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koban [17]
3 years ago
10

If the bus had a mass of 1,021kg how much does the bus weigh?

Physics
1 answer:
goblinko [34]3 years ago
4 0
Weight of the bus= mass x acceleration due to gravity = 1021x9.8 = 10005.8 N. Hope it helps.
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Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive
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Answer:

The mass of the massive object at the center of the Milky Way galaxy is 3.44\times10^{37}\ Kg

Explanation:

Given that,

Diameter = 10 light year

Orbital speed = 180 km/s

Suppose determine the mass of the massive object at the center of the Milky Way galaxy.

Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.

We need to calculate the radius of the orbit

Using formula of radius

r=\dfrac{d}{2}

r=\dfrac{15\times9.461\times10^{15}}{2}

r=7.09\times10^{16}\ m

We need to calculate the mass of the massive object at the center of the Milky Way galaxy

Using formula of mass

M=\dfrac{v^2r}{G}

Put the value into the formula

M=\dfrac{(180\times10^3)^2\times7.09\times10^{16}}{6.67\times10^{-11}}

M=3.44\times10^{37}\ Kg

Hence, The mass of the massive object at the center of the Milky Way galaxy is 3.44\times10^{37}\ Kg

5 0
3 years ago
A car driving on the highway is going 72mph After 3 hours how many miles will the car have driven ?
Sergeu [11.5K]

Answer:

216 miles

Explanation:

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Answer:

The general shape of a frequency distribution. For many data sets, statisticians use this information to determine whether there is a “normal” distribution of values. In normal distributions, the mean, median, and mode are the same. Whether the distribution is symmetrical or skewed in a certain direction. If the data is skewed to the right, this shows the mean will be greater than the median. Similarly, if the data is skewed left, the mean will be less than the median. The symmetry, or asymmetry, of the chart can help statisticians calculate probability. The modality of the data set. This means how many peaks exist in the data. For normal distributions, there will be one peak, or mode, in the data set.

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The body weighing 8kg moves in a straight line uniformly accelerated with an acceleration of 3m / s? on a horizontal surface, un
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Answer:

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3 0
3 years ago
A parallel-plate capacitor has square plates that are 7.20 cm on each side and 3.40 mm apart. The space between the plates is co
Lilit [14]

Answer:

U = 218 nJ

Explanation:

We are given;

Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m

Voltage across the capacitor; V = 96 V

Dimension of the square plates is 7.2cm x 7.2cm.

So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²

Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²

From relative permeability table;

Dielectric constant of Pyrex; k1 = 5.6

Dielectric constant of polystyrene; k2 = 2.56

Now, formula for capacitance of a capacitor with Dielectric is;

C = kC_o

Where, C_o = ε_o(A/d)

Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)

Since we have 2 capacitor, thus ;

C1 = k1*ε_o*(A/d)

C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C1 = 1.51 × 10^(-10) F

Similarly;

C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C2 = 0.691 × 10^(-10) F

For capacitors in series, formula for total capacitance(Cs) is;

1/Cs = (1/C1) + (1/C2)

Simplifying this, we have;

Cs = (C1*C2)/(C1 + C2)

Plugging in the relevant values ;

Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))

Cs = 0.474 × 10^(-10) F

The formula for energy stored in a capacitor with 2 Dielectrics is given as;

U = ½Cs*V²

So,

U = ½ × 0.474 × 10^(-10) × 96²

U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ

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3 years ago
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