First,
where 
is density, 
is mass, and 
is volume. We can compute the volume of the roll:
When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness 
. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).
So we have
where 
is the given area, so
If we're taking significant digits into account, the volume we found would have been 
, in turn making the thickness 
.
d =2.55.68m and t = 11.36s
In my opinion
The elastic potential energy of the spring is 0.31 J
Explanation:
The elastic potential energy of a spring is given by
where
k is the spring constant
x is the compression/stretching of the spring
For the spring in this problem, we have:
k = 500 N/m (spring constant)
x = 0.035 m (compression)
Substituting, we find the elastic potential energy:
Learn more about potential energy:
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Answer:
<h2>Radiation</h2>
Explanation:
<h3>Greetings !</h3>
The Sun reaches us by propagating through the vacuum of space. Sunlight reaches the Earth in about 8 minutes and 20 seconds. When this energy reaches the Earth's atmosphere, both conduction and convection play key roles to scatter it throughout the planet.
Answer: a. 198.6J b. - 198.6J
Explanation: Parameters given:
m = 15kg
g = 9.8m/s²
∅ = 12°
a. Work done by the force Fp on the cart if the ramp is 6.5m long.
Given the formula, Fp = Mgsin∅ = 15 x 9.8 x sin12° = 30.56N
Therefore Work done (Wp) = Fp x Ramp Length = 30.56 x 6.5 = 198.64Nm or 198.6J
b. The work done by the force mg on the cart.
Since the cart is being pushed upwards, it acts against gravity with its direction of motion. Taking into account the formula from the previous answer for Work Done (Wg) = Fmg x distance
= 15kg x -9.8m/s² x Sin12° x 6.5m
= - 198.6J
