Answer:
A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.
how much work is done on the monitor by (a) friction, (b) gravity
work(friction) = 453.5J
work(gravity) = -453.5J
Explanation:
Given that,
mass = 14kg
displacement length = 5.50m
displacement angle = 36.9°
velocity = 2.30cm/s
F = ma
work(friction) = mgsinθ .displacement
= (14) (9.81) (5.5sin36.9°)
= 453.5J
work(gravity)
= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)
= 126.9°
work(gravity) = (14) (9.81) (5.5cos126.9°)
= -453.5J
The circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.
<h3>
UF cable</h3>
UF cable is used as an underground feeder cable to distribute power from an existing building to outdoor equipment. UF cable can also be used as direct burial cable.
For the 24-volt branch circuit installed, the minimum burial depth will be 6 inches.
Thus, the circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.
Learn more about UF cable here: brainly.com/question/8591560
Answer: Noise above 70 dB can cause hearing damage
Explanation:
Answer:
The force required to move the quarterback with linebacker is <u>1215 N</u>
Explanation:
![\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Mass%20of%20linebacker%20%7D%20%5Cmathrm%7Bm%7D_%7B2%7D%3D150%20%5Cmathrm%7Bkg%7D)
![\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Mass%20of%20quarterback%20%7D%20%5Cmathrm%7Bm%7D_%7B2%7D%3D120%20%5Cmathrm%7Bkg%7D)
![\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Moved%20at%20an%20acceleration%20%7D%28a%29%3D4.5%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D%5E%7B2%7D)
Using Newton's second law, it is established that F = Ma
Where F is net force acting on the system, a is the acceleration and M is mass of the two object ![\left(m_{1}+m_{2}\right)](https://tex.z-dn.net/?f=%5Cleft%28m_%7B1%7D%2Bm_%7B2%7D%5Cright%29)
Now consider both
as a system, so net force acting on the system is ![\text { Force }=\left(m_{1}+m_{2}\right) a](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Force%20%7D%3D%5Cleft%28m_%7B1%7D%2Bm_%7B2%7D%5Cright%29%20a)
Substitute the given values in the above formula,
![\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Force%20%7D%3D%28150%2B120%29%20%5Cmathrm%7Bkg%7D%20%5Ctimes%204.5%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D%5E%7B2%7D)
![\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Force%20%7D%3D270%20%5Cmathrm%7Bkg%7D%20%5Ctimes%204.5%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D%5E%7B2%7D)
Force = 1215 N
<u>1215 N </u>is the force required to move the quarterback with linebacker.
Answer:
1160 ohm
Explanation:
We are given that
R'=580 ohm
Current=3 I
We have to find the resistance of the circuit.
Let R be the resistance of circuit.
In parallel
![\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Beq%7D%7D%3D%5Cfrac%7B1%7D%7BR_1%7D%2B%5Cfrac%7B1%7D%7BR_2%7D)
Using the formula
![\frac{1}{R_{eq}}=\frac{1}{580}+\frac{1}{R}=\frac{580+R}{580R}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Beq%7D%7D%3D%5Cfrac%7B1%7D%7B580%7D%2B%5Cfrac%7B1%7D%7BR%7D%3D%5Cfrac%7B580%2BR%7D%7B580R%7D)
![R_{eq}=\frac{580R}{580+R}](https://tex.z-dn.net/?f=R_%7Beq%7D%3D%5Cfrac%7B580R%7D%7B580%2BR%7D)
In parallel combination,Potential difference across each resistance remains same.
![V=IR](https://tex.z-dn.net/?f=V%3DIR)
Using the formula
![IR=3IR_{eq}](https://tex.z-dn.net/?f=IR%3D3IR_%7Beq%7D)
![IR=3I\times \frac{580R}{580+R}](https://tex.z-dn.net/?f=IR%3D3I%5Ctimes%20%5Cfrac%7B580R%7D%7B580%2BR%7D)
![580+R=3\times 580=1740](https://tex.z-dn.net/?f=580%2BR%3D3%5Ctimes%20580%3D1740)
![R=1740-580=1160\Omega](https://tex.z-dn.net/?f=R%3D1740-580%3D1160%5COmega)