When the frequency decreases the wavelength is further apart. When it increases its closer together. Think about a flat line when the frequency is low the wavelengths are wider. When its a high frequency the squiggly lines on the moniter are taller and thinner so the wavelengths are not as wide and not that far from each other depending on how high the frequency is.
(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.
(b) The work done in firing the projectile is 2,500 J.
<h3>
Kinetic energy of the projectile at maximum height</h3>
The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;
K.E = ¹/₂mv₀ₓ²
where;
- m is mass of the projectile
- v₀ₓ is the initial horizontal component of the velocity at maximum height
<u>Note:</u> At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.
K.E = (0.5)(2)(30²)
K.E = 900 J
<h3>Work done in firing the projectile</h3>
Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.
W = K.E(i) = ¹/₂mv²
where;
- v is the resultant velocity
v = √(30² + 40²)
v = 50 m/s
W = (0.5)(2)(50²)
W = 2,500 J
Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.
The work done in firing the projectile is 2,500 J.
Learn more about kinetic energy here: brainly.com/question/25959744
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Answer:
Choice A: approximately 
, assuming that the two pistons are connected via some confined liquid to form a simple machine.
Explanation:
Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:
.
By Pascal's Principle, because the first piston exerted a pressure of 
on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.
Assume that the second piston is part of that wall. The pressure on the second piston will also be . In other words:
.
To achieve a force of 
, the surface area of the second piston should be:
.
S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)
S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m
Distance double 720m*2=1440m
V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places
