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Naddika [18.5K]
3 years ago
7

On a day when the temperature reaches 50°F, the temperature in degrees Celsius is: 20°C

Physics
1 answer:
anzhelika [568]3 years ago
3 0

Answer:

10°C

Explanation:

To convert °F to °C, we use the formula:

°C =  (°F - 32) * ( 5/9)

So, to convert 50°F to the equivalent  in °C, we can proceed as follows:  

°C = ( 50 - 32 ) * (5/9)  

°C = ( 18 ) *  (5/9), which is, approximately,

°C = 9.999999999... ≈ 10 (5/9 ≈0.555555...)

So, 50°F is equivalent to 10°C.

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r-ruslan [8.4K]

<u>The correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge. </u>

<u> </u>

Further Explanation:

The electric field intensity at a point is the measure of the force exerted by a charge particle on another charge particle in the particular area of its strength.

The electric field intensity at a distance d due to a static charge having charge q is directly proportional to the amount of charge and inversely proportional to the square of the distance between them.

The Electric field intensity due to a charge is given as:

E = \dfrac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}

Here, E is the electric field intensity, q is the amount of charge and r is the distance of the charge from the point.

The above expression of electric field shows that the electric field intensity at a point depends on the amount of charge as well as the distance of the point from the charge.

<u>Thus, the correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge. </u>

<u> </u>

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Answer Details:

Grade: Senior school

Subject: Physics

Chapter: Electrostatics

Keywords:  Strength, electric field, charge, distance, electric field intensity, magnitude of charge, electrostatic, test charge, kq/r^2.

7 0
3 years ago
Read 2 more answers
If the amplitude of a simple harmonic oscillator is doubled, by what factor does the total energy increase?
dybincka [34]

Answer:

c)by a factor of four

Explanation:

The total energy of a simple harmonic oscillator is given by

E=\frac{1}{2}kA^2

where

k is the spring constant of the oscillator

A is the amplitude of the motion

In this problem, the amplitude of the oscillator is doubled, so

A' = 2A

Therefore, the new total energy is

E'=\frac{1}{2}k(2A)^2=4(\frac{1}{2}kA^2)=4E

So, the total energy increases by a factor 4.

6 0
3 years ago
In an energy pyramid, the total energy available for consumption from one trophic level to the next one it is
Flura [38]

Answer: Conserved

Explanation:

5 0
3 years ago
PLEASE HELP ME IM ON A TIMER
garri49 [273]

Answer:

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).

Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).

And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)

And the displacement will be defined by the folliwing vector operation:

A (0,0) = oi + 0j\\F (3,6) = 3i + 6 j\\Displacement vector = (3-0)i + (6-0)j = 3i + 6j

We know that the magnitude of the displacement vector is defined by the phytagoras theorem

Displacement = \sqrt{(3)^2+ (6)^2} \\Displacement = 6.70

And the angle will be defined by:

tan(beta)=3/6

beta = tan^-1(6/3)

beta = 63.43°

5 0
3 years ago
Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm
Ghella [55]

Answer:

Total electric potential, V=1.32\times 10^6\ volts

Explanation:

It is given that,

First charge, q_1=3\ \mu C=3\times 10^{-6}\ C

Second charge, q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C

Distance of first charge from origin, r_1=1.15\ cm=0.0115\ m

Distance of second charge from origin, r_2=2\ cm=0.02\ m

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}

V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})

V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})

V = 1321826.08 V

or

V=1.32\times 10^6\ volts

So, the total electric potential at the origin is 1.32\times 10^6\ volts. Hence, this is the required solution.

3 0
2 years ago
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