Answer:
675 Pa.
Explanation:
F = 5+2cos(15t) kN
Area (a) = 8*10-3 m2
Now at t =4 sec
F= 5+2cos(60)
= 5+2*0.5
= 6 kN
Now ,force efficiency is 90%.
Hence,the effectively transmitted force,
Fe = 0.90*6
= 5.4 kN
Hence,pressure is given as,
P = Fe/a
= 5.4*10^3/(8
*10^-3))
P = 675 Pa....answer
Answer:
Explanation:
Answer: Let ke = 1/2 IW^2 = 1/2 kMr^2 W^2 be Earth's rotational KE. W = 2pi/24 radians per hour rotation speed and k = 2/5 for a solid sphere M is Earth mass, r = 6.4E6 m.
Then ke = 1/2 2/5 6E24 (6.4E6)^2 (2pi/(24*3600))^2 = ? Joules. You can do the math, note W is converted to radians per second for unit consistency.
Let KE = 1/2 KMR^2 w^2 be Earth's orbital KE. w = 2pi/(365*24) radians per hour K = 1 for a point mass. Note I used 365 days, a more precise number is 365.25 days per year, which is why we have Leap Years.
Find KE/ke = 1/2 KMR^2 w^2//1/2 kMr^2 W^2 = (K/k)(w/W)^2 (R/r)^2 = (5/2) (365)^2 (1.5E11/6.4E6)^2 = 7.81E9 ANS
Explanation:
While studying the velocity of a cheetah over time in a spreadsheet program is given by :
y = 2.2 x + 1.2 ...(1)
We know that,
v=v₀+at ...(2)
v₀ is velocity when t = 0, v is velocity after time t, a is acceleration and t is time.
If we consider time t on x-axis and v on y axis, then only we can draw a plot of equation (2). On comparing equation (1) and (2) we get :
a = 2.2 (but it is not correct as we don't know about axes).
Hence, the correct option is (a) "We cant tell without knowing what was plotted on the horizontal and vertical axes"
