If the net force on a block is zero, the block will move at constant velocity
Explanation:
We can answer this question by applying Newton's second law of motion, which states that the net force on an object is equal to the product between its mass and its acceleration:
(1)
where
is the net force on the object
m is its mass
a is its acceleration
In this problem, we have a block, and the net force on it is zero:
According to eq.(1), this also implies that
So, the acceleration of the block is zero.
However, acceleration is the rate of change of velocity of a body:
where 
is the change in velocity in a time of 
. Since the acceleration is zero, this means that 
, and therefore the velocity of the object is constant.
Learn more about Newton's second law:
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A table would be the most appropriate because that way you can compare the data.
Answer:
a) the one with a lower orbit b) the one with a higher orbit
Explanation:
Let's consider orbital mechanics. To get an object in orbit, we need it to fall to earth parallel to the earth's surface. To understand it easily imagine a projectile thrown horizontally further and further away, at one point, the projectile hits the cannon from behind. Considering there is no wind resistance, that would be a projecile in orbit.
In other words, the circular orbits of some objects around a massive body are due to the equality between centrifugal acceleration and gravity acceleration.
.
so the velocity is
where "G" is the gravitational constant, "M" the mass of the massive body and "r" the distance between the object and the center of gravity of mass M. As you can note, if "r" increase, "v" decrease.
The orbital period of any object in orbit is
where "a" is length of semi-major axis (a = r in circular orbits). So if "r" increase, "T" increase.
Answer:
B = 1.1413 10⁻² T
Explanation:
We use energy concepts to calculate the proton velocity
starting point. When entering the electric field
Em₀ = U = q V
final point. Right out of the electric field
em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
q V = ½ m v²
v = 
we calculate
v = 
v = 
v = 25.15 10⁴ m / s
now enters the region with magnetic field, so it is subjected to a magnetic force
F = m a
the force is
F = q v x B
as the velocity is perpendicular to the magnetic field
F = q v B
acceleration is centripetal
a = v² / r
we substitute
qvB =1/2 m v² / r
B = v
we calculate
B = 
B = 1.1413 10⁻² T
Answer:
it does not mirror the image
