Answer:
(a) 1.257 x 10^5 J
(b) 1.456 Watt
Explanation:
Volume of blood, v = 7500 L = 7.5 m^3
Height, h = 1.63 m
density of blood, d = 1.05 x 10^3 kg/m^3
(a) work done = m x g x h
W = v x d x g x h = 7.5 x 1.05 x 1000 x 9.8 x 1.63 = 1.257 x 10^5 J
(b) time = 1 day = 24 x 60 x 60 s = 86400 seconds
Power = Work / time = 1.257 x 10^5 / 86400 = 1.456 Watt
Answer:
<h2>
44 m/s</h2>
Explanation:
In this problem we are expected to calculate the velocity of Georges movements.
Given data
Total distance covered by George= 850+250= 1100 meters
Time taken by George to cover the total distance= 25 seconds
We know that velocity is, v= distance/ time
Therefore substituting our data into the expression for velocity we have
v= 1100/ 25= 44 m/s
Hence the velocity in m/s is 44
Answer:
19.3m/s
Explanation:
Use third equation of motion

where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity
insert values to get answer
![v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s](https://tex.z-dn.net/?f=v%5E2-0%5E2%3D2%289.81m%2Fs%5E2%29%2838%2F2%29%5C%5Cv%5E2%3D9.81m%2Fs%5E2%20%2A38%5C%5Cv%5E2%3D372.78%5C%5Cv%3D%5Csqrt%5B%5D%7B372.78%7D%20%5C%5Cv%3D19.3m%2Fs)
Answer:
The intensity of the electric field is

Explanation:
The electric field equation is given by:

Where:
- k is the Coulomb constant
- q is the charge at 0.4100 m from the balloon
- d is the distance from the charge to the balloon
As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.
Then, let's use the equation written above.


I hope it helps you!
Answer:
Law 1. A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.
Law 2. A body acted upon by a force moves in such a manner that the time rate of change of momentum equals the force.
Law 3. If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction.