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2 years ago

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The moon’s rotational period and revolution period are equal. What is the result of this? Enter your answer in the space provide

d.

1 answer:

beks73 [17]2 years ago

5 0

Answer The Moon has synchronous rotation: it's rotation period is the same as its period of revolution

Explanation:

The Moon has synchronous rotation: it's rotation period is the same as its period of revolution

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A boat has a mass of 7660 kg. Its engines generate a drive force of 4080 N due west, while the wind exerts a force of 680 N due

makvit [3.9K]

Answer:

0.29 m/s due west.

Explanation:

According to newton's second law,

Net force acting on an object = mass×acceleration

From the question,

F+F₁+F₂ = ma................ Equation 1

Where F = The force generated from the engine, F₁ = Force exerted by the wind, F₂ = Force exerted due to the water, m = mass of the boat, a = acceleration of the boat.

Given: F = 4080 N , F₁ = -680 N(east), F₂ = -1160 N(east). m = 7660 kg

substitute into equation 1

4080-680-1160 = 7660(a)

2240 = 7660a

Therefore,

a = 2440/7660

a = 0.29 m/s due west.

8 0

2 years ago

Plz someone help me Asap

-Dominant- [34]

Answer:

all I know is C

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-KARL IS STOOPID

Explanation:

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2 years ago

A 64.8 kg astronaut is on a space walk when the tether line to the shuttle breaks. The astronaut is able to throw a 11.0 kg oxyg

Umnica [9.8K]

dfdfdfdddfdddddddddddddddd

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2 years ago

With detailed explaniation

belka [17]

Ø=37°
Initial velocity=u=20m/s
g=10m/s²

#A

$\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}$

$\\ \rm\Rrightarrow H_{max}=20sin^237$

$\\ \rm\Rrightarrow H_{max}=7.2m$

#B

$\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}$

$\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}$

$\\ \rm\Rrightarrow R=40sin74$

$\\ \rm\Rrightarrow R=38.5m$

#C

$\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}$

$\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}$

$\\ \rm\Rrightarrow T=4sin37$

$\\ \rm\Rrightarrow T=2.4s$

Now

$\\ \rm\Rrightarrow v=u-gt$

$\\ \rm\Rrightarrow v=20-10(2.4)$

$\\ \rm\Rrightarrow v=20-24$

$\\ \rm\Rrightarrow v=-4m/s$

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1 year ago

A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons

Vlada [557]

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

$K.E\geq P.E$

or

$\dfrac{1}{2}mv^2 \geq G\dfrac{Mm}{R}$

where $m$ is the mass of the asteroid, $R= 24,000,000\:m$ is its distance form earth's center, $M = 5.9*10^{24}kg$ is the mass of the earth, and $G = 6.7*10^{-11}m^3/kg\: s^2$ is the gravitational constant.

Solving for $v$ we get:

$v \geq \sqrt{\dfrac{2GM}{R} }$

putting in numerical values gives

$v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }$

$\boxed{v\geq 5739.5m/s}$

in kilometers this is

$v\geq5.7395m/s.$

Hence, the minimum speed required is 5.7395km/s.

5 0

2 years ago