Around 1915, henry sincosky of philadelphia suspended himself from a rafter by gripping the rafter with the thumb of each hand on one side and the fingers on the opposite side. sincosky’s mass was 79 kg. (after suspending himself, sincosky chinned himself on the rafter and then moved hand-over-hand along the rafter for about 6 feet.) if the coefficient of static friction between henry’s fingers and rafter was 0.70, what was the least magnitude of the normal force on the rafter from each thumb or opposing fingers to suspend henry via friction forces?

i think the answer is a, hope this is right, that would be my best guess.

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle