The given question is incomplete. The complete question is as follows.
The block has a weight of 75 lb and rests on the floor for which
= 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.
Determine the output of the motor at the instant
.
Explanation:
We will consider that equilibrium condition in vertical direction is as follows.
![\sum F_{y} = 0](https://tex.z-dn.net/?f=%5Csum%20F_%7By%7D%20%3D%200)
N - W = 0
N = W
or, N = 75 lb
Again, equilibrium condition in the vertical direction is as follows.
![\sum F_{x} = 0](https://tex.z-dn.net/?f=%5Csum%20F_%7Bx%7D%20%3D%200)
= 0
![T_{2} = \mu_{k} N](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%20%5Cmu_%7Bk%7D%20N)
= ![0.4 \times 75 lb](https://tex.z-dn.net/?f=0.4%20%5Ctimes%2075%20lb)
= 30 lb
Now, the equilibrium equation in the horizontal direction is as follows.
![\sum F_{x} = 0](https://tex.z-dn.net/?f=%5Csum%20F_%7Bx%7D%20%3D%200)
![T Cos (30^{o}) + T Cos (30^{o}) = T_{2}](https://tex.z-dn.net/?f=T%20Cos%20%2830%5E%7Bo%7D%29%20%2B%20T%20Cos%20%2830%5E%7Bo%7D%29%20%3D%20T_%7B2%7D)
![2T Cos (30^{o}) = T_{2}](https://tex.z-dn.net/?f=2T%20Cos%20%2830%5E%7Bo%7D%29%20%3D%20T_%7B2%7D)
or, T = ![\frac{T_{2}}{2 Cos (30^{o})}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B2%7D%7D%7B2%20Cos%20%2830%5E%7Bo%7D%29%7D)
= ![\frac{30}{2 Cos (30^{o})}](https://tex.z-dn.net/?f=%5Cfrac%7B30%7D%7B2%20Cos%20%2830%5E%7Bo%7D%29%7D)
= ![\frac{30}{1.732}](https://tex.z-dn.net/?f=%5Cfrac%7B30%7D%7B1.732%7D)
= 17.32 lb
Now, we will calculate the output power of the motor as follows.
P = Tv
= ![17.32 lb \times 6](https://tex.z-dn.net/?f=17.32%20lb%20%5Ctimes%206)
= ![103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}](https://tex.z-dn.net/?f=103.92%20%5Ctimes%20%5Cfrac%7B1%7D%7B550%7D%20%5Ctimes%20%5Cfrac%7Bhp%7D%7Bft%2Fs%7D)
= 0.189 hp
or, = 0.2 hp
Thus, we can conclude that output of the given motor is 0.2 hp.
Answer:
Light can pass through Object B faster than it can pass through Object A.
Explanation:
When light enters in the medium denser than air it bends towards the normal. Thus, if light enters in the medium with higher refracting index it slows down.
So, when light can pass through object B faster than that of object A as object A has higher refracting index than object B.
Speed of light is inversely proportional to the optical density. Optical density is directly proportional to refracting index. So, for higher optical density refracting index is higher and the speed of light is less.
Hence, optical density of object A is higher than that of object B.
Answer:
The magnitude of the change in gravitational potential energy of the earth-athlete system over the course of the race is 643,536Joules
Explanation
Potential energy is one of the form of mechanical energy and it is defined as the energy possessed by a body due to virtue of its position. When the body is under gravity, it possesses an energy called the gravitational potential energy.
Gravitational potential energy is expressed as shown:
GPE = mass × acceleration due to gravity × height
Given mass of athlete = 80kg
height covered = 820m
acceleration due to gravity = 9.81m/s
GPE = 80×9.81×820
GPE = 643,536Joules
Answer:
435 m
Explanation:
The precision with which the distance to the source of the earthquake can be estimated is equal to the difference in distance covered by the S- and P-waves in the time of 0.125 s.
The distance covered by each type of wave is given by
![d=vt](https://tex.z-dn.net/?f=d%3Dvt)
where
v is the speed of the waves
t is the time
For S-waves,
v = 3.74 km/s
t = 0.125 s
So the distance covered is
![d_s=(3.74)(0.125)=0.4675 km = 467.5 m](https://tex.z-dn.net/?f=d_s%3D%283.74%29%280.125%29%3D0.4675%20km%20%3D%20467.5%20m)
For P-waves,
v = 7.22 km/s
t = 0.125 s
So the distance covered is
![d_p=(7.22)(0.125)=0.9025 km = 902.5 m](https://tex.z-dn.net/?f=d_p%3D%287.22%29%280.125%29%3D0.9025%20km%20%3D%20902.5%20m)
So, the precision with which the distance can be determined is:
![\Delta d = 902.5 - 467.5 =435 m](https://tex.z-dn.net/?f=%5CDelta%20d%20%3D%20902.5%20-%20467.5%20%3D435%20m)
Answer:
Distance: 1600 m Displacement: 0
Explanation:
The distance is because He ran 400 meters 4 times getting 1600 m
4*400=1600
The displacement is 0 because displacement is the total distnce away from the starting point and since he ran laps around the track in the end he ended up in the same spot as last time.