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Margarita [4]
3 years ago
13

What happens when a object is dropped?

Physics
1 answer:
7nadin3 [17]3 years ago
5 0
B. it accelerates at a rate of 9.8m/s/
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What is the approximate size of the Earth's magnetic field? (dont ask me to specify thats what the question is and im as confuse
Olegator [25]

Answer:

The Earth's magnetic field intensity is roughly between 25,000 - 65,000 nT (.25 -.65 gauss).

Explanation:

<em>To measure the Earth's magnetism in any place, we must measure the direction and intensity of the field. The Earth's magnetic field is described by seven parameters. These are declination (D), inclination (I), horizontal intensity (H), the north (X), and east (Y) components of the horizontal intensity, vertical intensity (Z), and total intensity (F). The parameters describing the direction of the magnetic field are declination (D) and inclination (I). D and I are measured in units of degrees, positive east for D and positive down for me. The intensity of the total field (F) is described by the horizontal component (H), vertical component (Z), and the north (X) and east (Y) components of the horizontal intensity. These components may be measured in units of gauss but are generally reported in nanoTesla (1nT * 100,000 = 1 gauss). </em><em>The Earth's magnetic field intensity is roughly between 25,000 - 65,000 nT (.25 - .65 gauss). </em><em>Magnetic declination is the angle between magnetic north and true north. D is considered positive when the angle measured is east of true north and negative when west.  The magnetic inclination is the angle between the horizontal plane and the total field vector, measured positive into Earth. In older literature, the term “magnetic elements” is often referred to as D, I, and H.</em>

8 0
3 years ago
A piece of styrofoam has a charge of 0.002 mC and is placed 0.5 m from a grain of salt with a charge of 0.03 nC. How much electr
aleksklad [387]

Answer:

2.16×10⁻⁶ N

Explanation:

Applying,

F = kqq'/r² (coulomb's Law)....................... Equation 1

Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.

From the question,

Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m

Constant: k = 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.0×10⁻⁶)(3.0×10⁻¹¹)(8.99×10⁹)/0.5²

F = 2.16×10⁻⁶ N

5 0
2 years ago
In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this
soldier1979 [14.2K]

Answer:

1). v = - 2960526m/s

2). Toward us

3). v = - 493421m/s

4). Toward us

5). v = 1480263m/s

6).  Away from us

7). v = 3207236m/s

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm (\lambda_{0} = 121.6nm)

Redshift: \lambda_{measured} > \lambda_{0}

Blueshift: \lambda_{measured} < \lambda_{0}

Then, for this particular case it is gotten:

Star 1: \lambda_{measured} = 120.4nm

Star 2: \lambda_{measured} = 121.4nm

Star 3: \lambda_{measured} = 122.2nm

Star 4: \lambda_{measured} = 122.9nm

Star 1:

Blueshift: 120.4nm < 121.6nm

Toward us

Star 2:

Blueshift: 121.4nm < 121.6nm

Toward us

Star 3:

Redshift: 122.2nm > 121.6nm

Away from us

Star 4:

Redshift: 122.9nm > 121.6nm

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

v = c\frac{\Delta \lambda}{\lambda_{0}}  (1)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the wavelength at rest, v is the velocity of the source and c is the speed of light.

v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}}) (2)

<em>Case for star 1 \lambda_{measured} = 120.4 nm:</em>

<em></em>

v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})

v = - 2960526m/s

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 \lambda_{measured} = 121.4 nm:</em>

v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})

v = - 493421m/s

<em>Case for star 3 \lambda_{measured} = 122.2 nm:</em>

v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})

v = 1480263m/s

<em>Case for star 4 \lambda_{measured} = 122.9 nm:</em>

v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})

v = 3207236m/s

4 0
3 years ago
Which scientist proved experimentally that a shadow of the circular object illuminated 18. with coherent light would have a cent
Zina [86]

Answer:

Your answer is( D) - Arago

8 0
3 years ago
Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 2.6 m long, its mass is 0.5
Irina-Kira [14]

Answer:

15.13 m/s

Explanation:

The wave speed of the stretched rope can be calculated using the following formula

v = \sqrt{\frac{F_T}{\mu}}

where F_T = 44N is the tension on the rope and \mu = m/L = 0.5 / 2.6 = 0.1923 kg/m is the density of the rope per unit length

v = \sqrt{\frac{44}{0.1923}} = \sqrt{228.8} = 15.13 m/s

6 0
3 years ago
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