Answer:
313.92w
Explanation:
Formula for power:
P=W/∆t = Fv
Givens:
m=20kg
∆y=4.0m
∆t=2.5s
a=9.81m/s²
In order to find power, we first need to solve for work.
W=Fd (force*displacement), f=mg
W=mg∆y
W=(20kg)(9.81m/s²)(4.0m)
W=784.8J
P=W/∆t
P=784.8J/2.5s
P=313.92 watts
Answer:
you cant really give any answers because i cant tell u were to drag
Answer:
A body travels 10 meters during the first 5 seconds of its travel,and a total of 30 meters over the first 10 seconds of its travel
20miles / 5sec = 4miles /sec would be the average speed for the last 20 m
Explanation:
The answer is 4 m/s.
In the first 5 seconds, a body travelled 10 meters. In the first 10 seconds of the travel, the body travelled a total of 30 meters, which means that in the last 5 seconds, it travelled 20 meters (30m + 10m).
The relation of speed (v), distance (d), and time (t) can be expressed as:
v = d/t
We need to calculate the speed of the second 5 seconds of the travel:
d = 20 m (total 30 meters - first 10 meters)
t = 5 s (time from t = 5 seconds to t = 10 seconds)
Thus:
v = 20m / 5s = 4 m/s
PLEASE GIVE BRAINIEST!! HOPE THIS HELPS
Hello!
First one we can use that PE=mgh so we have
4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m
Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have
g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2
Hope this helps. Any questions please ask. Thank you.
<u>The two ways to find acceleration in non uniform motion are as follows:</u>
<u>Explanation:</u>
Non-uniform acceleration comprises the most common description of motion. Acceleration refers to the rate of changes of velocity per unit time. Basically, it implies that acceleration changes during motion. This variety can be communicated either as far as position (x) or time (t).
Accordingly, non-uniform acceleration motion can be carried out in 2 ways:
Calculus analysis is general and accurate, but limited to the availability of speed and acceleration expressions. It is not always possible to get the expression of motion attributes in the form "x" or "t". On the other hand, the graphic method is not accurate enough, but it can be used accurately if the graphic has the correct shapes.
The use of calculations involves differentiation and integration. Integration enables evaluation of the expression of acceleration of speed and expression of movement at a distance. Similarly, differentiation allows us to evaluate expression of speed position and expression speed to acceleration.
