Answer: 5 gm/cc
Explanation:
200 gm/40 cc
= 5 gm/cc
Explanation:
What is IEEE 802.11?
IEEE 802.11 is a set of WLAN standards for communication developed by the Institute for Electrical and Electronics Engineers (IEEE) and is unarguably most widely used WLAN technology.
Features: IEEE 802.11a
- The operating frequency band is 5 GHz.
- The maximum theoretical data rate is 54 Mbps, the typical throughput is around 25 Mbps and minimum data rate is 6 Mbps.
- It can support 64 users per access point.
Features: IEEE 802.11b
- The operating frequency band is 2.4 GHz.
- The maximum theoretical data rate is 11 Mbps but typical throughput is around 6 Mbps and minimum data rate is 1 Mbps.
- It can support 32 users per access point.
Wireless Coverage IEEE 802.11a Vs IEEE 802.11b:
- Signal coverage is one of the most important factors among users.
- The transmission range of IEEE 802.11a is not greater than 100 ft in indoor setting whereas IEEE 802.11b has a superior performance in this regard with transmission range up to 150 ft in indoor setting.
- The data rate has a direct relation with the access point coverage area, a higher data rate means less coverage area and a lower data rate results in increased coverage.
Let the distance between the towns be d and the speed of the air be s.
distance = speed * time
convert the minutes time into hours.
When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:
d
s−15
=
7
3
return trip is then :
d
s+15
=
7
5
Cross-multiplying both we get the two-variable system:
3d=7∗(s−15)5d=7∗(s+15)
3d=7s−1055d=7s+105
subtract first equation from second equation we get
2d=210d=105km
Substitute the value of d in the above equations for s.
5∗105=7s+1057s=420s=60km/hr
Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get:
