A) <u>Weight = mass × acceleration (due to gravity) </u>
= 60×9.8
= 588 N
<u>B) Potential energy = mass x gravity x change in height
</u>
1,000 = 60.0 x 9.8 x h
h = 1.7 m
<u>C) Kinetic energyF = potential energyI
</u>
KEF = 1/2mv2
PEI = mgh = 1,000 J
1/2mv2 = 1,000
1/2(60.0)v2 = 1,000
v2 = 33.33
v = 5.77 m/s
The answer is 0k because 143c equals nothing
Answer:
Explanation:
refractive index of ember = sin of angle of incidence / sin of angle of refraction
= sin 35 / sin24
= .5735 / .4067
= 1.41
This is refractive index of ember with respect to water
refractive index of ember with respect to water
= wμe = μe / μw
μe = wμe x μw
= 1.33 x 1.41
= 1.87
refractive index of ember with respect to air = 1.87 .
Answer:
Hi sorry for answering here but you didnt put the options there
Explanation:
I'll still try to answer though so maybe the mixture from one of the questions might be something like oil and water which don't mix and can be separated by decantation so something similar would work. Hope this helps
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>