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3 years ago

A rocket experiences a 45,000 N force as it accelerates at 30 m/s/s. What is the mass of the rocket?

Physics

1 answer:

Gre4nikov [31]3 years ago

6 0

Answer:

15,00kg

Explanation:

here's your solution

=> force = 45,000N

=> acceleration = 30m/s^2

=> mass = ?

=> Mass = force/acceleration

=> mass = 45,000/30

=> mass = 15,00kg

hope it helps

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A satellite is orbiting Earth in an approximately circular path. It completes one revolution each day (86,400 seconds). Its orbi

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For each revolution, the satellite covers the circumference
of its orbit.

The circumference of a circle is (pi) x (diameter).

Since the satellite is 22,500 miles from the center, that number
is the radius of the circle. The diameter is 45,000 miles.

The circumference is (pi) x (45,000 miles) = 141,372 miles

The satellite's average speed is 141,372 miles per day, or

(141,372 / 24) = 5,890 miles per hour

relative to the center of the Earth.

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How long does it take for the hubble telescope to orbit earth

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Answer:

95 minutes

Explanation:

According to N.A.S.A, the Hubble Space Telescope makes one orbit around Earth every 95 minutes.

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4 years ago

Read 2 more answers

What is kepler's law??

Elan Coil [88]

<h2>QUESTION:- </h2>

➜what is kepler's law??

$\huge\red{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W}\orange{ER}}}}$

Kepler gave the three laws or theorems of motion of the orbitals bodies

${\huge {\bold{ \red{ \star}}}}{ \blue{ \bold{FIRST \: \: \: LAW}}}$

This law state that the celestial bodies revolves around the stars in elliptical orbit and star as a single focus.

Example :- Earth revolves around the Sun as assuming it as single focus

This also shows that earth revolves around the sun in elliptical orbit.

${\huge {\bold{ \blue{ \star}}}}{ \green{ \bold{SECOND \: \: \: LAW}}}$

Area covered by the planet is equal in equal duration of time irrespective of the position of the planet.

It also states that Angular momentum is constant

As Angular momentum is constant it means areal velocity is also constant.

$\frac{ \triangle \: A}{ \triangle \: T} = \frac{L}{2m}△T△A=2mL$

where:-

A is the area.

T is the time.

L is the angular momentum.

M is the mass of the body.

${\huge {\bold{ \green{ \star}}}}{ \purple{ \bold{THIRD \: \: \: LAW}}}$

square of the time of the revolution is directly proportional to the cube of the distance between the planet and star in Astronomical unit.

${T}^{2} = {a}^{3}T2=a3$

where:-

T = time of revolution

a is the distance between the planet and star.

$\purple\star \: {Thanks \: And \: Brainlist} \blue \star \\ {\orange{ \star}}{if \: U \: Like d \: My \: Ans} {\green{ \star }}$

8 0

3 years ago

A 25 kg circular disk has a diameter of 2.5 feet and a thickness of 2.5 cm. Find the density of the disk in kg/m3. Next, find th

Gre4nikov [31]

Answer:

Assume that $\rm g= 9.81\; N\cdot kg^{-1}$ ; $\rho(\text{Water}) = \rm 1000\;kg\cdot m^{-3}$ .

Density of the disk: approximately $\rm 2.19\times 10^{3}\; kg\cdot m^{-3}$ .

Weight of the disk: approximately $\rm 245\;N$ .

Buoyant force on the disk if it is submerged under water: approximately $\rm 112\; N$ .

The disk will sink when placed in water.

Explanation:

Convert the dimensions of this disk to SI units:

Diameter: $d = \rm 25\; inches = (25\times 0.3048)\; m = 0.762\;m$ .
Thickness $h = \rm 2.5\; cm = (2.5\times 0.01)\; m = 0.025\;m$ .

The radius of a circle is 1/2 its diameter:

$\displaystyle r = \rm \frac{1}{2}\times 0.762\;m = 0.381\; m$ .

Volume of this disk:

$V(\text{disk}) = \pi\cdot r^{2}\cdot h = \pi\times 0.381^{2}\times 0.025 \approx 0.0114009\; m^{3}$ .

Density of this disk:

$\displaystyle \rho(\text{disk}) = \frac{m}{V} = \rm \frac{25\; kg}{0.0114009\; m^{3}} = 2.19\times 10^{3}\;kg\cdot m^{-3}$ .

$\rho(\text{disk}) >\rho(\text{water})$ indicates that the disk will sink when placed in water.

Weight of the object:

$W(\text{disk}) = m\cdot g = \rm 25\times 9.81 = 245.25\; N$ .

The buoyant force on an object in water is equal to the weight of water that this object displaces. When this disk is submerged under water, it will displace approximately $\rm 0.0114009\; m^{3}$ of water. The buoyant force on the disk will be:

$\begin{aligned}F(\text{buoyant force}) &= W(\text{Water Displaced}) \\& = \rho\cdot V(\text{Water Displaced})\cdot g\\ & = \rm 1\times 10^{3}\; kg\cdot m^{-3}\times 0.0114009\; m^{3}\times 9.81\; N\cdot kg^{-1}\\ &\approx \rm 112\; N\end{aligned}$ .

The size of this disk's weight is greater than the size of the buoyant force on it when submerged under water. As a result, the disk will sink when placed in water.

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3 years ago