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Kobotan [32]
2 years ago
6

As a freely falling object picks up downward speed, what happens to the power supplied by the gravitational force?.

Physics
1 answer:
siniylev [52]2 years ago
5 0

Answer:

As a freely falling object picks up downward speed. What happens to the power supplied by the gravitational force? Does it increase, decrease or stay the same? The power will increase because (Power=work/time; work=f(d); and F x d/t; FV).

Explanation:

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A robin flies a distance of 45963 cm. How far has it flown in kilometers?
alina1380 [7]

Answer:

0.46km

Explanation:

45963cm/100cm=459.63m/1000m=0.45963 or 0.46km

6 0
2 years ago
When you look at the backside of a shiny teaspoon held at arm's length, do you see yourself upright or upside down? (b) When you
igor_vitrenko [27]

Answer:

a) The back spoon gives a right image (upright)

b) the front gives an inverted image

Explanation:

The spoon is a curved metallic object, when we see ourselves from the back we have a convex mirror, in this type of mirror when the law of reflection is applied the rays diverge therefore the eye-brain system forms the image with the prolongation of the rays, therefore the image is straight and smaller than the object.

When we look through the deep side of the spoon, we have a concave mirror and as the object (we) is further away than the distance, the rays converge to a point, so the image is real, inverted smaller than the object.

In summary.

a) The back spoon gives a right image (upright)

b) the front gives an inverted image

3 0
2 years ago
In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
Bezzdna [24]

Answer with Explanation:

We are given that

r=0.053 nm=0.053\times 10^{-9} m

1 nm=10^{-9} m

Charge on proton,q=1.6\times 10^{-19} C

a.We have to find the electric  potential of the proton at the position of the electron.

We know that the electric potential

V=\frac{kq}{r}

Where k=9\times 10^9

V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

V=27.17 V

B.Potential energy of electron,U=\frac{kq_e q_p}{r}

Where

q_e=-1.6\times 10^{-19} c=Charge on electron

q_p=q=1.6\times 10^{-19} C=Charge on proton

Using the formula

U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

U=-4.35\times 10^{-18} J

8 0
3 years ago
A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot
denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

P_r = 345 Watt

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

P_r = 345 Watt

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

\Delta V = E + i r

here we have

E = 12 V

i = 56 A

r = 0.11

now we have

\Delta V = 12 + (0.11)(56) = 18.16 V

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

P_r = i^2 r

P_r = 56^2 (0.11)

P_r = 345 W

Part c)

Rate of energy conversion into EMF is given as

P_{emf} = i E

P_{emf} = (56)(12)

P_{emf} = 672 Watt

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

V = E - i r

V = 12 - (56)(0.11)

V = 12 - 6.16 = 5.84 V

Part e)

now the rate of energy dissipation is given as

P_r = i^2 r

P_r = 56^2 (0.11)

P_r = 345 W

7 0
3 years ago
How are waves that are created by a musical instrument and waves that
Wittaler [7]

Answer:

both caused  by physical vibrations

7 0
3 years ago
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