What do you mean? I'm confused... You need to put the rest of the question
Answer:
1317.4 m
Explanation:
We are given that
Angle=
Initial speed =
We have to find the horizontal distance covered by the shell after 5.03 s.
Horizontal component of initial speed=
Vertical component of initial speed=
Time=t=5.03 s
Horizontal distance =
Using the formula
Horizontal distance=
Horizontal distance=1317.4 m
Hence, the horizontal distance covered by the shell=1317.4 m
Answer:
Please find the answer in the explanation.
Explanation:
Given that two porters are available to carry a long timber wood.out of them one is weak. how do you make less load to the weak one?
We can make the weak one to carry less load through two different ways or means.
First, if we can locate the centre of gravity and centre of mass of the long timbre wood, the week one can carry the other end away from the center of gravity and centre of mass.
Second, the strong porter can carry the long timbre wood almost to the fulcrum and allow the weak one to support at the other end. By doing this, the weak one will only carry light portion of the load.
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:
Capacitance is 0.572×10⁻¹⁰ Farad
Explanation:
Radius = R₁ = 6.25 cm = 6.25×10⁻² m
Radius = R₂ = 15 cm = 15×10⁻² m
Dielectric constant = k = 4.8
Electric constant = ε₀ = 8.854×10⁻¹² F/m
ε/ε₀=k
ε=kε₀

∴ Capacitance is 0.572×10⁻¹⁰ Farad