Answer : The average speed of the sprinter is, 34.95 Km/hr
Solution :
Average velocity : It is defined as the distance traveled by the time taken.
Formula used for average velocity :
where,
= average velocity
d = distance traveled = 200 m
t = time taken = 20.6 s
Now put all the given values in the above formula, we get the average velocity of the sprinter.
conversion :
(1 Km = 1000m)
(1 hr = 3600 s)
Therefore, the average speed of the sprinter is, 34.95 Km/hr
Answer and Explanation:
Parallax method is used for finding the distance of objects in space there are two types of parallax method that is stellar parallax and trigonometric parallax.The disadvantage of using parallax method is that it can can not reach so far in the Galaxy due to this reason parallax method is generally not used for measuring distance in galaxy.
Answer:
The answer is 218
Explanation:
Weight = mass * gravitational acceleration
weight is represented by F
F = 25kg (8.7)
(I'm pretty sure that you don't have to include the meters per second/per second thing)
<span>3.36x10^5 Pascals
The ideal gas law is
PV=nRT
where
P = Pressure
V = Volume
n = number of moles of gas particles
R = Ideal gas constant
T = Absolute temperature
Since n and R will remain constant, let's divide both sides of the equation by T, getting
PV=nRT
PV/T=nR
Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation
P1V1/T1 = P2V2/T2
where
P1, V1, T1 = Initial pressure, volume, temperature
P2, V2, T2 = Final pressure, volume, temperature
Now convert the temperatures to absolute temperature by adding 273.15 to both of them.
T1 = 27 + 273.15 = 300.15
T2 = 157 + 273.15 = 430.15
Substitute the known values into the equation
1.5E5*0.75/300.15 = P2*0.48/430.15
And solve for P2
1.5E5*0.75/300.15 = P2*0.48/430.15
430.15 * 1.5E5*0.75/300.15 = P2*0.48
64522500*0.75/300.15 = P2*0.48
48391875/300.15 = P2*0.48
161225.6372 = P2*0.48
161225.6372/0.48 = P2
335886.7441 = P2
Rounding to 3 significant figures gives 3.36x10^5 Pascals.
(technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>
