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3 years ago

14

The shuttles main engine provides 154,360 kg of thrust for 8 minutes. If the shuttle accelerated at 29m/s/s, and fires for at le

ast 8 minutes, then how far does the shuttle go

1 answer:

Vinil7 [7]3 years ago

3 0

Answer:

The answer to the question is

3340800 m far

Explanation:

To solve the question, we note that acceleration = 29 m/s²

Time of acceleration = 8 minutes

Then if the shuttle starts from rest, we have

S = u·t+0.5·a·t² where u = 0 m/s = initial velocity

S = distance traveled, m

a = acceleration of the motion, m/s²

t = time of travel

S = 0.5·a·t² = 0.5×29×(8×60)² = 3340800 m far

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The mass of the Sun is 1. 99 × 1030 kg. Jupiter is 7. 79 × 108 km away from the Sun and has a mass of 1. 90 × 1027 kg. The gravi

german

The gravitational force is s type of force that has the ability to attract any two objects having mass. The gravitational force will be $4.16\times10^{23}$ .

<h3>What is the gravitational force?</h3>

The gravitational force is s type of force that has the ability to attract any two objects with mass. Gravitational force tries to pull two masses towards each other.

$F= G\frac{m_1m_2}{r^{2} }$

Given,

mass of the sun ( $m_1$ )= $1.99\times10^{23}$ kg

mass of Jupiter( $m_2$ )= $7.79\times10^{8}$ kg

distance between the sun and Jupiter (r)= $1.90\times10^{27}$ m

$F= G\frac{m_1m_2}{r^{2} }\\\\\\F=4.16\times10^{23}\times\frac{1.99\times10^{23}\times7.79\times10^{8}}{({1.90\times10^{27})^2} }$

$F= 4.16\times(10)^{23}$ Newton

Hence the gravitational force between the sun and Jupiter will be $4.16\times10^{23}$

To learn more about gravitational force refer to the link:

brainly.com/question/24783651

4 0

2 years ago

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

Lubov Fominskaja [6]

Answer:

The moment of inertia is $I=\frac{M}{12} a^{2}$

Explanation:

The moment of inertia is equal:

$I=\int\limits^a_b {r^{2} } \, dm$

If r is $-\frac{a}{2}$

and $dm=\frac{M}{a} dr$

$I=\int\limits^a_b {r^{2}\frac{M}{a} } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}$

$I=\frac{M}{a} \int\limits^a_b {r^{2} } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\ I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}$

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3 years ago

An object of irregular shape has a characteristic length of L = 0.5 m and is maintained at a uniform surface temperature of Ts =

goblinko [34]

Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

Characteristic length = 0.5 m

Surface temperature = 400 K

Atmospheric temperature = 300 K

Velocity = 25 m/s

Air velocity = 5 m/s

Characteristic length of second object = 2.5 m

We have same shape and density of both objects so the reynold number will be same,

We need to calculate the value of the average convection coefficient

Using formula of reynold number for both objects

$R_{1}=R_{2}$

$\dfrac{u_{1}L_{1}}{\eta_{1}}=\dfrac{u_{2}L_{2}}{\eta_{2}}$

$\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}$

Here, $k_{1}=k_{2}$

$h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}$

$h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}$

Put the value into the formula

$h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}$

$h_{2}=20\ W/Km^2$

Hence, The value of the average convection coefficient is 20 W/Km².

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3 years ago

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inessss [21]

The right answer is B. hope this helps you :)

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2 years ago

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3 years ago

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