Answer: 1.5×10^10 N/C
Explanation:
E= F/q
Where E= magnitude of the electric field
F= force of attraction
q= charge of the given body
Given F= 6.5×10^-8 N
q= 4.3× 10^-18 C
Therefore, E = 6.5×10 ^-8/ 4.3×10^-18
E = 1.5×10^10 N/C
The model bridge captures all the structural attributes of the real bridge, at a reduced scale.
Part a.
Note that volume is proportional to the cube of length. Therefore the actual bridge will have 100^3 = 10^6 times the mass of the model bridge.
Because the model bridge weighs 50 N, the real bridge weighs
(50 N)*10^6 = 50 MN.
Part b.
The model bridge matches the structural characteristics of the actual bridge.
Therefore the real bridge will not sag either.
Answer:
The energy is 
(a) is correct option
Explanation:
Given that,
Energy = 4480 j
Weight of nitrogen = 20 g
Boil temperature = 77 K
Pressure = 1 atm
We need to calculate the internal energy
Using first law of thermodynamics


Put the value into the formula



We need to calculate the number of molecules in 20 g N₂
Using formula of number of molecules

Put the value into the formula


We need to calculate the energy
Using formula of energy

Put the value into the formula


Hence, The energy is 
Answer:
A 75.1 N and a direction of 152° to the vertical.
B 85.0 N at 0° to the vertical.
Explanation:
A) The interaction partner of this normal force has what magnitude and direction?
The interaction partner of this normal force is the component of the weight of the crate perpendicular to the ramp. <u>It has a magnitude of 85cos28° = 75.1 N and a direction of 180° - 28° = 152° to the vertical(since it is directed downwards perpendicular to the ramp).</u>
B) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?
Since this force has to balance the weight of the crate, its magnitude is 85.0 N. Its direction has to be vertically opposite to that of the weight.
Since the weight is 180° to the vertical (since it is directed downwards), this force is 0° to the vertical.
<u>So, this force has a magnitude of 85.0 N and a direction of 0° to the vertical.</u>