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zalisa [80]
3 years ago
9

What the suns mass in scientific notation????

Physics
2 answers:
Allushta [10]3 years ago
6 0

Explanation:

In scientific notation the Sun's mass is: =1.989 x 10 ^30 kg

Gala2k [10]3 years ago
6 0

Answer:

I think in scientific notation the Sun's mass becomes: <u>M Sun = 1.989 x 10 30 kg.</u> The number above the ten, called the power of ten or exponent, stands for the number of decimal places. If it is positive, as in the mass of the Sun, the decimal places are in front of the decimal point.

I hope this help you!:)

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A flock of ducks is trying to migrate south for the winter, but they keep being blown off course by a wind blowing from the west
Minchanka [31]

The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}

(see the attached graphic)

We have

  • ducks (relative to wind) = 7.0 m/s in some direction <em>θ</em> relative to the positive horizontal direction, or

\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)

  • wind (relative to Earth) = 5.0 m/s due East, or

\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)

  • ducks (relative to earth) = some speed <em>v</em> due South, or

\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)

Then by setting components equal, we have

\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0

\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v

We only care about the direction for this question, which we get from the first equation:

\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}

\cos\theta=-\dfrac57

\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies <em>both</em> equations. We want

\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v

which means <em>θ</em> must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

8 0
3 years ago
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length t
gladu [14]

The question is incomplete. The complete question is :

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

Solution :

Given that :

The predicted range is 0.3503 m

Time of the fall is :

$t=\sqrt{\frac{2H}{g}}$

v_1t= 0.35  ...........(i)

v_0t= 1.09  ...........(ii)

Dividing the equation (ii) by (i)

$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$

∴ v_0=3.11  \ v_1

Now loss of energy  = change in the kinetic energy

$W=\frac{1}{2} m [v_0^2-v_1^2]$

$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$

$W=7.307\times 10^{-3} \ v_0^2$

If f is average friction force, then

(f)(L) = W

(f) (1) = $7.307\times 10^{-3} \ v_0^2$

(f)  = $7.307\times 10^{-3} \ v_0^2$

3 0
3 years ago
Which of the following experiments could be used to determine the inertial mass of a block? A. Place the block on a rough horizo
valentinak56 [21]

Answer:

D, using a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force

Explanation:

For this you would use the net force equation acceleration=net force * mass however you will want to isolate mass so it would be acceleration/ net force to get mass. Then process of elimination comes to play.

3 0
3 years ago
About how much of Earth's water is available for humans to use?
lesya692 [45]

Answer:

0.3 %

Explanation:

Earth cleans and replenishes the water supply through the hydrologic cycle. The earth has an abundance of water, but unfortunately, only a small percentage, is even usable by people.

5 0
3 years ago
A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of
tekilochka [14]

The largest mass of cargo the balloon can lift is 791.06 kg

First, we need to calculate the mass of helium.

Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.

The volume of the balloon also equals the volume of helium present.

Now, the mass of helium m = density of helium, ρ × volume of helium, V

m = ρV

Since ρ = 0.179 kg/m³

m = ρV

m = ρ4πr³/3.

m = 0.179 kg/m³ × 4π(7.15 m)³/3

m = 0.179 kg/m³ × 4π(365.525875 m³)/3

m = 0.179 kg/m³ × 1462.1035π m³/3

m = 261.7165265π/3 kg

m = 822.207/3 kg

m = 274.07 kg

Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.

The weight of this mass W = Mg where g = acceleration due to gravity.

The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.

W' = m'g

Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'

We know that the volume of air displaced, V' = volume of balloon, V

So, V' = V = 4πr³/3.

Since the density of air, ρ' = 1.29 kg/m³,

m' = ρ'V

m = 1.29 kg/m³ × 4π(7.15 m)³/3

m = 1.29 kg/m³ × 4π(365.525875 m³)/3

m = 1.29 kg/m³ × 1462.1035π m³/3

m = 1886.113515π/3 kg

m = 5925.4/3 kg

m = 1975.13 kg

So, the net weight W" that the balloon can lift is W" = W' - W = m'g - Mg = (m' - M )g = (1975.13 kg - 1184.07 kg)g = 791.06g.

So, the net mass m" = W"/g = 791.06g/g = 791.06 kg

This net mass is the largest mass of cargo that the balloon can lift.

Thus, the largest mass of cargo the balloon can lift is 791.06 kg

Learn more about balloons here:

brainly.com/question/21890581

8 0
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