Answer:
50 m/s opposite direction to the motion of the truck
Explanation:
From the question,
Applying the law of conservation of momentum
mu+m'u' = V(m+m')...….. Equation 1
Where m = mass of the truck, u = initial velocity of the truck, m' = mass of the car, u' = initial velocity of the car, V = Final velocity after collision
Given: m = 2500 kg, u = 20 m/s, m' = 1000 kg, V = 0 m/s (both car stop after collision)
Substitute these values into equation 1
2500(20)+1000(u') = 0(2500+1000)
2500(20)+1000(u') = 0
Solve for u'
u' = -[2500(20)]/1000
u' = -50 m/s
The negative sign shows that the car travels in opposite direction to the truck
Hence the car initial velocity before collision is 50 m/s in opposite direction to the motion of the truck
You divide 120 from 950 and that's the number of hours, 7.9
Answer:
Explanation:
Earliest standards were dependent on a single frequency/channel to both send and receive. This shared medium creates the same problem as half-duplex coax cable. Because receivers had to wait for the signal before sending a response, this reduced the overall bandwidth.
Other factors affect wireless signal propagation, too, including RF interference, antenna choice, and obstacles such as walls, trees, and even weather (precipitation, for example).
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
