The metals are the elements that more easily can lose electrons to fill their outermost shell. And among the metals the alkalyne are most likely than alkalyne earth and these are most likely than other metals.
As less electrons the metal has in its outermost shell the more likely it will lose an electron fo fill its outermost shell.
So, Li, Na, K, Rb, Cs, and Fr have one electron in their outermost shell, so they are more likely to lose an electron to fill their outermost shell than Be, Mg, Ca, Sr, Ba, Ram which have two electrons in their outermost shell.
Using the same reasoning, Be, Mg, Ca, Sr, Ba, Ra are more likely to lose an electron to fill its outermost shell than Al, Ga, In, Tl.
Al, Ga, In, Tl are more likely to lose an electron than Si, Ge, Sn, Pb.
Si, Ge, Sn, Pb are more likely to lose an electron than O, S, Se, Te, Po
O, S, Se, Te, Po are more likely to lose an electron than F, Cl, Br, I, At
F, Cl, Br, I, At (halogens) and He, Ne, Ar, Kr, Xe, Rn (noble gases) will not likely lose electrons.
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Answer: A proton would experience the same magnitude of force.
Explanation: The equation relating Force F charge Q and Electric Field E, is given as
E = F/Q
The magnitude of the force felt would be the same since + and - charge as same magnitude from the question (-1C and +1C). But the direction of force of the proton which as a positive charge would be towards/same direction as the Electric Field.
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