A. Velocity of planet and gravity between sun and planet
A is correct according to below calculation.
m₁v₀₁+m₂v₀₂=m₁v₁+m₂v₂
((m₁v₀₁+m₂v₀₂)-m₁v₁)/m₂=v₂
v₂=((.5*12-.75*16)+(.5*21.6))/.75
v₂=6.4 m/s
Answer:
P /K = 1,997 10⁻³⁶ s⁻¹
Explanation:
For this exercise let's start by finding the radiation emitted from the accelerator
= 
the radius of the orbit is the radius of the accelerator a = r = 0.530 m
let's calculate
\frac{dE}{dt} = [(1.6 10⁻¹⁹)² 0.530²] / [6π 8.85 10⁻¹² (3 108)³]
P= \frac{dE}{dt}= 1.597 10⁻⁵⁴ W
Now let's reduce the kinetic energy to SI units
K = 5.0 10⁶ eV (1.6 10⁻¹⁹ J / 1 eV) = 8.0 10⁻¹⁹ J
the fraction of energy emitted is
P / K = 1.597 10⁻⁵⁴ / 8.0 10⁻¹⁹
P /K = 1,997 10⁻³⁶ s⁻¹
Answer:
15 V
Explanation:
From the question,
For series connection: (i) Both resistor have a common current flowing through the (ii) The combined resistance = R1+R2
Rt = R1+R2.................. Equation 1
Given: R1 = 5 ohms, R2 = 7.5 ohms.
Rt = 5+7.5 = 12.5 ohms.
Applying Ohm's law,
V = IRt................... Equation 2
Where V = Voltage, I = current.
make I The subject of the equation
I = V/Rt.............. Equation 3
Given: V = 25 V, Rt = 12.5 ohms.
Substitute into equation 3
I = 25/12.5
I = 2 A.
Now,
Voltage drop across the 7.5 ohms resistor = R2×I
Voltage drop across the 7.5 ohms resistor = 7.5×2
Voltage drop across the 7.5 ohms resistor = 15 V
