Answer:
Explanation:
According to Equations of Projectile motion :

vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec
(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec

(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m

(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m
Work is the amount of energy transferred
Explanation:
In physics, work is a measure of the energy transfer occurring in a process. Typically, we talk about work when energy is converted from one form into another.
For instance, work is done when a force is applied on an object. The work done on the object is given by:

where
where
F is the magnitude of the force
d is the displacement
is the angle between the direction of the force and of the displacement
We notice the following:
- No work is done when the force is perpendicular to the displacement (
) - The work is maximum when the force is parallel to the displacement
Whenever work is done, there is also an energy transfer taking place. For instance, in the previous example, when the force is applied to the object, the object will accelerate (assume there is no friction), and will gain kinetic energy: therefore, there is a transfer of energy to the object.
Learn more about work:
brainly.com/question/6763771
brainly.com/question/6443626
#LearnwithBrainly
<span>The absolute magnitude of a star is how bright it would appear to us
if it were located ten parsecs (about 32.6 light years) from us. So it's
a way of treating all stars equally ... on a "level playing field" ... and it
describes each star's actual brightness. </span>
<span>g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:
R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km
here, the speed of the satellite is:
v = sqrt(R*9.00m/s^2) = 7730 m/s
the time it would take the satellite to complete one full rotation is:
T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h
Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu! :)</span>
Answer:
The correct answer is "6666.67 N".
Explanation:
The given values are:
Mass,
m = 0.100
Relative speed,
v = 4.00 x 10³
time,
t = 6.00 x 10⁻⁸
As we know,
⇒ 
On substituting the given values, we get
⇒ 
⇒ 