Gravity on the surface = 4 m/s^2
Now, the acceleration due to centripetal motion, a = v^2/R
Where,
v= 10^3 m/s, R = 10^6 m
Then,
a = (10^3)^2/(10^6) = 1 m^2/s
The net gravitational acceleration = 4-1 = 3 m/s^2
The reading on the spring scale = ma = 40*3 = 120 N
Distance (s)=100m
time
T₁=10
T₂=10.2
T₃=10.4
T₄=10.6
by v=

we get
V₁=

V₁=10m/s
V₂=

V₂=9.8m/s
V₃=

V₃=9.61 m/s
V₄=

V₄=9.43 m/s
V₁:V₂:V₃:V₄ = 10 : 9.8 : 9.61 : 9.43
The value of the thrust developed by the engine of a boeing 777 in N and Kgf are ;
i) 376616N
ii) 38430Kgf
<h3>What is force?</h3>
Force is a push or a pull. The reactive force always serve to balance the applied force. We are here asked to convert the the thrust developed by the engine of a boeing 777 which is about 85400 lbf to the following units;
i) N
ii)kgf
Thus;
1 Ib = 0.45 Kg
1 lbf = 0.45 Kg * 9.8 m/s^2 = 4.41 N
We know that;
1 lbf = 4.41 N
85400 lbf = 85400 lbf. * 4.41 N/1 lbf
= 376616N
Again;
1 lbf = 0.45 Kgf
85400 lbf = 85400 lbf * 0.45 Kgf/1 lbf
= 38430Kgf
Learn more about force:brainly.com/question/13191643
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A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. The change in the electric potential energy of the proton-field system when the proton travels to x = 2.50m is -3.40 × 10⁻¹⁶ J (Option B)
<h3 /><h3>
How is the change in electric potential energy of the proton-field system calculated?</h3>
- Work done on the proton =Negative of the change in the electric potential energy of the proton field
- In the given case, W = -qΔV
- -W = qΔV
- = qEcosθ
- Therefore, work done on the proton = -e(8.50×
N/C)(2.5m)(1) - = -3.40×
J - Any change in the potential energy indicates the work done by the proton.
- Therefore the positive sign shows that the potential energy increases when the proton does the work.
- The negative sign shows that the potential energy decreases when the proton does the work.
To learn more about electric potential energy, refer
brainly.com/question/14306881
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