Answer:
(a) -202 m/s²
(b) 198 m
Explanation:
Given data
- Initial speed (v₀): 283 m/s
- Final speed (vf): 0 (rest)
(a) The acceleration (a) is the change in the speed over the time elapsed.
a = (vf - v₀)/t = (0 - 283 m/s)/ 1.40s = -202 m/s²
(b) We can find the distance traveled (d) using the following kinematic expression.
y = v₀ × t + 1/2 × a × t²
y = 283 m/s × 1.40 s + 1/2 × (-202 m/s²) × (1.40 s)²
y = 198 m
Gravitational acceleration, g = GM/r^2. Additionally, for a satellite in a circular orbit, g = v^2/r
Where, G = Gravitational constant, M = Mass of earth, r = distance from the center of the earth to the satellite, v = linear speed of the satellite.
Equating the two expressions;
v^2/r = GM/r^2
v = Sqrt (GM/r);
But GM = Constant = 398600.5 km^3/sec^2
r = Altitude+Radius of the earth = 159+6371 = 6530 km
Substituting;
v = Sqrt (398600.5/6530) = 7.81 km/sec = 781 m/s
The initial temperature of the bar is 25. To get to the t temperature you need to add (t-25) degrees Celsius.
for 1 degree................... 7 Joules
y given degree........ p Joules
p=7y
In our case y=(t-25) .
h(t) = 7(t-25) which is the final answer.
