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3 years ago

6

A spacecraft in outer space increases velocity by firing its rockets. How can hot gases escaping from its rockets change the vel

ocity of the craft when there is nothing in space for the gases to push against?

1 answer:

Elena-2011 [213]3 years ago

3 0

Answer:

Momentum is conserved.

Explanation:

-Momentum is conserved.

-By Newton's third law (For every action, there is an equal and opposite reaction.)-the change in momentum of gases in one direction must be balanced by an equal change in momentum of the spacecraft in the opposite direction.

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As a city planner, you receive complaints from local residents about the safety of nearby roads and streets. One complaint conce

WINSTONCH [101]

Answer:

a) x₁ = 290.50 feet , x₂ = 169.74 feet , b) v_max= 41 mph

Explanation:

For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis fr = m a

Y Axis N-W = 0

N = W = mg

The force of friction has the expression

fr = μ N

We replace

μ mg = ma

a = μ g

g = 32 feet / s²

Let's calculate the acceleration for each coefficient and friction

μ a (feet / s2)

0.599 19.168

0.536 17,152

0.480 15.360

0.350 11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

v² = v₀² - 2 a x

When the speed stops it is zero

x₁ = v₀² / 2 a₁

Let's reduce speed

v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

x₁ = 80.667² / (2 11.2)

x₁ = 290.50 feet

The minimum braking distance

x₂ = 80.667² / (2 19.168)

x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

0 = v₀² - 2 a x

v₀ = √2 a x

We calculate the speed for the two accelerations

v₀₁ = √ (2 11.2 155)

v₀₁ = 58.92 feet / s

v₀₂ = √ (2 19.168 155)

v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

5 0

3 years ago

A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline

balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

$\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\$

Where:

$\mu$ is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

$\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2$

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

$gsin\theta - \mu g cos\theta = a_x\\$

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

5 0

3 years ago

6. Протон с кинетична енергия Ek = 1,6.10 J, който се

anzhelika [568]

Answer:

Tahukah kamu, mengapa motor berhenti di depan lampu merah?

karena motornya direm. Coba kalau nggak direm bisa bahaya tuh.

Tebak binatang apa yang jago renang?

Bebek. Kalau ikan bukan renang tapi menyelam

Kenapa di rel kereta api ditaruh batu?

Soalnya kalau ditaruh duit nanti pada diambil.

3 0

2 years ago

Although these quantities vary from one type of cell to another, a cell can be 1.9 μm in diameter with a cell wall 60 nm thick.

DaniilM [7]

Answer: $2(10)^{-9} mg$

Explanation:

We know the total diameter of the cell (assumed spherical) is:

$d=1.9\mu m=1.9(10)^{-6} m$

Then its total radius $r=\frac{d}{2}=\frac{1.9(10)^{-6} m}{2}=9.5(10)^{-7} m$

On the other hand, we know the thickness of the cell wall is $r_{t}=60 nm= 60(10)^{-9} m$ and its density is the same as water ( $\rho=997 kg/m^{3}$ ).

Since density is the relation between the mass $m$ and the volume $V$ :

$\rho=\frac{m}{V}$

The mass is: $m=\rho V$ (1)

Now if we are talking about this cell as a thin spherical shell, its volume will be:

$V=\frac{4}{3}\pi R^{3}$ (2)

Where $R=r-r_{w}=9.5(10)^{-7} m - 60(10)^{-9} m$

Then:

$V=\frac{4}{3}\pi (9.5(10)^{-7} m - 60(10)^{-9} m)^{3}$ (3)

$V=2.952(10)^{-18} m^{3}$ (4)

Substituting (4) in (1):

$m=(997 kg/m^{3})(2.952(10)^{-18} m^{3})$ (5)

$m=2.94(10)^{-15} kg$ (6)

Knowing $1 kg=1000 g$ and $1 mg=0.001 g$ :

$m=2.94(10)^{-15} kg=2(10)^{-9} mg$

7 0

3 years ago

Calculate the difference between<br>the points B (-2, 3) and C (-5,8)

zzz [600]

Explanation:

the answer to this question would be mj.us 1 and minus 3

3 0

2 years ago