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vodomira [7]
2 years ago
8

2.2.4 Quiz: Conservation of Energy

Physics
1 answer:
dexar [7]2 years ago
7 0

Answer:

it is the judicious use of energy to prevent wastage

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Q6) A speed skater moving to the left across frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadil
77julia77 [94]

Answer:

-2.8 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity The S. I unit of acceleration is m/s²

Using the equation of motion,

v² = u² + 2as................... Equation 1

Where v = Final velocity, u = initial velocity, a = acceleration, s = distance,

Given: v = 6.0 m/s, u = 8.0 m/s, s = 5.0 m.

Substituting into equation 1

6² = 8²+2(a)5

36 = 64 + 10a

10a = 36-64

10a = -28

10a/10 = -28/10

a = -2.8 m/s²

Note: a is negative because because the skater decelerate on the rough ice

Hence the magnitude of her acceleration is  = -2.8 m/s²

6 0
3 years ago
Please Help On These 2 Questions!!!!! I severely need help!!!!!!
lys-0071 [83]
3. In a uniform electric field, the equation for the magnitude of the magnetic field is E=(V/d). V= voltage d= distance. If the magnetic field magnitude is constant , as stated in your problem, then the voltage must stay the same otherwise the value of "E" would change". And the problem already told us the "E" is uniform and so, not changing. Does that make sense?

4a. If the magnetic field lines are equally spaced apart, in other words share the same density. Then we know that the magnitude of the magnetic field is unchanging. This is because the density of of the magnetic field lines(how many are in a certain area) is related to the magnitude being expressed by the electric field. Greater magnitude is expressed by the presence of more lines (higher line density) 

4b. The electric potential is measured in Volts(V) and is uniform along  the same equipotential line. What is an equipotential line(gray)? It is a line drawn perpendicular(forms a right angle with) to the magnetic field lines(black) to show the changes in electric potential. One space where electric potential will always be the same because it will always be equal to 0 Volts is exactly in between a positive and negative charges of equal charge value I have pointed to this line with a purple arrow in my picture.

I really hope this makes sense to you and that my pictures help! :)

3 0
2 years ago
At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravit
Over [174]

Answer:

v=32.9m/s

Explanation:

The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation a=v^2/r

This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).

We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:

v=\sqrt{ar} = \sqrt{12.5gr}=\sqrt{(12.5)(9.8m/s)(8.84m)}=32.9m/s

7 0
2 years ago
Real life examples of atomic physics
Flauer [41]

Answer:

magnets

static electricity

heat

7 0
2 years ago
*WILL MARK BRAINLIEST FOR RIGHT ANSWER* How much current must be applied across a 60 Ω light bulb filament in order for it to co
sp2606 [1]

Answer: The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

I = \sqrt{\frac{P}{R} }

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }

Then, simplify the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} }

Rationalize the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}

Simplify the numerator by finding its factors:

I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}

The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

8 0
3 years ago
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