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inn [45]
4 years ago
10

A 675-pound trailer is sitting on an exit ramp inclined at 40o on Highway 35. How much vertical force is required to keep the tr

ailer from rolling back down the exit ramp?
Physics
1 answer:
Phantasy [73]4 years ago
8 0

Answer:

433.88 lbs

Explanation:

We are given that mass of trailer =675 pound

Angle=40^{\circ}

We have to find the vertical force required to keep the trailer from rolling back down the exit ramp.

The force required to keep  the trailer from rolling back down the exit ramp=wsin\theta

Where w=675 pounds and \theta=40^{\circ}

Substitute the values then we get

F=675 sin40^{\circ}

sin40^{\circ}=0.642787

Substitute the value then we get

F=675\times 0.642787=433.88 lbs

Hence, the force required to keep the trailer from rolling back down the exits ramp=433.88 lbs

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  • The total resistance in the circuit is 16 Ohms.
  • The total current in the circuit is 0.5 Ampere.
  • The current at R_1 is 0.5 Ampere.
  • The current at R_3 is 0.5 Ampere.
  • The voltage drop atR_1 is 4 volts.
  • The voltage drop at R_2 is 2.5  volts.
  • The voltage drop at R_3 is 1.5 volts.
  • The total power consumed by the circuit is 4watts
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Given:

The voltage across the circuit = V = 8 V

The resistors connected are in series:

R_1=8 \Omega, R_2=5\Omega ,R_3=3 \Omega

To find:

The values of from 1 to 10.

Solution

The voltage across the circuit = V = 8 V

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R_{eq}=R_1+R_2+R_3\\=8 \Omega +5 \Omega + 3\Omega =16\omega

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So, the current in R_1, R_2 \&R_3:

I_1= I_2= I_3=I=0.5 A\\

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The current across  R_1 = I = 0.5 A

V_1=I\times R_1\\\\=0.5A\times 8\Omega = 4 V

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The current across  R_2 = I = 0.5 A

V_2=I\times R_2\\\\=0.5A\times 5\Omega = 2.5 V

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The current across  R_3 = I = 0.5 A

V_3=I\times R_3\\\\=0.5A\times 3\Omega = 1.5 V

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P= V\times I \\\\= 0.5 A\times 8 V = 4 watt

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Learn more about, current, voltage, resistance, and power of the circuit here:

brainly.com/question/11683246?referrer=searchResults

brainly.com/question/1430450?referrer=searchResults

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