Muscle surround your body

A region of space in which a test charge experiences a force is referred to as

Alecsey [184]

Answer: The correct answer is- Field.

Explanation -

The modified space around a charged particle is known as electric field.

The strength of electric field at a point is calculated as the force experienced by a unit test charge present at that point.

Thus, a region of space in which a test charge experiences a force is referred to as Field.

4 0

3 years ago

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Convert 15 litre into cubic metre

atroni [7]

Answer:

0.015m^3

Explanation:

1 m^3 = 1000 liters

x m^3 = 15 liters

Cross multiply

xm^3 x 1000 l = 15 l

Divide both sides by 1000

xm^3 x1000/1000 = 15/1000

xm^3 = 0.015m^3

Therefore 15 liter = 0.015m^3

5 0

3 years ago

You have a string with a mass of 0.0135 kg. You stretch the string with a force of 8.29 N, giving it a length of 1.83 m. Then yo

Sphinxa [80]

The wavelength of the standing wave at fourth harmonic is; λ = 0.985 m and the frequency of the wave at the calculated wavelength is; f = 36.84 Hz

Given Conditions:

mass of string; m = 0.0133 kg

Force on the string; F = 8.89 N

Length of string; L = 1.97 m

1. To find the wavelength at the fourth normal node.

At the fourth harmonic, there will be 2 nodes.

Thus, the wavelength will be;

λ = L/2

λ = 1.97/2

λ = 0.985 m

2. To find the velocity of the wave from the formula;

v = √(F/(m/L)

Plugging in the relevant values gives;

v = √(8.89/(0.0133/1.97)

v = 36.2876 m/s

Now, formula for frequency here is;

f = v/λ

f = 36.2876/0.985

f = 36.84 Hz

Read more about Harmonics of standing waves at; brainly.com/question/10274257

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7 0

2 years ago

Consider two planets of mass m and 2m,

Rzqust [24]

Answer:

Part a)

$\frac{F_1}{F_2} = 10.125$

Part b)

$\frac{v_1}{v_2} = \sqrt{\frac{4.5r}{r}} = 2.12$

Part c)

$\frac{T_1}{T_2} = 9.54$

Explanation:

Part a)

As we know that the gravitational force is given as

$F = \frac{GMm}{r^2}$

so we will have to find the ratio of force on two planets due to star

so here we have

$\frac{F_1}{F_2} = \frac{m_1r_2^2}{m_2r_1^2}$

$\frac{F_1}{F_2} = \frac{m (4.5r)^2}{(2m) r}$

$\frac{F_1}{F_2} = 10.125$

Part b)

Orbital speed is given as

$v = \sqrt{\frac{GM}{r}}$

so the ratio of two orbital speed is given as

$\frac{v_1}{v_2} = \frac{r_2}{r_1}$

$\frac{v_1}{v_2} = \sqrt{\frac{4.5r}{r}} = 2.12$

Part c)

Time period is given as

$T = 2\pi\sqrt{\frac{r^3}{GM}}$

so the ratio of two time period is given as

$\frac{T_1}{T_2} = \sqrt{\frac{r_1^3}{r_2^3}}$

$\frac{T_1}{T_2} = \sqrt{\frac{4.5r^3}{r^3}}$

$\frac{T_1}{T_2} = 9.54$

8 0

3 years ago

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The emission spectra you obtained from a star show a hydrogen spectrum that is shifted toward the blue end of the electromagneti

kvasek [131]

Answer:

The correct option is;

The star is moving toward Earth.

Explanation:

The shifting of the wavelength of light wave toward the blue end of the electromagnetic spectrum is termed blue shift.

A blue-shift of an electromagnetic wave corresponds to the wavelength decrease of the wave, which is equivalent increase in energy, resulting in an increase in the observed frequency of the wave.

Astronomers make use of the shifting of the wavelength of a wave to understand the relative motion of galaxies.

The wavelength of an approaching electromagnetic ave shifts towards the blue end of the electromagnetic spectrum because the wavelength is shorter.

The opposite of phenomenon is red-shift.

4 0

3 years ago