The answer is tropical rainforest
Answer: False
Explanation:
It's received in elliptical cavities.
Answer:
(a) The distance-time graph for an object with uniform speed is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram
(b) The distance-time graph for an object with non-uniform speed is giving by a curved line sloped graph with varying gradient as shown in the attached diagram
(c) The velocity-time graph for a car with uniform motion is giving by a horizontal line graph at the speed of constant motion with a zero gradient as shown in the attached diagram
(d) The velocity-time graph for a car moving with uniform acceleration is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram
(e) The velocity-time graph for a car moving with non-uniform acceleration is giving by a curved line sloped graph with varying gradient as shown in the attached diagram
(f) According to Newton's first law of motion, an object at rest will remain at rest with no motion unless acted by a force, an therefore, will have no motion with time
Explanation:
Answer:
2.04 s
Explanation:
v = at + v₀
(-30.0 m/s) = (-9.8 m/s²) t + (-10.0 m/s)
t = 2.04 s
Answer:![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
Explanation:
Given
Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge
Force due to the charge placed at diagonally opposite end on -q charge

where
Distance between the two charges

negative sign indicates that it is an attraction force
Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

The magnitude of force by both the charge is same but at an angle of 
thus combination of two forces at 2 and 3 will be

Now it will add with force due to 1 charge
Thus net force will be
![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)