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chubhunter [2.5K]
2 years ago
6

Witch SI units are commonly used to measure mass and weight

Physics
1 answer:
Minchanka [31]2 years ago
8 0

Explanation: Mass is Kilogram

Weight is Newton

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What law describes light
Darina [25.2K]

The law of reflection

6 0
2 years ago
Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

  $=0.822 \times 10^{-2} \ km/s$

 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

  = 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

3 0
3 years ago
A student in Denver (altitude = 1 mile = 1609 m above sea level) brings a physics book of mass 1.3 kg to the top of a ten story
Nesterboy [21]

Answer:

Explanation:

Potential energy is the energy of a body due to is virtue of rest.

Potential energy is given as mgh

g is a constant and it is 9.81m/s²

And also the mass of the body is given as 1.3kg

Now the height of the body is

He took a book to a storey building of height 26m

He still holds the book 151 cm (1.51m) above the house.

The house is on an altitude of 1609m from the sea level.

Total Ug with out the sea level is

Ug=mgh

Ug=1.3 × 9.81 ×(26+1.51)

Ug=350.84J

Then, the potential energy due to the sea level is given as

Ug=mgh

Where g = 1/6371 m/s²

Therefore

Ug=mgh

Ug=1.3 × 1/6371 ×1609

Ug=0.328J

Total energy = 0.328+350.84

Ug=351.17J

8 0
3 years ago
Why do planetary scientists hypothesize that the moon formed with a molten surface
Verdich [7]

The oldest lunar rock samples are approximately 4.4 billion years old and composed of anorthosite, a mineral that crystallizes and rises to the top of a lava ocean.

5 0
3 years ago
A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
3 years ago
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