Explanation:
Formula for angle subtended at the center of the circular arc is as follows.
where, S = length of the rod
r = radius
Putting the given values into the above formula as follows.
= 
= 
= 
Now, we will calculate the charge density as follows.
= 
= 
Now, at the center of arc we will calculate the electric field as follows.
E = 
= 
= 34.08 N/C
Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.
Answer:
S = V0 t + 1/2 a t^2
S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)
S = 1500 m + .6 * 90000 m = 55,500 m
Check: V0 = 5 m/s
V2 = V0 + a t = 5 + 1.2 * 300 = 365 m/s
Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s (note uniform motion)
S = 185 * 300 = 55,500 m
We calculated V2 above at 365 m/s the speed after 300 sec
Rubbing both pieces cause each piece to have a negative charge.
When two parts have the same they repel each other, so holding one piece up tot he end of the other piece would push it away.
Because one piece is held in the middle by a string, it would rotate the piece in a circle.
If they held the piece to the other end of the one held by a string it would start to rotate in the opposite direction.
Answer:
There are total eight planets in the solar system and the average distance from the sun to each planet in increasing order is given below.
Explanation:
The average distance from the sun is listed below in increasing order.
1. Mercury - It is the most closet planet to Sun, 57 million km
2. Venus - 108 million km
3. Earth - 150 million km
4. Mars - 228 million km
5. Jupiter - 779 million km
6. Saturn - 1.43 billion km
7. Uranus - 2.88 billion km
8. Neptun - It is the most farthest from the Sun, 4.50 billion km
To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.
By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore
Here,
m = mass
g =Gravitational energy
The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore
Remember the expression for which you can determine the relationship between mass, volume and density, in which
In this case the density would be that of the object, replacing
Since the displaced volume of water is 0.429 we will have to
The density of water under normal conditions is 
, so
The density of the object is 
