Answer:
Av = 25 [m/s]
Explanation:
To solve this problem we must use the definition of speed, which is defined as the relationship between distance over time. for this case we have.
where:
Av = speed [km/h] or [m/s]
distance = 180 [km]
time = 2 [hr]
Therefore the speed is equal to:
![Av = \frac{180}{2} \\Av = 90 [km/h]](https://tex.z-dn.net/?f=Av%20%3D%20%5Cfrac%7B180%7D%7B2%7D%20%5C%5CAv%20%3D%2090%20%5Bkm%2Fh%5D)
Now we must convert from kilometers per hour to meters per second
![90[\frac{km}{h}]*1000[\frac{m}{1km}]*1[\frac{h}{3600s} ]= 25 [m/s]](https://tex.z-dn.net/?f=90%5B%5Cfrac%7Bkm%7D%7Bh%7D%5D%2A1000%5B%5Cfrac%7Bm%7D%7B1km%7D%5D%2A1%5B%5Cfrac%7Bh%7D%7B3600s%7D%20%5D%3D%2025%20%5Bm%2Fs%5D)
During cytokinesis, the cytoplasm of the cell is divided in half, and the cell membrane grows to enclose each cell, forming two separate cells as a result. The end result of mitosis and cytokinesis is two genetically identical cells where only one cell existed before.
Answer:
1.73 m/s²
Explanation:
Given:
Δx = 250 m
v₀ = 0 m/s
t = 17 s
Find: a
Δx = v₀ t + ½ at²
250 m = (0 m/s) (17 s) + ½ a (17 s)²
a = 1.73 m/s²
Draw a diagram to illustrate the problem as shown below.
The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s
Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s
The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
= 21.675 m
Answer: The height is 21.7 m (nearest tenth)
Answer:
0.65 kg*m/s and 0.165 kg*m/s
Explanation:
Step one:
given data
mass m= 0.5kg
initial velolcity u=1.3m/s
final velocity v= 0.97m/s
Required
The change in momentum
Step two:
We know that the expression for impulse is given as
Ft= mv
Ft= 0.5*1.3
Ft= 0.65 kg*m/s
The expression for the change in momentum is given as
P= mΔv
substitute
Pt= 0.5*(1.3-0.97)
Pt= 0.5*0.33
Pt=0.165 kg*m/s
