<h2>
Answer:</h2>
(c) 5m/s²
<h2>
Explanation:</h2>
Total acceleration (a) of a particle in a circular motion is the vector sum of the radial or centripetal acceleration (
) of the particle and the tangential acceleration (
) of the particle and its magnitude can be calculated as follows;
a = 
---------------------(i)
<em>But;</em>
= 
------------------------------(ii)
Where;
v = instantaneous velocity
r = radius of the circular path of motion
<em>From the question;</em>
v = 30m/s
r = 300m
(i) First let's calculate the centripetal acceleration by substituting the values above into equation (ii) as follows;
= 
= 
= 3m/s²
(ii) From the question, the velocity of the particle is increasing at a constant rate of 4m/s² and that is the tangential acceleration , of the particle. i.e;
= 4m/s²
(iii) Now substitute the values of and into equation (i) as follows;
a = 
a = 
a = 
a = 5m/s²
Therefore, the magnitude of its total acceleration a, is 5m/s²
Please post in English so i or someone else can help you.
Answer: 9938.8 km
Explanation:
1 pound-force = 4.48 N
30.0 pounds-force = 134.4 N
The force of gravitation between Earth and object on the surface of is given by:
Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.
At height, h above the surface of the Earth, the weight of the object:
we need to find "h"
taking the ratio of two:
Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
where
is the charge density
is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
