I'm pretty sure the answer is b 28n hope helps :)
Answer:
Explanation:
The variables we know and are given are:
time, t = 20s
Charge, Q = 3x1-^-6 electrons, which is just 3x10^-6C (C stands for Coulombs, which is the unit for Charge)
We need to find the current, I, and since we know Q and t we can substitute these values into the given equation:
I=Q/t (which if you look at what the RHS is saying, its Charge over time, or more literally means the amount of charge passing a point over a period of time)
If we substitute these values, we will get I as:
I = Q / t
I = 3x10^-6 / 20
I = 1.5x10^-7 A
Hope this helps!
Answer:
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Explanation:
We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is
w² = mg d / I
In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow
d = L / 2
The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated
I = ¼ m r2 + ⅓ m L2
I = m (¼ r2 + ⅓ L2)
now let's use the concept of density to calculate the mass of the system
ρ = m / V
m = ρ V
the volume of a cylinder is
V = π r² L
m = ρ π r² L
let's substitute
w² = m g (L / 2) / m (¼ r² + ⅓ L²)
w² = g L / (½ r² + 2/3 L²)
L >> r
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Answer:
distance cover is = 102.53 m
Explanation:
Given data:
speed of object is 17.1 m/s
from equation of motion we know that
where d_1 is distance covered in time t1
so
=
where d_2 is distance covered in time t2
distance cover is = 213.31 - 110.78 = 102.53 m
Answer:
734.215N
Explanation:
First we calculate the angle that corresponds to a 5% slope using the Tan-1 function
then we use the component that corresponds to the direction parallel to the road, additionally we must multiply by the gravity value to find the weight(g=9.81m/s^2)
Wx=M*g*sen(2.86)=1500kg*9.81*sen(2.86)=734.215N
