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1 year ago

3.2 Define conservation of energy.

Physics

1 answer:

Free_Kalibri [48]1 year ago

5 0

Answer:

A principle stating that energy cannot be created or destroyed, but can be altered from one form to another.

Explanation:

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A 2,000 kg car starts from rest and coasts down from the top of a 5.00 m long driveway that is sloped at an angel of 20o with th

ehidna [41]

Answer:

The speed at the bottom of the driveway is3.67m/s.

Explanation:

Height,h= 5sin20°= 1.71m

Potential energy PE=mgh= 2000×9.8×1.71

PE= 33516J

KE= PE- Fk ×d

0.5mv^2= 33516 - (4000×5)

0.5×2000v^2= 33516 - 20000

1000v^2= 13516

v^2= 13516/1000

v =sqrt 13.516

v =3.67m/s

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3 years ago

Mary walked north from her home to Sheila's home, which is 4.0 kilometers away. Then she turned right and walked another 3.0 kil

natka813 [3]

Well her speed and velocity are the same 8 kilometers per hour<span />

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2 years ago

Read 2 more answers

Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o

KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹ N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹ N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea = +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹ N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb = -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴ + -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

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3 years ago

If a cell phone is dropped from a very tall building, how far has the phone fallen after 2.7 seconds, neglecting air resistance?

Zielflug [23.3K]

The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
$g=9.81 m/s^2$

So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by
$y(t) = \frac{1}{2}gt^2$
And if we substitute t=2.7 s, we find the distance covered:
$y(t)= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2=35.8 m$

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2 years ago

Dust particles in air have a typical mass of 5.0 x 10-16 kg. They undergo irregular motion due to collisions with air molecules.

Mars2501 [29]

Answer:

Rms speed of the particle will be $38.68\times 10^8m/sec$