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2 years ago

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Using the mass and radius of the Sun given under the "Vital Statistics" section of Chapter 10 in your textbook, calculate the Su

n's average density. Note: Remember that the radius is half the diameter and be sure to convert from kilometers to meters before using in the calculations. Use the following equations: LaTeX: V=\frac{4}{3}\pi r^3 V = 4 3 π r 3 where V = volume (m3) and r = radius (m) LaTeX: \rho=\frac{m}{V} rho = m V where LaTeX: \rho rho = density (kg/m3), m = mass (kg), and V = volume (m3)

1 answer:

atroni [7]2 years ago

5 0

Answer:

The density of the sun is 4434kg/m³

This was found by dividing the mass (1.989 ×10³⁰kg) by the volume (4.486×10²⁶ m³) which was calculated using V = 4/3×pi ×r³

Explanation:

See attachment below.

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When a pendulum is pulled back from its equilibrium position by 10∘, the restoring force is 1.0 N. When it is pulled back to 30∘

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Answer: B

Explanation: I said B because if you pull something back what is going to be more of a force pulling back or letting it go for a rubier band yes it will have more force if you let it go

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2 years ago

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A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy

GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

$v=\sqrt{v_1^2+v_2^2}$

$v=\sqrt{5^2+8^2}=9.43\ m/s$

Let E is the kinetic energy of the plane. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2$

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

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2 years ago

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If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it

Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

$\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

Explanation:

From the question we are told that

The mass of the rocket is m = 6300 kg

The velocity at gas is being ejected is u = 2000 m/s

The initial acceleration desired is $a = 27.0 \ m/s$

The time taken for the gas to be ejected is t = 1 s

Generally this desired acceleration is mathematically represented as

$a = \frac{u * \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}$

Here $\frac{\Delta m}{\Delta t }$ is the rate at which gas is being ejected with respect to time

Substituting values

$27 = \frac{2000 * \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}$

=> $170100 -27* \frac{\Delta m}{\Delta t} = 2000 * \frac{\Delta m}{\Delta t}$

=> $170100 = 2027 * \frac{\Delta m}{\Delta t}$

=> $\frac{\Delta m}{\Delta t} = \frac{170100}{2027}$

=> $\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

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2 years ago

As shown in the figure below, Justin walks from the house to his truck on a windy day. He walks 20 m toward

juin [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is $v =0.333 \ m/s$ in positive x -direction

The speed is $s = 0.733 \ m/s$

Explanation:

From the question we are told that

The distance from the house to truck is D = 20 m

The distance traveled back to retrieve wind-blown hat is d = 15

The distance from the wind-blown hat position too the truck is k = 20 m

The total time taken is t = 75 s

Generally when calculating the displacement the Justin's backward movement to collect his wind - blown hat is taken as negative

Generally Justin's displacement is mathematically represented as

$L = 20 - 15 + 20$

=> $L = 25 \ m$

Generally the average velocity is mathematically represented as

$v = \frac{L}{t}$

=> $v = \frac{25}{75}$

=> $v =0.333 \ m/s$

Generally the distance covered by Justin is mathematically represented as

$R = D+ d + k$

=> $R = 20 + 15 +20$

=> $R = 55 \ m$

Generally Justin's average speed over a 75 s period is mathematically represented as

$s = \frac{R}{ t}$

=> $s = \frac{55}{ 75}$

=> $s = 0.733 \ m/s$

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2 years ago

The engine on a fighter airplane can exert a force of 105,840 N (24,000 pounds). The take-off mass of the plane is 16,875 kg. (I

FrozenT [24]

Answer:

The acceleration you can get with that engine in your car is around 70,56 $(\frac{m}{s^{2} })$ or 7,26 $(\frac{ft}{s^{2} } )$ using 1500kg of mass or 3306 pounds

Explanation:

Using the equation of the force that is:

$F=m*a$

So, you notice that you know the force that give the engine, so changing the equation and using a mass of a car in 1500 kg or 3306 pounds

$a=\frac{F}{m} =\frac{105840 N }{1500 (kg) }$

$a=\frac{105840 (\frac{kg*m}{s^{2} } )} {1500 kg }$

<em> Note: N or Newton units are: $\frac{kg * m}{s^{2} }$ </em>

$a= 70,54 \frac{m}{s^{2} }$

Also in pounds you can compared

$a= \frac{2400 lf }{3 306 lf}$

Note: lf in force units are: $\frac{lf*ft}{s^{2} }$

$a=7,26 \frac{ft}{s^{2} }$

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2 years ago