A wave that transfers energy through vibrating In a medium
We can solve the problem by using the law of conservation of energy:
- at the beginning, all mechanical energy of the object is just kinetic energy: , where m is the mass and v is the velocity
- at the point of maximum height, all mechanical energy of the object is just gravitational potential energy: , where h is the maximum height
Therefore, the conservation of energy becomes:
Re-arranging, we find the maximum height:
On Earth, solar radiation at the frequencies of visible light largely passes through the atmosphere to warm the planetary surface. The surface itself emits energy at the lower frequencies of infrared thermal radiation. Infrared radiation is absorbed by greenhouse gases in the atmosphere. These gases also radiate energy, some of which is directed to the surface and lower atmosphere. The mechanism is named after the effect of solar radiation passing through glass and warming a greenhouse, but the way it retains heat is fundamentally different as a greenhouse works by reducing airflow, isolating the warm air inside the structure so that heat is not lost by convection
Answer:
Explanation:
Energy stored in a capacitor
= 1/2 C₁V²
capacity of a capacitor
c = εK A / d
k is dielectric and d is distance between plates .
When the distance between the plates is halved and then filled with a dielectric (κ = 4.3)
capacity becomes 4.3 x 2 times
New capacity
C₂ = 8.6 C₁
Energy of modified capacitor
1/2 C₂ V²= 1/2 x 8.6 c x V²
Energy becomes
8.6 times.
Energy stored = 8.6 x 10⁻⁴ J
a) create an expression
for the ball's initial horizontal velocity, V0x, in terms of the variables
given in the problem statement.
v0x = vf * cos(Θf)
<span>
b) calculate the ball's initial vertical velocity, V0y, in
m/s</span>
v0x = 32.4m/s * cos(-25.5º)
= 29.2 m/s <span>
tanΘ = v1y / v0x → tan(-25.5) = v1y / 29.2m/s → v1y = -13.93
m/s
the vertical velocity when the ball was caught.
(v0y)² = (v1y)² + 2as = (-13.93m/s)² + 2 * 9.8m/s² * 5.5m = 301.78
m²/s²
v0y = 17.37 m/s
c) calculate the magnitude of the ball's initial velocity,
v0, in m/s</span>
v0 = sqrt (v0y^2 +
v0x^2)
v0 = sqrt (17.37^2 + 29.2^2)
m/s
v0 = 33.98 m/s
<span>
d) find the angle, theta0, in degrees above the horizontal at
which which the ball left the bat.</span>
tan Θ = v0y/v0x
<span>Θ = arctan(17.37/29.2) =
30.75º above horizontal</span>