First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg.
F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N
Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m
Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356 N
Ratio = 0.356 N/589.18 N
<em>Ratio = 6.04</em>
Answer:
25.82 m/s
Explanation:
We are given;
Force exerted by baseball player; F = 100 N
Distance covered by ball; d = 0.5 m
Mass of ball; m = 0.15 kg
Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.
We should note that work done is a measure of the energy exerted by the baseball player.
Thus;
F × d = ½mv²
100 × 0.5 = ½ × 0.15 × v²
v² = (2 × 100 × 0.5)/0.15
v² = 666.67
v = √666.67
v = 25.82 m/s
Part b is equal to F in standards of society and it’s quality of math during the 1900s
(That was a bit of Social Studies lol)
Answer:
18 W
Explanation:
Applying,
P = V²/R.................. Equation 1
Where P = Power of both glowing bulbs, V = Voltage, R = Combined Resistance of both bulbs
Since: It is a series circuit,
Then,
R = R1+R2............. Equation 2
Where R1= Resistance of the first bulb, R2 = Resistance of the second bulb
Given: R1 = R2 = 8 Ω
Substitute into equation 1
R = 8+8
R = 16 Ω
Also Given: V = 12 V
Substitute into equation 1
P = 12²/8
P = 144/8
P = 18 W
KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required
to maintain a specified surface temperature for water and air flows.
FIND: Convection coefficients for the water and air flow convection processes, hw and ha,
respectively.
ASSUMPTIONS: Flow is cross-wise over cylinder which is very long in the direction
normal to flow.
The convection heat rate from the cylinder per unit length of the cylinder has
the form
q' = h*(pi*D)*(Ts-Tinf)
and solving for the heat transfer convection coefficient, find
Water
hw = q'/((pi*D)*(Ts-Tinf))
hw = (38*10^3 W/m) / ((pi*(0.030m))*(80-25)C)=
7330.77314 W/m^2K
Air
ha = (400W/m) / ((pi*(0.030m))*(80-25)C)=<span>
77.166033 </span> W/m^2K
COMMENTS: Note that the air velocity is 10 times that of the water flow, yet
hw ≈ 95 × ha.
These values for the convection coefficient are typical for forced convection heat transfer with
liquids and gases
Watter is a better convective heat transfer media than air
