B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
Answer:
I just toughed it out and talked with friends
Explanation:
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.
