Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
Answer:
0.108 rad/s².
Explanation:
Given that
Initial angular velocity ,ωi = 0 rad/s
Final angular velocity ωf= 0.5 rev/s
We know that
1 rev/s = 6.28 rad/s
ωf= 3.14 rad/s
t= 28.9 s
We know that (if acceleration is constant)
ωf=ωi + α t
α=Angular acceleration
3.14 = 0 + α x 28.9
Therefore the acceleration will be 0.108 rad/s².
Therefore the answer will be 0.108 rad/s².
Answer:
Frquency=3,994Hz
Explanation:
Tension =967N
Density of string (μ)=0.023g/cm
Length of the stretched spring=308cm
Fundamental frequency for nth harmonic :
Fn=n/2L(√T/μ)
Substituting the given values to find the frequency :
f1=1/2(308cm) *(0.01m/1cm)[(√967N)/(0.023g/cm)(0.1kg)/(0.1kg/m)/(1g/cm)]
=6.16m[(√967N)/0.0023kg/m)]
=3,994.20Hz
Approximately,
The frequency will be =3,994Hz
Answer:
63360 mi/h
Explanation:
<u>How to find the speed of an object</u>
Calculate speed, distance, or time using the formula d = st, distance equals speed times time. The Speed Distance Time Calculator can solve for the unknown SDT value given two known values.
Time can be entered or solved for in units of second S (s), minutes (min), hours (hr), or hours and minutes and seconds (hh:mm: ss). See shortcuts for time formats below.
To solve for distance use the formula for distance D = st, or distance equals speed times time.
distance = speed x time
Rate and speed are similar since they both represent some distance per unit time like miles per hour or kilometers per hour. If rate r is the same as speed s, r = s = d/t. You can use the equivalent formula d = rt which means distance equals rate times time.
distance = rate x time
To solve for speed or rate use the formula for speed, s = d/t which means speed equals distance divided by time.
speed = distance/time
To solve for time use the formula for time, t = d/s which means time equals distance divided by speed.
time = distance/speed
Therefore, the speed = 63360 miles per hour
= 63360 mi/h
