In the united states a standard letter sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for
1 answer:
You might be interested in
Answer:
(a)Average velocity ,v =128.74 Km/hr
(b)Kinetic Energy , K=958546.875 Joule
(c)Distance, s=268.8m
(d)Acceleration, a= - 2.38
<u>Explanation</u>:
<u>Given</u>:
Distance travelled = 40 miles
Time taken = 30 minutes.
(A) The average velocity in kilometres/hour
Converting 40 miles into km ,
we know that,
1 mile = 1.60934
40 miles = 40 x 1.60934
so 40 miles = 64.3738 Km
similarly converting 30 minutes into hours
1 minute =
30 minute =
30 minute =
Now
Average velocity =
Substituting Values,
Average velocity =
Average velocity =
Average velocity =128.74 Km/hr
(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?
Converting 1.5 tons into kg we get
1 ton = 1000 kg
so 1.5 ton =1500 kg
converting velocity to m/s
=>35.75 m/s----------------------------------------------------------(1)
kinetic energy K=
Substituting the values,
K=
K=
K=
K=
K=958546.875 Joule---------------------------------------------(2)
(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping
Lets use Distance formula,
Substituting the known values,
-------------------------------------(3)
(D) What is the average acceleration of the car (in m/s2) during braking?
Using the formula
v=u +at
re arranging the formula we get,
Substituting the values
a= - 2.38 ----------------------------------------(4)
Now substituting 4 in 3 we get
s=268.8m--------------------------------------------------------------(5)
Answer:
v_{f} = 115.95 m / s
Explanation:
This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions
Thrust =
-v₀ = v_{e}
where v_{e} is the velocity of the gases relative to the rocket
let's apply these expressions to our case
the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units
M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg
The final mass is the mass of the engines + the mass of the rocket
M_{f} = 25.5 +54.5 = 80 g = 0.080 kg
thrust and duration of ignition are given
thrust = 5.26 N
t = 1.90 s
Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear
thrust = v_{e}
v_{e} = thrust
v_{e} = 5.26
v_{e} = - 786.93 m / s
the negative sign indicates that the direction of the gases is opposite to the direction of the rocket
now we look for the final speed of the rocket, which as part of rest its initial speed is zero
v_{f}-0 = v_{e}
we calculate
v_{f} = 786.93 ln (0.0927 / 0.080)
v_{f} = 115.95 m / s