The speed of light is: c
= 3x10^8 m/s <span>
or
c = 186,000,000 miles/sec = 1.86x10^8 mi/s
1 furlong = 0.125 mile
1 fortnight = 2 weeks(7d/wk)(24h/d)(3600s/h)
= 1209600s = 1.2096x10^6 s
Therefore,
c =1.86x10^8 mi/s(1furl/0.125mi)(1.2096x10^6s/fort)
<span>c = 18x10^14 furlong/fortnight = 18x10^8 Mfurlong/fortnight</span></span>
Answer:
1. -8.20 m/s²
2. 73.4 m
3. 19.4 m
Explanation:
1. Apply Newton's second law to the car in the y direction.
∑F = ma
N − mg = 0
N = mg
Apply Newton's second law to the car in the x direction.
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Given μ = 0.837:
a = -(9.8 m/s²) (0.837)
a = -8.20 m/s²
2. Given:
v₀ = 34.7 m/s
v = 0 m/s
a = -8.20 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx
Δx = 73.4 m
3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.
d = v₀t
d = (34.7 m/s) (0.56 s)
d = 19.4 m
Speed is v = d/t
Or speed is distance over time
So...
40min / 60min = 0.6667 or 2/3 --> Finding what proportion 40 minutes is to an hour or 60 minutes as we need the units of hours to match up
45km/h = d/0.6667h
d = (45)(0.667)
d = 30.0015 or 30km
Answer:
a) about 20.4 meters high
b) about 4.08 seconds
Explanation:
Part a)
To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.


In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

To solve for "t" in this quadratic equation, we can factor it out as shown:

Therefore there are two possible solutions when each of the two factors equals zero:
1) t= 0 (which is not representative of our case) , and
2) the expression in parenthesis is zero:

The lines can be traced out with a compass. The needle is like a permanent magnet and the north indicator is the north end of a magnet.