Explanation:
given solution
h=45m v^2=u^2+2gh
g=10m/s^2 v^2=0^2+2×10m/s^2×45m
vi=0 v^2=900m^2/s^2
The charge of the copper nucleus is 29 times the charge of one proton:

the charge of the electron is

and their separation is

The magnitude of the electrostatic force between them is given by:

where

is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
Answer:
1.41 m/s^2
Explanation:
First of all, let's convert the two speeds from km/h to m/s:


Now we find the centripetal acceleration which is given by

where
v = 12.8 m/s is the speed
r = 140 m is the radius of the curve
Substituting values, we find

we also have a tangential acceleration, which is given by

where
t = 17.0 s
Substituting values,

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

Answer:

Explanation:
The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.
Also, we know that the centripetal force of an object describing a circular motion is given by:

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.
Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So
and
(Since
). Then, we get:

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).
Answer:
Deltoid Force, 
Additional Information:
Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.
Explanation:
The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.
1. The torque about the point in the shoulder for the deltoid muscle,
2. The torque of the arm,
Assuming the arm is just being stretched and there is no rotation going on,
= 0
= 0
⇒ 

Where,
is radius of the deltoid
is the force of the deltiod
is the angle of the deltiod
is the radius of the arm
is the force of the arm ,
which is the mass of the arm and acceleration due to gravity
is the angle of the arm
The force of the deltoid muscle is,

but
,
∴ 