Answer:
b.it depends on the distance it falls
<span>The question is asking for the reason why water is considered a polar molecule. The correct option is A. Water molecule is a polar molecule because it has one side which is positively charged and the other side is negatively charged. Each molecule of water has two hydrogen atom and one oxygen atom. The slight negative charge near the oxygen atom attract nearby hydrogen atoms from water and the slightly positive charge of the hydrogen atom attract oxygen atom. The attraction set up by these charges are responsible for the polarity of water molecules. </span>
An interesting problem, and thanks to the precise heading you put for the question.
We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.
Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m
Assume g = -9.81 m/s^2
initial velocity, v m/s (to be determined)
Solution:
(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation,
S(T)=800+(vy)T+(1/2)aT^2 ....(1)
Where S is height measured from ground.
substitute values in (1): S(20)=800+(0.8v)T+(-9.81)T^2 =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s for T=20 s
(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s
Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m
(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m
(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T =>
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)
vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s (magnitude)
in direction theta = atan(43.575,138.1)
= 17.5 degrees with the vertical, downward and forward. (direction)
Okay, haven't done physics in years, let's see if I remember this.
So Coulomb's Law states that
![F = k \frac{Q_1Q_2}{d^2}](https://tex.z-dn.net/?f=F%20%3D%20k%20%5Cfrac%7BQ_1Q_2%7D%7Bd%5E2%7D)
so if we double the charge on
![Q_1](https://tex.z-dn.net/?f=Q_1)
and double the distance to
![(2d)](https://tex.z-dn.net/?f=%282d%29)
we plug these into the equation to find
<span>
![F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}](https://tex.z-dn.net/?f=F_%7Bnew%7D%20%3D%20k%20%5Cfrac%7B2Q_1Q_2%7D%7B%282d%29%5E2%7D%3Dk%20%5Cfrac%7B2Q_1Q_2%7D%7B4d%5E2%7D%20%3D%20%5Cfrac%7B2%7D%7B4%7D%20%5Ccdot%20k%20%5Cfrac%7BQ_1Q_2%7D%7Bd%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%20F_%7Bold%7D)
</span>
So we see the new force is exactly 1/2 of the old force so your answer should be
![\frac{1}{2}F](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DF)
if I can remember my physics correctly.
Answer:
Newton's second law is F=ma. And since inertia is the resistance an object has against a change in velocity or force, you must push a lot harder to start moving an object, but once it starts to move, it is a lot easier, until you go to stop it.
Explanation: