Answer:
Force of 37.8 × 10^(6) N attracts the two charges
Explanation:
The force between two charges is given by
F = k*q1*q2/r²
Where q1 and q2 are 0.06 C and 0.07 C.
r is the distance between q1 and q2 which is equal to 3 m
k is a constant = 9 × 10^(9) N.m²/C²
F = (9 × 10^(9) × 0.06 × 0.07)/3²
F = 37.8 × 10^(6) N
Answer:
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Answer:
7 orbitals are allowed in a sub shell if the angular momentum quantum number for electrons in that sub shell is 3.
Explanation:
We that different values of m for a given l provide the total number of ways in which a given s, p,d and f sub shells in presence of magnetic field can be arranged in space along x, y ,z- axis or total number of orbitals into which a given subshell can be divided.
Range for given l lies between -l to +l .
The possible values of m are -3 , -2 , -1 , 0 , 1 ,2 , 3 .
Total number of orbitals are 7.
Answer:
d. None of the above.
Explanation:
In a parabolic motion, you have that in the complete trajectory the component velocity is constant and the vertical component changes in time. Then, the total velocity vector is not zero.
In the complete trajectory the gravitational acceleration is always present. Then, the grasshopper's acceleration vector is not zero.
At the top of the arc the grasshopper is not at equilibrium because the gravitational force is constantly acting on the grasshopper.
Then, the correct answer is:
d. None of the above.
Answer:
a) F = 4.9 10⁴ N, b) F₁ = 122.5 N
Explanation:
To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height
1) pressure is defined by the relation
P = F / A
to lift the weight of the truck the force of the piston must be equal to the weight of the truck
∑F = 0
F-W = 0
F = W = mg
F = 5000 9.8
F = 4.9 10⁴ N
the area of the pisto is
A = pi r²
A = pi d² / 4
A = pi 1 ^ 2/4
A = 0.7854 m²
pressure is
P = 4.9 104 / 0.7854
P = 3.85 104 Pa
2) Let's find a point with the same height on the two pistons, the pressure is the same
where subscript 1 is for the small piston and subscript 2 is for the large piston
F₁ = 
the force applied must be equal to the weight of the truck
F₁ =
F₁ = (0.05 / 1) ² 5000 9.8
F₁ = 122.5 N