And what finish the question
Answer: minimum speed of launch must be 7.45m/s
Explanation:
Given the following:
Height or distance (s) = 2.83m
The final velocity(Vf) at maximum height = 0
Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2
From the 3rd equation of motion:
V^2 = u^2 - 2gs
Where V = final velocity
u = initial velocity
Therefore, u = Vi
u = √Vf^2 - 2gs
u = √0^2 - 2(-9.8)(2.83)
u = √0 + 55.468
u = √55.468
u = 7.4476 m/s
u = 7.45m/s
Answer:
the car have travelled 0.31 mile during that time
Explanation:
Applying the Equation of motion;
s = 0.5(u+v)t
Where;
s = distance travelled
u = initial speed = 0 mph
v = Final speed = 50 mph
t = time taken = 3/4 min = 3/4 ÷ 60 hours = 1/80 hour
Substituting the given values into the equation;
s = 0.5(0+50)×(1/80)
s = 0.3125 miles
s ~= 0.31 mile
the car have travelled 0.31 mile during that time
Answer:
E- The star becomes a red giant (LATEST STAGE)
F- The surface of the star becomes brighter and cooler
C- Pressure from the star's hydrogen-burning shell causes the non burning envelope to expand
A- The shell of hydrogen surrounding the star's nonburning helium core ignites.
D- The star's non burning helium core starts to contract and heat up
B- Pressure in the star's core decreases (EARLIEST STAGE)
(A star moves away from the main sequence once its core runs out of hydrogen to fuse into helium. The energy once supplied by hydrogen burning reduces and the core starts to compress under the force of gravity. This contraction allows the core and surrounding layers to heat up. Finally, the hydrogen shell around the core becomes hot enough to ignite hydrogen burning.
