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sdas [7]
3 years ago
11

Question 1 of 20

Physics
2 answers:
ad-work [718]3 years ago
8 0

Answer

D. move a small magnet back and forth within a section of the coiled wire.

Explanation:

i put that for the test and i got it right

Tanzania [10]3 years ago
8 0

Answer:

d

Explanation: got it right

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HEL ME PLEASE I HAVE NO IDEA OF WHAT TO DO!!!!!!
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b, the fastest the car has traveled out of those options is 4-5 seconds.

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3 years ago
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Which of the following statements describes electric fields?
tia_tia [17]
C)Electric Charges Produce Electric Fields
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4. Magnets with one pole are now routinely made.<br> True or False
weqwewe [10]

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5 0
3 years ago
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
hammer [34]
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N
therefore, the work done by the weight is
W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J

8 0
3 years ago
The normal force acts ________________________ to the surface and the friction force act _______________ to the surface.
Agata [3.3K]
The first one is perpendicular and second one is opposite
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4 years ago
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