There is still air inside of a house, which is pushing the roof upwards, so the forces are equal and the roof is not crushed.
Potential energy is first transformed into kinetic energy as she pedals, then gravitational as she coasts down the hill.
Assume no air resistance, and g = 9.8 m/s².
Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°
The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s
The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m
Answer: h = 110.4 m
Answer:
(a) Negative Q
(b) Positive Q
Explanation:
Charge is the inherent property of matter due to the transference of electrons.
There are three methods of charging a body.
(i) Charging by friction: When two uncharged bodies rubbed together, then one body gets positive charged and the other is negatively charges it is due to the transference of electrons form one body to another.
(ii) Conduction: when a charged body comes in contact with the another uncharged body, the uncharged body gets the same charge and the charge is distributed equally.
(iii) Induction: When a uncharged body keep near the charged body, the uncharged body gets the same amount of charge but opposite in sign.
(a) When a small tack of charge Q is lowered into the hole, then due to the process of induction, the charge on the inner surface of the shell is - Q.
(b) Due to the process of conduction, the charge on the outer surface of the shell is Q.
The momentum of the red cart before the collision is 0.2 kgm/s and the blue cart is 0.
The momentum of the red cart after the collision is 0.05 kgm/s and the blue cart is 0.15 kgm/s.
The change in momentum of the system of the carts is 0.
<h3>
Initial momentum of the carts before collision</h3>
The momentum of the carts before the collision is calculated as follows;
P(red) = 0.5 kg x 0.4 m/s = 0.2 kgm/s
P(blue) = 1.5 x 0 = 0
<h3>Momentum of the carts after collision</h3>
The momentum of the carts after the collision is calculated as follows;
P(red) = 0.5 x 0.1 = 0.05 kgm/s
P(blue) = 1.5 0.1 = 0.15 kgm/s
<h3>Change in momentum of the carts</h3>
ΔP = (0.05 + 0.15) - (0.2)
ΔP = 0
Learn more about momentum here: brainly.com/question/7538238
