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Travka [436]
3 years ago
5

With a force of 5 Newton's Amanda pushes the stacks of books to the right. At the same time Jeremih her little brother pushes th

e stacks of books to the left with a force of 10 Newton's what is the direction and magnitude of the resulting force
Physics
2 answers:
nignag [31]3 years ago
5 0

Answer:

5n to the left

Explanation:

STUISLAND 2020

Fynjy0 [20]3 years ago
4 0

Alright, so solving for the net force is a rather simple and easy step. I want to briefly explain how you determine the net force, but before I do, let's examine the problem together. What has been provided to us? What direction is it going? Well, we can look at the information!

Amanda's Force: 5 N

Jeremiah's Force: 10 N

They're both pushing on the books, but in different directions. Left and right. If Amanda is pushing the books to the right 5 newtons, and Jeremiah is pushing the books to the left with 10 newtons, that means the net force is 5. The book is being pushed to the left! We can say that because Jeremiah's force is much larger than his sisters.

We can determine the net force from the following;

Net Force → 10 N - 5 N = 5 N

In conclusion, your answer should be a total of five (5) newtons

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8 0
3 years ago
If you wanted to
VLD [36.1K]

Answer:

Option C

Explanation:

According to the formula

  • \\ \boxed{\sf R=\rho\dfrac{\ell}{A}}

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5 0
2 years ago
Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.
vagabundo [1.1K]

Answer:

R_total = 14.57 Ω ,  V_C = 1.176 V

Explanation:

To solve this circuit we are going to find the equivalent resistance of each branch, let's remember

* Serial resistance  

         R_{eq} = ∑ R_{i}

* For resistance in parallel

        1 / R_{eq} = ∑ 1/R_{i}

We solve the two branches of the wheatstone bridge

Series resistors

Branch B

         R_B = Rb + R4

         R_B = 2 + 18

         R_B = 20 Ω

Branch C

         R_C5 = Rc + R5

         R_C5 = 3 + 12

         R_C5 = 15 Ω

Resistance in parallel R_B and R_C5

         1 / R_BC = 1 / R_B + 1 / R_C5

          1 / R_BC = 1/20 + 1/15 = 0.116666

          R_BC = 8.57 Ω

Now we have a single branch, we solve the series resistance

          R_total = R_A + R_BC

          R_total = 6 + 8.57

          R_total = 14.57 Ω

b) they ask us for the voltage in the resistance R_C

Let's remember that the voltage in a series circuit is the sum of the voltages

           10 = V_a + V_BC

           10 = i R_a + i R_BC = i (R_a + R_BC)

           i = 10 / (R_a + R_BC)

           i = 10 / (14.57)

           i = 0.6863 A

The current in the series circuit is constant

          V_BC = i R_BC

          V_BC = 0.6863 8.57

          V_BC = 5.8819 V

This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.

     Branch C

             V_BC = i R_C5

             i = V_BC / R_C5

             i = 5.8819 / 15

             i = 0.39213 A

In this branch we have two resistors in series, let's remember that the current in a series circuit is constant

             V_C = i R_C

              V_C = 0.39213 3

              V_C = 1.176 V

3 0
3 years ago
Helppppp pleaseee :(
Elza [17]

1) push down on the end of the lever, and 2) 3/4 of the way from the fulcrum

7 0
3 years ago
David rowed a boat upstream for three miles and then returned to point he started from. The entire journey took four hours. The
frez [133]
Upstream speed = S - 1
Downstream speed = S + 1

Average speed = total distance / total time

Average speed = (S - 1) + (S + 1) / 2
= S

S = 6 miles / 4 hours
S = 1.5 miles per hour
4 0
3 years ago
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