It is the acceleration of an object in free fall
Explanation:
When an object is in free fall, it is subjected only to one force: the force of gravity, which pulls the object downward, with a magnitude (near the Earth's surface) which is given by
where
m is the mass of the object
is the acceleration due to gravity
We can apply Newton's second law to the object in free fall:
where
F is the net force on the object
a is the acceleration of the object
m is the mass
However, since there is only the force of gravity acting on the object, the net force is equal to the force of gravity: so we can equate the two equations, obtaining that
Which means that the acceleration of an object in free fall (acted upon the force of gravity only) is equal to the acceleration due to gravity, .
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Answer:
No number answer; don't want to pull out a calculator lol
Explanation:
Capacitors are added in parallel; opposite of resistors.
So for 9, add 10 and 2.5 then do ((1/12.5)+(1/.3))^-1.
For 10, add 0.75 and 15 first, then the rest is the same idea as 9.
False because your deltoids are in your shoulders not your back
(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.
(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.
<h3>
Magnitude of electric field </h3>
The magnitude of electric field is given by the following equation.
F = qE
But F = mg
mg = qE
E = mg/q
where;
- E is the electric field
- m is mass of the particle
- g is acceleration due to gravity
- q is charge of the particle
<h3>For an electron</h3>
E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)
E = 5.57 x 10⁻¹¹ N/C
<h3>For proton</h3>
E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)
E = 1.02 x 10⁻⁷ N/C
Thus, the required vertical electric field is greater when the charge is proton.
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