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2 years ago

Calculate the average force with which the Sun pulls the Earth. It will help to know

Physics

1 answer:

algol132 years ago

4 0

Answer:

7.9 x 10^21 pound-force

Explanation:

The average distance between the Earth and sun is 150 trillion meters, or 1.5 x 10^11 meters. The mass of the sun is 1.99 x 10^30 kilograms, while the Earth weighs in at 6.0 x 10^24 kilograms. The gravitational constant is 6.67 x 10^-11 meter^3 / (kilogram - second^2). So the Earth and sun pull on each other with a force equal to 3.52 x 10^22 newtons. The newton is a unit of force equal to a kilogram-meter/second^2. One newton is equal to 0.22 of the rarely used English unit called pound-force, so 3.52 x 10^22 newtons is 7.9 x 10^21 pound-force.

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4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the p

Komok [63]

Answer:

14657.32 J

Explanation:

Given Parameters ;

Number of moles mono atomic gas A , n 1 = 4 .2 mol

Number of moles mono atomic gas B , n 2 = 3.2mol

Initial energy of gas A , K A = 9500 J

Thermal energy given by gas A to gas B , Δ K = 600 J

Gas constant R = 8.314 J / molK

Let K B be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2)n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2)n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J

6 0

3 years ago

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

3 years ago

The primary gases of the atmosphere are _____. nitrogen and carbon dioxide nitrogen, oxygen, and argon nitrogen and oxygen nitro

Paladinen [302]

<span>The primary gases of the atmosphere are _____:
</span>
nitrogen, oxygen, and carbondioxide

6 0

3 years ago

Read 2 more answers

You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfo

dolphi86 [110]

Answer:

Explanation:

To convert gram / centimeter³ to kg / m³

gram / centimeter³

= 10⁻³ kg / centimeter³

= 10⁻³ / (10⁻²)³ kg / m³

= 10⁻³ / 10⁻⁶ kg / m³

= 10⁻³⁺⁶ kg / m³

= 10³ kg / m³

So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

2.33 gram / cm³

= 2.33 x 10³ kg / m³ .

3 0

3 years ago

When a charge of 8 C flows past any point along a circuit in 2 seconds, the current is ________ A?

RoseWind [281]

The answer would be 4 A.

7 0

3 years ago