Answer:
a)54L/min
b)0.845
Explanation:
a) A x V=
where suffix 1,2,3 refers to the three pipes.
=27L/min+16L/min+11 L/min
=54L/min
b) A x V=54L/min => x v
d= 2 cm
x v = 54
v= x
-> x =27L/min => x
= 1.3cm
x = 27
= x
Next is to find the ratio of speed i.e
x / x =>
= 0.845
Answer: The magnitude of force per length that each wire exert on the other wire is 2.67×10^-5 N/m.
The force is repulsive.
Explanation: Please see the attachments below
Answer:
Distance, d = 0.1 m
It is given that,
Initial velocity of meson,
Finally, the meson is coming to rest v = 0
Acceleration of the meson, (opposite to initial velocity)
Using third equation of motion as :
s is the distance the meson travelled before coming to rest.
So,
s = 0.1 m
The meson will cover the distance of 0.1 m before coming to rest. Hence, this is the required solution.
Answer: 459.14 N
Explanation:
from the question, we have
diameter = 10 m
radius (r) = 5 m
weight (Fw) = 670 N
time (t) = 8 seconds
Circular motion has centripetal force and acceleration pointing perpendicular and inwards of the path, therefore we apply the equation below
∑ F = F c = F w − Fn ..............equation 1
Fn = Fw − Fc = mg − (mv^2 / r) ...................equation 2
substituting the value of v as (2πr / T) we now have
Fn = mg − (m(2πr / T )^2) / r
Fn= mg − (4(π^2)mr / T^2) ..........equation 3
Fw (mass of the person) = mg
therefore m = Fw / g
m = 670 / 9.8 = 68.367 kg
now substituting our values into equation 3
Fn = 670 - ( (4 x (π^2) x 68.367 x 5 ) / 8^2)
Fn = 670 - 210.86
Fn = 459.14 N
Answer:
350J
Explanation:
Given parameters:
Weight of bag = 20N
Distance moved horizontally = 35m
Force applied = 10N
Unknown:
Work done on the bag = ?
Solution:
Work done is the force applied to move a body through given distance.
Work done = Force applied x distance
So;
Work done = 10 x 35 = 350J