Answer:8.28 km
Explanation:
Given
First it drifts 
2.5 km
Secondly it drifts 
4.70 km
After that it drifted along east direction 5.1 km
After that it drifts 
7.2 km
After that it drifts 
2.8 km
=![\left [ 2.5cos45+4.7cos60+5.1-7.2cos55-2.8cos5\right ]\hat{i}](https://tex.z-dn.net/?f=%5Cleft%20%5B%202.5cos45%2B4.7cos60%2B5.1-7.2cos55-2.8cos5%5Cright%20%5D%5Chat%7Bi%7D)
+![\left [ 2.5sin45-4.7sin60-7.2sin55+2.8sin5\right ]\hat{j}](https://tex.z-dn.net/?f=%5Cleft%20%5B%202.5sin45-4.7sin60-7.2sin55%2B2.8sin5%5Cright%20%5D%5Chat%7Bj%7D)
for direction
south of east
Answer:
Time, t = 13.34 seconds.
Explanation:
Given the following data;
Initial velocity, u = 85km/hr to meters per seconds = 85*1000/3600 = 23.61 m/s
Final velocity, v = 45km/hr to meters per seconds = 45*1000/3600 = 12.5 m/s
Acceleration, a = -3 km/hr/sec to meters per seconds square = -3*1000/3600 = -0.833m/s²
To find the time;
Acceleration = (v - u)/t
-0.833 = (12.5 - 23.61)/t
-0.833t = -11.11
t = 11.11/0.833
Time, t = 13.34 seconds.
Nicholas Copernicus correctly assumed that the planets revolved around the sun but he incorrectly assumed that the planets followed a perfect circle orbit around the sun. It was later on discovered by Johannes Kepler that the planets moved around the sun following an elliptical orbit.
Answer:
799.54 ft
Explanation:
Linear thermal expansion is:
ΔL = α L₀ ΔT
where ΔL is the change in length,
α is the linear thermal expansion coefficient,
L₀ is the original length,
and ΔT is the change in temperature.
Given:
α = 1.2×10⁻⁵ / °C
L₀ = 800 ft
ΔT = -17°C − 31°C = -48°C
Find: ΔL
ΔL = (1.2×10⁻⁵ / °C) (800 ft) (-48°C)
ΔL = -0.4608
Rounded to two significant figures, the change in length is -0.46 ft.
Therefore, the final length is approximately 800 ft − 0.46 ft = 799.54 ft.
Answer:
g = 5 m/s square
Explanation:
Weight(W), Mass(m), Gravity(g)
W = mg
1,000N = 200g
g = 1000/200
g = 5 m/s square
