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2 years ago

6

An astronomical unit (A.U.) is 1 point A) a term for defining the luminosity of a star B) the average distance from the Earth to

the sun C) the average distance of any given planet to the sun D) equal to a light year

1 answer:

Elenna [48]2 years ago

7 0

Answer:

B) the average distance from the Earth to the Sun

Explanation:

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Dana is on a train traveling at a speed of 20 km/h. Dana walks from the front of the train to the back of the train at a speed o

Maslowich

Answer:

16km/h

Explanation:

Vt=20km/h ---train speed

Vd=4km/h

Donas speed relative to ground is:

Vrd=Vt-Vd

Donas is moving in opposite direction of train .

Vrd=20km/h-4km/h

Vrd=16km/h

7 0

3 years ago

7. How do Newton’s laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can pass a

Arisa [49]

The first law states that “objects at rest and objects in motion remain in motion in a straight line unless acted upon by an unbalanced force”. Keeping the ice smooth will make sure there is not friction, friction would slow the puck down

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2 years ago

Read 2 more answers

Which equation correctly relates mechanical energy, thermal energy and total of energy when there is friction present in the sys

agasfer [191]

Answer:

E total = ME + E thermal

Explanation:

APEX

4 0

3 years ago

A bald eagle is flying to the left with a speed of 34 meters

Shtirlitz [24]

Answer:

the speed after 3 seconds is 10 m/s

Explanation:

The computation of the speed is shown below:

As we know that

V = U + at

Here,

U = 34 m/s

a = - 8 m/s²

t = 3 Sec

V = velocity after 3 sec

V = 34 + (-8)3

= 34 - 24

V = 10 m/s

Hence, the speed after 3 seconds is 10 m/s

4 0

2 years ago

Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage

agasfer [191]

$v = 2.45×10^3\:\text{m/s}$

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

$F = \dfrac{d{p}}{d{t}}$ (1)

Assuming that the velocity remains constant then

$F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}$

Solving for $v,$ we get

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)$

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

$F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}$

The exhaust rate dm/dt is

$\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}$

$\;\;\;\;\;= 1.36×10^4\:\text{kg/s}$

Therefore, the velocity at which the exhaust gases exit the engines is

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}$

$\;\;\;= 2.45×10^3\:\text{m/s}$

6 0

1 year ago