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dusya [7]
3 years ago
10

A 0.250 kg car rest rolls down a frictionless incline from a starting height of 0.75 m what is the final velocity of the car? SH

OW ALL WORK.
Also what difference would it make if it is not frictionless? Please and thank you!
Physics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

v = 3.83 m/s

Explanation:

Given that,

The mass of a car, m = 0.250

It rolls down a frictionless incline from a starting height of 0.75 m

We need to find the final velocity of the car. Let v be the velocity of the car. It can be calculated using the third equation of motion i.e.

v^2-u^2=2as

Here,

u = 0 and a = g

v=\sqrt{2gs} \\\\v=\sqrt{2\times 9.8\times 0.75} \\\\v=3.83\ m/s

So, the final velocity of the car is 3.83 m/s.

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At t = 0, a car registers at 30 miles/hr. Forty seconds later, the car’s velocity is now at 50 miles/hr. Assuming constant accel
ratelena [41]
D.

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7 0
2 years ago
A barbel weighing 215N is raised to a height of 2.0 M above the ground. It is then dropped from that height.
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Answer:

  • 430 J
  • 6.26 m/s

Explanation:

A. The kinetic energy is the same as the initial potential energy:

  PE = mgh = (215 N)(2.0 M) = 430 J

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B. The velocity achieved by falling from a height h is given by ...

  v = √(2gh)

  v =  √(2·9.8 m/s^2·2 m) = √(39.2 m^2/s^2)

  v ≈ 6.26 m/s

8 0
3 years ago
Consider a current carrying a wire coming out of your computer screen towards you. Which statement below correctly describes the
Ugo [173]

Answer:

1. The magnetic field encircles the wire in a counterclockwise direction

Explanation:

When we have a current carrying wire perpendicular to the screen in which the current flows out of the screen then by the Maxwell's right-hand thumb rule we place the thumb of our right hand in the direction of the current and curl the remaining fingers around the wire, these curled fingers denote the direction of the magnetic field which is in the counter-clock wise direction.

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5 0
3 years ago
Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r
cricket20 [7]

Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

                                         E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

                                 E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

                                  E_net = E_2 + E_4

                                  E_2 = E_4

                                  E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

                                  r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

                                  r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

                                  E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

                                 E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

Hence,

                                  E_net = 2*E_part(a)*cos(Q)

- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:

                                  Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

- Now, compute the net electric field E_net:

                                  E_net = 2*(99.518)*cos(27.12)

                                  E_net = 177.151 N / C

               

5 0
3 years ago
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