Answer:
Explanation:
Given that,
Height of the bridge is 20m
Initial before he throws the rock
The height is hi = 20 m
Then, final height hitting the water
hf = 0 m
Initial speed the rock is throw
Vi = 15m/s
The final speed at which the rock hits the water
Vf = 24.8 m/s
Using conservation of energy given by the question hint
Ki + Ui = Kf + Uf
Where
Ki is initial kinetic energy
Ui is initial potential energy
Kf is final kinetic energy
Uf is final potential energy
Then,
Ki + Ui = Kf + Uf
Where
Ei = Ki + Ui
Where Ei is initial energy
Ei = ½mVi² + m•g•hi
Ei = ½m × 15² + m × 9.8 × 20
Ei = 112.5m + 196m
Ei = 308.5m J
Now,
Ef = Kf + Uf
Ef = ½mVf² + m•g•hf
Ef = ½m × 24.8² + m × 9.8 × 0
Ef = 307.52m + 0
Ef = 307.52m J
Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw
<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>
Answer:
label A= radio waves, label C= infrared, Label D= visible Light, Label G= gamma rays.
Explanation:
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The photon can be absorbed and the energy of the photon is exactly equal to the energy-level difference between the ground state and the level d.
Answer:
T₂ =602 °C
Explanation:
Given that
T₁ = 227°C =227+273 K
T₁ =500 k
Gauge pressure at condition 1 given = 100 KPa
The absolute pressure at condition 1 will be
P₁ = 100 + 100 KPa
P₁ =200 KPa
Gauge pressure at condition 2 given = 250 KPa
The absolute pressure at condition 2 will be
P₂ = 250 + 100 KPa
P₂ =350 KPa
The temperature at condition 2 = T₂
We know that
T₂ = 875 K
T₂ =875- 273 °C
T₂ =602 °C
