Answer:
My answer is 7.2 km
Explanation:
When Stephen goes to the south and then to the east, he is drawing a right triangle, where the 4 km and 6 km sides are the cathetus of a right triangle.
Then we use the Pithagorean theorem to solve this problem. We need to find the hypotenuse.
c² = a² + b²
c² = 4² + 6²
c² = 16 + 36
c² = 52
c = 7.2 km
Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
Answer:
answer a: a large front gear with a small back gear
answer b: a small front gear with a large back gear
Explanation:
just simple gearing ratios
Answer:
Time zone is one important factor in difference in location and this in turn affects the result of the resolution and rotation of shadow produced from the sun or other illumination.
Therefore someone at a place might see a clear large shadow due to shinny sun reflection and another a small or dull Shadow at same time if the intensity of the sun or lighting source is going down.
Explanation:
The closer a body/object is to a lighting source the larger the shadow it produces, and the farther the body the smaller the shadow produced.
Answer:
1.1775 x 10^-3 m^3 /s
Explanation:
viscosity, η = 0.250 Ns/m^2
radius, r = 5 mm = 5 x 10^-3 m
length, l = 25 cm = 0.25 m
Pressure, P = 300 kPa = 300000 Pa
According to the Poisuellie's formula
Volume flow per unit time is
V = 1.1775 x 10^-3 m^3 /s
Thus, the volume of oil flowing per second is 1.1775 x 10^-3 m^3 /s.
