Answer:
10.032 N.
Explanation:
From the question given above, the following data were obtainedb
Mass (m) = 2.5 Kg
Final length = 10.0 cm
Original length = 21.4 cm
Spring constant (K) = 88 N/m
Force (N) =?
Next, we shall determine the compression of the spring. This can be obtained as follow:
Final length = 10.0 cm
Original length = 21.4 cm
Compression (e) =?
e = Original length – final length
e = 21.4 – 10
e = 11.4 cm
Next, we shall convert 11.4 cm to m. This can be obtained as follow:
100 cm = 1 m
Therefore,
11.4cm = 11.4 cm × 1 m / 100 cm
10 cm = 0.114 m
Finally, we shall determine the force. This can be obtained as illustrated below:
Compression (e) = 0.114 m
Spring constant (K) = 88 N/m
Force (N) =?
F = Ke
F = 88 × 0.114
F = 10.032 N
Thus, the force in the spring is 10.032 N
If I've understood it correctly, the right one should look like this: In most cultures, the difference between the majority viewpoint and the minority viewpoint is that the majority group is made to feel normal and the minority group is made to feel divergent.
Answer:
156.1 rad/s = 24.8 rev/s
Explanation:
Torque = Momentum of inertial × radial acceleration = Iα
τ = 26.6 N.m
I = 0.162 kg.m²
26.6 = 0.162 × α
α = 164.2 rad/s²
Using equations of motion,
θ = 11.8 rev = 11.8 × 2π = 74.2 rad
w₀ = 0 rad/s (since the grindstone starts from rest)
w = ?
α = 164.2 rad/s²
w² = w₀² + 2(α)(θ)
w² = 0² + (2×164.2)(74.2)
w = 156.1 rad/s = 24.8 rev/s
Hope this Helps!!!
Answer:
4 times greater
Explanation:
<u>Step 1:</u> Calculate light-collecting area of a 20-meter telescope (A₁) by using area of a circle.
Area of circle = π*r² =
Where d is the diameter of the circle = 20-m
A₁ = 314.2 m²
<u>Step 2:</u> Calculate light-collecting area of a 10-meter Keck telescope (A₂)
Where d is the diameter of the circle = 10-m
A₂ = 78.55 m²
<u>Step 3</u>: divide A₁ by A₂
= 4
Therefor, the 20-meter telescope light-collecting area would be 4 times greater than that of the 10-meter Keck telescope.
Answer:
Explanation:
Given that,
Charge density is λ = 12 nC/m
And radius 3cm
r=0.03m.
The charge density of a along a circular arc is given as
λ= Q/πr
Then, Q=πrλ
Q=π×0.03×12×10^-9
Q=1.131×10^-9 C
Q=1.131 nC
Then, electric field along x axis is symmetrical and if cancels out
Now, Ey is in the negative direction
Electric field is given as,
Ey=-2kQ/πr²
K is constant =9×10^9Nm²/C²
Ey=-2×9×10^9×1.131×10^-9/(π ×0.03²)
Ey=7200 N/C.
The direction is negative direction of y axis, check attachment for diagram.
b. Electric potential at the origin is given as
V=Ed
d=r=0.03
V=7200×0.03
V=216V
