V o = 6 m/s,
t = 2 s
v = 10 m/s
v = v o + a t
a t = v - v o
a = ( v - v o ) / t
a = ( 10 m/s - 6 m/s ) / 2 s = 4 m/s / 2 s = 2 m/s²
Answer:
The runner`s acceleration is 2 m/s².
Explanation:
Formula for steady flow energy equation for the flow of fluid is as follows.
![m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w](https://tex.z-dn.net/?f=m%5Bh_%7B1%7D%20%2B%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2%7D%5D%20%2B%20z_%7B1%7Dg%5D%20%2B%20q%20%3D%20m%5Bh_%7B1%7D%20%2B%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2%7D%20%2B%20z_%7B1%7Dg%5D%20%2B%20w)
Now, we will substitute 0 for both
and
, 0 for w, 334.9 kJ/kg for
, 2726.5 kJ/kg for
, 5 m/s for
and 220 m/s for
.
Putting the given values into the above formula as follows.
![1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0](https://tex.z-dn.net/?f=1%20%5Ctimes%20%5B334.9%20%5Ctimes%2010%5E%7B3%7D%20J%2Fkg%20%2B%20%5Cfrac%7B%285%20m%2Fs%29%5E%7B2%7D%7D%7B2%7D%20%2B%200%5D%20%2B%20q%20%3D%201%20%5Ctimes%20%5B2726.5%20%5Ctimes%2010%5E%7B3%7D%20%2B%20%5Cfrac%7B%28220%20m%2Fs%29%5E%7B2%7D%7D%7B2%7D%20%2B%200%5D%20%2B%200)
q = 6597.711 kJ
Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.
Kinetic energy is directly proportional to the square of their velocities... so if velocity has been increased 3 times, increase in K.E would be: 3^2 = 9
In short, Your Answer would be: KE2 = 9 * KE1
Hope this helps!
Answer:
7.5 x 10⁻⁸N
Explanation:
Given parameters:
Mass 1 = 60kg
Mass 2 = 75kg
Distance between the bodies = 2m
Unknown:
Gravitational fore = ?
Solution:
The gravitational force between the two bodies can be derived using;
F =
G is the universal gravitation constant = 6.67 x 10⁻¹¹m³kg⁻¹s⁻²
Insert the parameters and solve;
F =
= 7.5 x 10⁻⁸N
Explanation:
There are two forces acting upon the book. One force - the Earth's gravitational pull - exerts a downward force. The other force - the push of the table on the book (sometimes referred to as a normal force) - pushes upward on the book.