Answer:
The kangaroo was 1.164s in the air before returning to Earth
Explanation:
For this we are going to use the equation of distance for an uniformly accelerated movement, that is:
Where:
x = Final distance
xo = Initial point
Vo = Initial velocity
a = Acceleration
t = time
We have the following values:
x = 1.66m
xo = 0m (the kangaroo starts from the floor)
Vo = 0 m/s (each jump starts from the floor and from a resting position)
a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)
t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.
Now replace the values in the equation
It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is
t = 0.582s + 0.582s
t = 1.164s
The kangaroo was 1.164s in the air before returning to Earth
Some of the most fundamental concepts of fluid properties are temperature, density, and composition.
Answer:
New pressure (P2) = 22.92 Kpa (Approx)
Explanation:
Given:
Old pressure (P1) = 22 Kpa
Temperature (T1) = 287 K
Temperature (T2) = 299 K
Find:
New pressure (P2)
Computation:
P1 / T1 = P2 / T2
22 / 287 = P2 / 299
New pressure (P2) = 22.92 Kpa (Approx)
A is the answer mass and force
Explanation:
The law of repulsion is given by Coulomb. The mathematical form of Coulomb law is given by :
...............(1)
Where
F is the force
k is the electrostatic constant
are electric charges
r is the distance between charges
The Newton's law of universal gravitation is given by :
..............(2)
G is the universal gravitational constant
From equation (1) and (2) it is clear that both law obeys inverse square law and both are of same type. So, the law of repulsion by Coulomb agrees with the Newton's law of universal gravitation.
