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gladu [14]
3 years ago
11

Solve for the tension in the left rope, TL, in the special case that x=0. Be sure the result checks with your intuition. Express

your answer in terms of W and the dimensions L and x. Not all of these variables may show up in the solution.

Physics
1 answer:
Tatiana [17]3 years ago
3 0

Answer:

at x=0

T_{l}=W*L/L

and

T_{l}=W-T_{r}

Explanation:

Solve for the tension in the left rope, TL, in the special case that x=0. Be sure the result checks with your intuition. Express your answer in terms of W and the dimensions L and x. Not all of these variables may show up in the solution.

moments is the product of force and the perpendicular distance in line of the action of the given force

from the principle of moments which states that the sum of clockwise moments ,must be equal to the sum of anticlockwise moments.

also, sum of upward forces must be equal to sum of downward forces

Going by the aforementioned,

taking moments about T_{r}

W*(l-x)=T_{l}*(L)

T_{l}=W*(L-x)/(L)..............1

at x=0

T_{l}=W*L/L

also

T_{l}+ T_{r}=W

T_{l}=W-T_{r}

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A car driving at 20 m/s accelerates continuously 2m/s2​ ​ for 3 seconds. What is its final velocity
Lesechka [4]

Answer: V= u+ at

V= final velocity

u=initial velocity

a=acceleration

t=time taken

V= 20 + 2*3

V= 26m/s

Explanation:

5 0
3 years ago
It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass
vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0
3 years ago
If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w
marshall27 [118]

Answer:

P(3600)=593.247W

Explanation:

First, let's find the voltage through the resistor using ohm's law:

V=IR=20*8=160V

AC power as function of time can be calculated as:

P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)  (1)

Where:

\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency

Because of the problem doesn't give us additional information, let's assume:

\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W

5 0
4 years ago
I will mark as the brainliest answer<br><br>plz 8,9,10​
blagie [28]

Answer:

8.  acceleration = \dfrac{d(velocity)}{d(time)}  = 1 unit .

9. acceleration = \dfrac{d(velocity)}{d(time)}  = -1 unit.

10. acceleration = \dfrac{d(velocity)}{d(time)}  = 0 units.

Explanation:

8. i) acceleration = velocity / time

  ii) In this figure velocity = time

  iii) therefore acceleration = \dfrac{d(velocity)}{d(time)} = 1 unit .

9. i) acceleration = velocity / time

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     therefore velocity = (-0.5 \times time) + 5

  iii) therefore acceleration = \dfrac{d(velocity)}{d(time)}  = -1 units.

10.) velocity is constant at 2

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5 0
3 years ago
Can i please get some help with this.
Ivenika [448]

the answer is D cuz electricity is a conductive

5 0
4 years ago
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