Answer:
The take-off speed is 41.48 
Explanation:
Given :
Range 
m
Projectile angle 
13°
From the formula of range,
Find the velocity from above equation,
( ∵ 
)
Therefore, the take-off speed is 41.48
Answer:
Push or Pull Forces - example
When you push against a wall the force that you exert is an example of a push force. When you pull a trolley car the force that you exert is an example of pull force.
<h2>
Answer: either way</h2>
The balloon contains neutral charge atoms, that is, it has the same number of electrons (negative charge), protons (positive charge) and neutrons (no charge).
Then, when two objects come into contact, the electrons of one of them can become part of the other.
Thus, by bringing the balloon closer to the wall, the wall, which is also made up of atoms, will reorder its charges in such a way that its electrons or protons become part of the balloon, charging it.
Answer:
%
%
Explanation:
From the exercise we know two information. The real speed and the experimental measured by the speedometer
Since the speedometer is only accurate to within 0.1km/h the experimental speed is
Knowing that we can calculate Kinetic energy for the real and experimental speed
Now, the potential error in her calculated kinetic energy is:
%
%
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
