Answer:
The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.
Explanation:
Given;
mass of the object, m = 2 kg
weigh of the object, W = 20 N
tension on the rope, T = 12 N
The acceleration of the object is calculated by applying Newton's second law of motion as follows;
T = F + W
T = ma + W
ma = T - W
(the negative sign indicates deceleration of the object)
The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.
radio waves bc they have the longest wave lenthgs in a magnetic spectrum
Answer: hello your question is incomplete below is the missing part
A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.
answer:
- q
Explanation:
Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero
given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q
It is 10.20 m from the ground.
<u>Explanation:</u>
<u>Given:</u>
m = 0.5 kg
PE = 50 J
We know that the Potential energy is calculated by the formula:
where m is the is mass in kg; g is acceleration due to gravity which is 9.8 m/s and h is height in meters.
PE is the Potential Energy.
Potential Energy is the amount of energy stored when an object is stationary.
Here, if we substitute the values in the formula, we get
50 = 0.5 × 9.8 × h
50 = 4.9 × h
h = 10.20 m
Answer:
![r_{cm}=[12.73,12.73]cm](https://tex.z-dn.net/?f=r_%7Bcm%7D%3D%5B12.73%2C12.73%5Dcm)
Explanation:
The general equation to calculate the center of mass is:
Any differential of mass can be calculated as:
Where "a" is the radius of the circle and λ is the linear density of the wire.
The linear density is given by:
So, the differential of mass is:
Now we proceed to calculate X and Y coordinates of the center of mass separately:
Solving both integrals, we get:
Therefore, the position of the center of mass is:
