
The most effective forces on the object are the backward force of air resistance relatively very small in magnitude, and the force of gravity. Because the spiral path of the satellite is not perpendicular to the gravitational force, one element of the gravitational force pulls forward. at the satellite to do fantastic work & make its speed increase.
<h3>What is called gravitational force?</h3>
Gravity, additionally referred to as gravitation, is a force that exists amongst all material gadgets withinside the universe. For any objects or particles having nonzero mass, the force of gravity tends to draw them in the direction of each other. Gravity operates on objects of all sizes, from subatomic particles to clusters of galaxies.
To learn more about gravitational force, visit;
brainly.com/question/9266911
#SPJ4
Answer:
the velocity of the bullet-wood system after the collision is 2.48 m/s
Explanation:
Given;
mass of the bullet, m₀ = 20 g = 0.02 kg
velocity of the bullet, v₀ = 250 m/s
mass of the wood, m₁ = 2 kg
velocity of the wood, v₁ = 0
Let the velocity of the bullet-wood system after collision = v
Apply the principle of conservation of linear momentum to calculate the final velocity of the system;
Initial momentum = final momentum
m₀v₀ + m₁v₁ = v(m₀ + m₁)
0.02 x 250 + 2 x 0 = v(2 + 0.02)
5 + 0 = v(2.02)
5 = 2.02v
v = 5/2.02
v = 2.48 m/s
Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s
Answer:

Explanation:
If we have a periodic wave we need to satisfy the following basic relationship:

From the last formula we see that the velocity is proportional fo the frequency.
For this case we have the following info given by the problem:

We know that the frequency is the reciprocal of the period so we have this formula:

And if we replace we got:

Now since we have the value for the wavelength we can find the velocity like this:

And if we convert this into cm/s we got:

Answer:
you add 67 to 1+1 is this clear also chicken nuggets
For Ethernet, if an adapter determines that a frame it has just received is addressed to a different adapter
a. it discards the frame without sending an error message to the network layer
b. it sends a NACK (not acknowledged frame) to the sending host
c. it delivers the frame to the network layer, and lets the network layer decide what to do
d. it discards the frame and sends an error message to the network layer
Answer:
Option A
Explanation:
The nodal address has to match the signal message address for it to function well but if the it doesn't match the nodal receiver address, it disregards it.