The equator is the closest part of out planet that is closest to the sun, which means that the continents closest to the equator will in turn be warmer.
Incomplete question as time is missing.I have assumed some times here.The complete question is here
Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.
Explanation:
Given data
Vi=10 m/s
S=70 m
(a) t₁=0.5 s
(b) t₂=1 s
(c) t₃=1.5 s
(d) t₄=2 s
(e) t₅=2.5 s
To find
Displacement S from t₁ to t₅
Velocity V from t₁ to t₅
Solution
According to kinematic equation of motion and given information conclude that v is given by

Also get the equation of displacement

These two formula are used to find velocity as well as displacement for time t₁ to t₅
For t₁=0.5 s

For t₂
For t₃

For t₄

For t₅

Answer:
Explanation:
given that
Radius =0.75m
Cnet=0.13nC
a. Electric field inside the sphere located 0.5m from the center of the sphere.
The electric field located inside the sphere is zero.
b. The electric field located 0.25m beneath the sphere.
Since the radius is 0.75m
Then, the total distance of the electric field from the centre of the circle is 0.75+0.25=1m
Then
E=kq/r2
K=9e9Nm2/C2
q=0.13e-9C
r=1m
Then,
E= 9e9×0.13e-9/1^2
E=1.17N/C. Q.E.D
Answer:
The magnitude of gravitational force between two masses is
.
Explanation:
Given that,
Mass of first lead ball, 
Mass of the other lead ball, 
The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m
We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

So, the magnitude of gravitational force between two masses is
. Hence, this is the required solution.
Answer:
5833.33
Explanation:
= Angular acceleration
= Number of revolutions
= Initial angular speed = 0
t = Time taken = 2 s
Final angular speed

From the equation of rotational motion we have


The number of revolutions is 5833.33