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zimovet [89]
3 years ago
13

Think about the motion of a roller coaster car, after completing a trip around the track

Physics
1 answer:
Alenkinab [10]3 years ago
6 0

Answer:

it should be B let me know if i am wrong

Explanation:

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A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering
Black_prince [1.1K]
<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)

Where:

y is the height of the stone at 6s (the value we want to find)

y_{o}=925ft is the initial height of the stone

V_{o}=20ft/s is the initial velocity of the stone

t=6s is the time  at which we need to find the height

g=32ft/s^{2} is the acceleration due to gravity

Having this clear, let's find y from (1):

y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2} (2)

Finally:

y=469ft This is the height of the stone at t=6s

4 0
3 years ago
an incandescent lightbulb has an efficiency of 2.1% and a power of 60 w. how much light energy does the lighbulb produce in 1 se
AveGali [126]

Explanation:

Power output of the bulb:

0.021 × 60 W = 1.26 W

Energy produced by the bulb in 1 second:

E = Pt

E = (1.26 W) (1 s)

E = 1.26 J

Round as needed.

5 0
3 years ago
If the distance between two masses is tripled, the gravitational force between changes by a factor of:_______
adell [148]

Answer:

option A

Explanation:

given,

distance between two masses is doubled

new distance, r' = 3 r

using gravitational force equation

F = \dfrac{GMm}{r^2}............(1)

new gravitational force

F' = \dfrac{GMm}{r'^2}

now from the given condition

F' = \dfrac{GMm}{(3r)^2}

F' = \dfrac{GMm}{9r^2}

F' = \dfrac{1}{9}\dfrac{GMm}{r^2}

now, from equation (1)

F' = \dfrac{1}{9}F

\dfrac{F'}{F} = \dfrac{1}{9}

now, the change in gravitational force factor is equal to \dfrac{1}{9}

Hence, the correct answer is option A

3 0
3 years ago
Point charge μC is located at x =, y = , point charge is located at x = 0m. What are (a)the magnitude and (b)direction of the to
pishuonlain [190]

Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0

Explanation:

  • a) First of all find the distance between the two charges;
  • x = 0, y = 0.30  and x = 0.40 m, y = 0
  • r = √( 0.4² + 0.3²)
  • = 0.5m

hence, the force F = 2Kq1q2cosθ /r²...............equation 1

but cosθ = y/r = 0.3/0.5

cosθ = 0.6

plugging back to equation 1;

F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2

F = 540 x 10^-3

Magnitude of Force = 0.54N

b) Direction is at angle 90

6 0
3 years ago
Why is electrical energy considered as potential energy as its the energy of moving electrons?
Veseljchak [2.6K]

Answer:

Explanation: When the electrons move in another direction, they convert this chemical potential energy to electricity in the circuit, thus discharging the battery. So, the battery is all potential energy.

8 0
2 years ago
Read 2 more answers
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