Answer:
Given that
ωo = 4.7 rad/s
α = −0.25 rad/s²
θ₀ = 0
a)
For maximum turn angle ,the final angular (ω)speed of wheel should be zero.
ω² = ω²o - 2 α θ
0² = 4.7² - 2 x 0.25 x θ
θ=44.18 rad ≅44 rad ( max)
b) and c)
We know that
θ = 1 /2 θmax
θ = 22 rad
By solving above equation we get
t= 32.12 s
t=5.47 s
d) and e)
θ = -9.9 rad
By solving above equation we get
t= - s
t= 39.6 s
Answer:
The force felt by charge 3 is F=(-5.6*10⁻⁶,3.36⁻⁵)N
Explanation:
As the superposition principle applies to static charges, we can find the net electric force as the sum of the two forces felt by q3.
Looking at the drawing and knowing that they form an equilateral triangle of lenght 4 we can conclude that each internal angle is 60°.
So, the positions in our coordinate system are:
Now using Coulomb's force:
Where d=4, q1 = -7.8*10⁻⁹C, q2 = -15.6 *10⁻⁹C, q3 = 8.0 *10⁻⁹C, k=8.98*10⁹, e0=8.8*10¹⁰:
Replacing we get 2 equations:
To work with the sam
F=∑F_i=3.5*10⁻⁴(0.023,0.032)+7*10⁻⁴(-0.016,0.032)=
=((3.5*10⁻⁴-7*10⁻⁴)*0.016,(3.5*10⁻⁴+7*10⁻⁴)*0.032)=
F=(-5.6*10⁻⁶,3.36⁻⁵)N
Answer:
-No
Explanation:
Given that
Mass of box = 100 kg
Coefficient of friction ,μ= 0.23
We know that friction force depends on the normal force acts on the box
Fr= μ N
When we pull the box then :
Normal force N= mg
Friction force Fr= μ mg
When we push the box :
Lets take pushing force = F
θ =Angle make by pushing force from the vertical line
Normal force
N = mg + F cosθ
Fr= μ ( mg + F cosθ )
The friction force is more when we push the box.That is why this is not easier to push the box.
Therefore answer is ---No
