answer: derived physical quantities are those quantities that are obtained from the basic physical quantities by multiplication or division and area is one of them
A & C
Explanation:
The combination of the earth's weak gravity and its closeness to the sun does not allow it to hold hydrogen and helium gases in its atmosphere. Its relative closeness to the sun means it is hot enough such that the helium and hydrogen molecules would have high kinetic energy. Remember that gravity acts strongly on larger masses, therefore it would require very strong gravity to have an influence on lighter gas molecules like hydrogen and helium let alone when they have a high kinetic energy. This means these molecules can easily escape the earth’s atmosphere into space.
Planets that are larger (meaning they have a stronger gravity) and farther from the sun (meaning these molecules won't have a very high kinetic energy) are able to hold these lighter gases in their atmosphere. Examples of such planets are Jupiter.
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Answer:
a) 520m
b) 10.30 s
c) 100,95 m/s
Explanation:
a) According the given information, the rocket suddenly stops when it reach the height of 520m, because the engines fail, and then it begins the free fall.
This means the maximum height this rocket reached before falling was 520 m.
b) As we are dealing with constant acceleration (due gravity) 
we can use the following formula:
(1)
Where:
is the initial height of the rocket (at the exact moment in which it stops due engines fail)
is the final height of the rocket (when it finally hits the launch pad)
is the initial velocity of the rocket (at the exact moment in which it stops the velocity is zero and then it begins to fall)
is the acceleration due gravity
is the time it takes to the rocket to hit the launch pad
Clearing :
(2)
(3)
(4)
(5) This is the time
c) Now we need to find the final velocity 
for this rocket, and the following equation will be perfect to find it:
(6)
(7)
(8) This is the final velocity of the rocket. Note the negative sign indicates its direction is downwards (to the launch pad)
Answer:
Largest
[q1=+1nC, q2=-1nC, q3=-1nC] & [q1=-1nC, q2=+1nC, q3=+1nC]
[q1=+1nC, q2=-1nC, q3=+1nC]
[q1=+1nC, q2=+1nC, q3=-1nC]
Smallest
[q1=+1nC, q2=+1nC, q3=+1nC] & [q1=-1nC, q2=-1nC, q3=-1nC]
Explanation:
The force required to punch a hole 1 cm in diameter in a steel sheet is 0.176 x 10¹³ N.
<h3>What is force?</h3>
The force is defined as the shear stress or pressure applied per unit area.
F = P/A
Given is a steel sheet whose thickness t = 5mm = 5x 10⁻³ m, shearing strength P = 2.76 × 10⁸ N/m², diameter of hole d = 1cm = 1x 10⁻² m
The area to be punched A = πd xt
A = π x 1x 10⁻² x5x 10⁻³
A = 15.7 x 10⁻⁵ m²
The force required to punch a hole is
F = 2.76 × 10⁸ / 15.7 x 10⁻⁵
F = 0.176 x 10¹³ N.
Thus, the punching force is 0.176 x 10¹³ N.
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