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JulijaS [17]
3 years ago
7

A student uses a compressed spring of force constant 22 N/m to shoot a 0.0075 kg eraser across a desk. The magnitude of the forc

e of friction on the eraser is 0.042 N. How far along the horizontal desk will the eraser slide if the spring is initially compressed 0.035m
Physics
1 answer:
muminat3 years ago
3 0
The initial elastic potential energy stored in the spring is:
U= \frac{1}{2}k x^2 =  \frac{1}{2}(22 N/m)(0.035 m)^2=0.0135 J

Then, the spring is released, all this potential energy is converted into kinetic energy of the eraser:
U= K=0.0135 J

Then, the force of friction does a work to stop the eraser in this motion, and this work is equal to
W=Fd
where F is the magnitude of the friction force and d is the  distance covered by the eraser.

For energy conservation, the work  done by the friction force must be equal to the initial energy of the eraser, so:
K=W=Fd
and so we find d:
d= \frac{K}{F}= \frac{0.0135 J}{0.042 N}=0.32 m
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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

               

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The glass in a window is 35 inches wide and 20 inches tall, and standard atmospheric pressure is 14.7 pounds per square inch. Wh
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Answer:103 pounds

Explanation:

Given

width of window b=35 in.

height of window h=20 in.

standard atmospheric pressure P_{outside}=14.7 psi

Also P_{inside}=1.01P_{outside}

Thus Net Force on the window will be the algebraic sum of Force due to outside and inside Pressure .

F_{net}=(P_{inside}-P_{outside})\cdot A

F_{net}=P_{outside}(1.01-1)\times 35\times 20

F_{net}=14.7\times 0.01\times 35\times 20

F_{net}=102.9\ pounds\approx 103\ pounds            

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Answer: Asteroids, meteoroids, and comets are remnants of the early solar system. This Statement is TRUE.

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METEOROID: these are small rocky or metallic objects found in outer space.

ASTEROIDS: these are also known as minor planets of the inner solar system. They are irregularly shaped object in space that orbits the Sun.

COMETS: these are dusty chunk of ice, that moves in a highly elliptical orbit about the sun.

Asteroids, meteoroids, and comets as remnants of the early solar system was further proved in nebular hypothesis

initially proposed in the eighteenth century by German philosopher Immanuel Kant and French mathematician Pierre-Simon Laplace. (The word nebula means a gaseous cloud.) According to the modern version of the theory, about 4.5 to 5 billion years ago the solar system developed out of a huge cloud of gases and dust floating through space. These materials were at first very thin and highly dispersed.

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3 years ago
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