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PtichkaEL [24]
3 years ago
14

Complete the following statement:When work is done on a positive test charge by an external force to move it from one location t

o another, electric potential _________.( increase or decrease).
Physics
1 answer:
Lena [83]3 years ago
4 0

Answer:

increase

Explanation:

Electric potential is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field. Taking a test positive charge from one point to another means that work is done against the field hence electric potential increases.

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How long will it take a car to accelerate from 15.2 to 23.5 m/s if the car has an average acceleration of 3.2 m/s?
Nuetrik [128]
 a =  ( V2 - V1)/( t2 - t1)
 3.2 =  ( 23.5m/s - 15.2m/s)/(t - 0)
3.2m/s = 8.3/t
t(3.2) = 8.3
t = 8.3/3.2
t = 2.59 seconds
7 0
3 years ago
Read 2 more answers
b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
Nat2105 [25]

Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
What is the frequency of a photon with an energy of 4. 56 x 10^-19 j
Sauron [17]

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma  = 6.88 \times 10^{14}\ s^{-1}

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

Learn more about Frequency:

brainly.com/question/5102661

#SPJ4

5 0
2 years ago
Read 2 more answers
Convert 2 kg to cg give you answer in SI
BARSIC [14]

1 kg=100000 cg

2 kg=200000 cq

If mass is the quantity then kg is the S.I

2 kg=2kg

6 0
3 years ago
The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he
schepotkina [342]

Answer:

The change in volume is 6.885\times 10^{- 5}\

Solution:

As per the question:

Coefficient of linear expansion of Copper, \alpha = 17\times 10^{- 6}\ K^{- 1}

Initial Temperature, T = 0^{\circ} = 273 K

Final Temperature, T' = 100^{\circ} = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = 30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}

V' = V(1 + \gamma \Delta T)

\gamma = 3\alpha

V' = V(1 + 3\alpha \Delta T)

where

V' = Final volume

V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))

\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\

7 0
3 years ago
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