Answer:
<h2>
650W/m²</h2>
Explanation:
Intensity of the sunlight is expressed as I = Power/cross sectional area. It is measured in W/m²
Given parameters
Power rating = 6.50Watts
Cross sectional area = 100cm²
Before we calculate the intensity, we need to convert the area to m² first.
100cm² = 10cm * 10cm
SInce 100cm = 1m
10cm = (10/100)m
10cm = 0.1m
100cm² = 0.1m * 0.1m = 0.01m²
Area (in m²) = 0.01m²
Required
Intensity of the sunlight I
I = P/A
I = 6.5/0.01
I = 650W/m²
Hence, the intensity of the sunlight in W/m² is 650W/m²
Answer:
The travel would take 6.7 years.
Explanation:
The equation for an object moving in a straight line with acceleration is:
x = x0 + v0 t + 1/2a*t²
where:
x = position at time t
x0 = initial position
v0 = initial velocity
a = acceleration
t = time
In a movement with constant speed, a = 0 and the equation for the position will be:
x = x0 + v t
where v = velocity
Let´s calculate the position from the Earth after half a year moving with an acceleration of 1.3 g = 1.3 * 9.8 m/s² = 12.74 m/s²:
Seconds in half a year:
1/2 year = 1.58 x 10⁷ s
x = 0 m + 0 m/s + 1/2 * 12.74 m/s² * (1.58 x 10⁷ s)² = 1.59 x 10¹⁵ m
Now let´s see how much time it takes the travel to the nearest star after this half year.
The velocity will be the final velocity achived after the half-year travel with an acceleration of 12.74 m/s²
v = v0 + a t
Since the spacecraft starts from rest, v0 = 0
v = 12.74 m/s² * 1.58 x 10⁷ s = 2.01 x 10 ⁸ m/s
Using the equation for position:
x = x0 + v t
4.1 x 10¹⁶ m = 1.59 x 10¹⁵ m + 2.01 x 10 ⁸ m/s * t
(4.1 x 10¹⁶ m - 1.59 x 10¹⁵ m) / 2.01 x 10 ⁸ m/s = t
t = 2.0 x 10⁸ s * 1 year / 3.2 x 10 ⁷ s = 6.2 years.
The travel to the nearest star would take 6.2 years + 0.5 years = 6.7 years.
Explanation:
<em>hopes </em><em>it's </em><em>h</em><em>elped </em><em>uh </em><em>mate</em><em>!</em>
Answer:
Explanation:
Angular speed of the motion ( SHM )
ω = √k/m
= √(580/.23 )
= 50.20 radian /s
a ) Rate of doing work
= power = force x velocity
At the equilibrium position force becomes zero so
rate of doing work is zero.
b )
If a be the amplitude
1/2 k a² = 170
a = .7655 m
kinetic energy at equilibrium = 1/ 2 m v₀²
1/ 2 m v₀² = 170
.5 x 23 v₀² = 170
v₀ = 3.84 m /s which is the maximum velocity.
Given x = .66 where rate of doing work is to be calculated.
Force at x = ω² x
= 50.20² x .66 =
= 1663.22 N
Velocity v = v₀ √( a² - x² )
= 3.84 √( .7655² - .66 )
= 3.84 x .387
= 1.486 m/s
power = force x velocity
= 1663.22 x 1.486
= 2471.55 W .
