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blagie [28]
3 years ago
6

Will pushing on a car always change the car's mechanical energy?

Physics
1 answer:
tamaranim1 [39]3 years ago
7 0

I assume you mean that the car's motor is not running ... the car is just
sitting there.

If that's so, then the car's mechanical energy is just like the mechanical
energy of any other object.  It has potential energy if it's in a high place
from which it can roll or fall, and it has kinetic energy if it's moving.

-- If you make the car move by pushing it, then you gave it kinetic energy
that it didn't have while it was just sitting there.

-- If it's already moving slowly, and you're able to make it move faster by
pushing, then you increased its kinetic energy.

-- If you're able to push it up a hill, no matter how small the hill is but just
to any higher place, then you gave it more gravitational potential energy
than it had before you came along.

In all of these cases, if you exert a force and keep exerting it through some
distance while the car moves, then you have done "work", which is just
another name for mechanical energy, and your work adds to the mechanical
energy of the car.

But if you didn't move the car, then no matter how hard you pushed, no work
was done, and the car's mechanical energy didn't change.


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Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
4 years ago
Read 2 more answers
From what is known about spring tides and neap tides, you can conclude that:_________.
mixas84 [53]

Answer:

D

Explanation:

Because neap and spring tides happen twice a month during new moon and full moon .It occurs twice in a month which is 14 days

7 0
3 years ago
the figure shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angl
Cerrena [4.2K]

Answer:

(a) 1.054 m/s²

(b) 1.404 m/s²

Explanation:

0.5·m·g·cos(θ) - μs·m·g·(1 - sin(θ))  - μk·m·g·(1 - sin(θ))  = m·a

Which gives;

0.5·g·cos(θ) - μ·g·(1 - sin(θ)   = a

Where:

m = Mass of the of the block

μ = Coefficient of friction

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration of the block

θ = Angle of elevation of the block = 20°

Therefore;

0.5×9.81·cos(20°) - μs×9.81×(1 - sin(20°)  - μk×9.81×(1 - sin(20°) = a

(a) When the static friction μs = 0.610  and the dynamic friction μk = 0.500, we have;

0.5×9.81·cos(20°) - 0.610×9.81×(1 - sin(20°)  - 0.500×9.81×(1 - sin(20°) = 1.054 m/s²

(b) When the static friction μs = 0.400  and the dynamic friction μk = 0.300, we have;

0.5×9.81·cos(20°) - 0.400×9.81×(1 - sin(20°)  - 0.300×9.81×(1 - sin(20°) = 1.404 m/s².

3 0
3 years ago
What is santa's average speed when visiting houses if he visits 800 million houses over 48 hours?
goldenfox [79]

Well, "speed" normally = (total distance covered) / (time to cover the distance).

In this case, we don't know the total distance that Santa has traveled, so the question can be answered only if we're willing to consider some other meaning of "speed".

Maybe we could say that in this case, "speed" means the average rate at which he visits houses. Then ...

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Speed of visits = (800 million houses) / (48 hours)

Speed of visits = 16.67 million houses per hour.

That's 277.8 thousand houses per minute

That's <em>4,630 houses per second</em>

8 0
3 years ago
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AleksAgata [21]

Explanation:

Net force on the car= mass of the car × acceleration

F=1×10^3×4.5

=4.5×10^3 N

3 0
3 years ago
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