:
In 5 minutes, they went 10 miles at both 2, 3, and 4 checkpoints. The bus then starts to speed up.
Hope this helps!
Calculate the magnitude and direction of the resultant of the following forces
Answer and I will give you brainiliest
Answer:
250N same direction
Explanation:
100+150 = 250N
same direction
I’ll mark brainless please hurry
Answer:
Covalent bonds can form between similar atoms.
Explanation:
A long, straight wire has a uniform constant charge with linear charge density, - 3.60 nC/m. The wire is surrounded by a long nonconducting, thin-walled cylindrical shell that is charged on its outside surface, such that the electric field outside the shell is zero. The shell has a radius of 1.50 cm.
Required:
What uniform area charge density rho is needed on the shell for the electric field to be zero outside the shell?
Answer:
Uniform area charge density rho is needed is 3.82*10^-8 C.m^-2
Explanation:
See the attached files.
To find the rho, I used Gauss law for cylindrical shell which is equation 1 and Gauss law for the rod which is equation 4.
Note that in equation 4, Lamda is the charge per length while L is the length if the rod. Also R is the radius of the shell.
The final answer is 3.82*10^-8 C.m^-2 which is the uniform area charge density rho is needed.
A skier pushes off the top of a hill with an initial speed of 3.30 m/s. How fast will she be moving after dropping 5.00 meters in elevation if friction is negligible?
Answer:
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A constant electric field of 5.00 N/C points along the positive x-direction. An electron, initially at rest, moves a distance of 2.00 m in this space. How fast is the electron moving after its 2.00 m journey
Answer:
1.875 x 10⁶ m /s .
Explanation:
Force on electron = E e where E is electric field and e is charge on electron
acceleration generated = Ee / m where m is mass of the electron .
Putting the values
acceleration generated = 5 x 1.6 x 10⁻¹⁹ / 9.1 x 10⁻³¹
= .879 x 10¹² m /s²
v² = u² + 2 as , initial velocity u = 0 , displacement s = 2 m
v² = 0 + 2 x .879 x 10¹² x 2
v = 1.875 x 10⁶ m /s .