Answer:
The temperature is 2541.799 K
Explanation:
The formula for black body radiation is given by the relation;
Q = eσAT⁴
Where:
Q = Rate of heat transfer 56.6
σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)
A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²
e = emissivity = 0.288
T = Temperature
Therefore, we have;
T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴
T = 2541.799 K
The temperature = 2541.799 K.
Before the skydiver opens the parachute, his velocity would be increasing greatly as much as 9.8 m/s². Opening the parachute would increase the surface area to which air may cause resistance. The skydiver then reaches his terminal velocity.
Answer:
-0.912 m/s
Explanation:
When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.
where 
are the mass of the child, the boat and the package, respectively. 
are the velocity of the package and the boat after throwing.
I think F= mv²/r
And F=ma
So, ma = mv²/r
a = v²/r
a = 100/5
a = 20 m/s
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