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RoseWind [281]
4 years ago
5

While on vacation in the tropical Bahamas, Suzy decides to try out a water jet pack. She places the jet pack on her back (like y

ou would wear a backpack), pushes the buttons in her hands, and water comes shooting out of the bottom of the jetpack, exerting a force on the ocean below. Why does Suzy fly up into the air? In your answer, be sure to state which of Newton's laws of motion this pertains to, describe this law, and explain how it causes Suzy to fly up into the sky.
Physics
1 answer:
Veronika [31]4 years ago
5 0

Answer:

-Suzy flies up into the sky because of the upward force on the jetpack.

-Newton's third law of motion can be ascribed to this.

Explanation:

We know that in momentum, Force is described as the rate of change of momentum. Thus, for us to understand the forces, we will need to establish how the momentum of the water changes.

Suppose the mass flow rate of the water is denoted as M kg/sec and the upward velocity is denoted as v_u.

Now, since water is lying stationary at the sea, the initial momentum of the water will be zero. If every second the boat pumps upwards with mass flow rate M at velocity vu, the momentum per second would be;

Δp = F_water = Mv_u

Now, from Newton's third law the boat/pump would experience a downwards force with the formula:

F_boat = −Mv_u

Furthermore at the jetpack, the water does a reverse turn in direction and is now again directed back down at a different velocity v_d. Therefore the final momentum per second is Mv_d and therefore change in momentum is now;

Δp = M(v_d - v_u)

The velocity v_d will be negative since the boat/pump would experience a downwards force. Thus, the force on the water will be downwards and F_water ≤ 0 .

We will finally use Newton's third law to conclude that the upward force on the jetpack will be;

F_jetpack = M(v_u - v_d)

Thus upward force is what makes Suzy to fly up in the sky.

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Consider a helium balloon on a string tied to the seat of your stationary car. The windows are closed, so there is no air motion
timama [110]

Answer:

The balloon will move forward.

The density of the air will be greater at the back of the balloon; similar

to the density of air being greater at lower altitudes due to gravitational

attraction because of the weight of the air in an air column.

A block of wood in water rises because of the difference in pressures

on the top and bottom of the block.

6 0
3 years ago
While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at
Mashcka [7]

Answer:

The speed of the moving walkway is 1.50 m/s

Explanation:

The position of the child can be calculated using the following equation:

x = x0 + v · t

Where :

x = position of the child at time t.

v = velocity of the child.

t = time.

When the child runs in the same direction as the walkway, the velocity of the child will be its  velocity relative to the walkway plus the velocity of the walkway. Then, if we place the origin of the frame of reference at the start of the walkway:

x = x0 + v · t

25 m = 0 m + (2.8 m/s + v) · t₁

Where v is the velocity of the walkway

On its way back, the velocity of the child relative to the walkway is in the opposite direction to the velocity of the walkway. Then:

x = x0 + v · t

0 m = 25 m + (-2.8 m/s + v) · t₂

We also know that t₁ + t₂ = 25 s

Then: t₁ = 25 - t₂

So, we can write the following system of equations:

25 m = (2.8 m/s + v) · (25 s - t₂)

-25 m = (-2.8 m/s + v) · t₂

Let´s take the second equation and solve it for t₂

-25 m / (-2.8 m/s + v) = t₂

Now, let´s replace t₂ in the first equation:

25 m = (2.8 m/s + v) · (25 s + 25 m / (-2.8 m/s + v))

Let´s sum the fraction: 25 s + 25 m / (-2.8 m/s + v)

25 m = (2.8 m/s + v) · (25 s ·(-2.8 m/s + v) + 25 m) / (-2.8 m/s + v)

multiply by (-2.8 m/s + v) both sides of the equation:

25 m(-2.8 m/s + v) = (2.8 m/s + v) · (-70 m + 25 s · v + 25 m)

Apply distributive property:

-70 m²/s +25 m·v = -196 m²/s +70 m·v +70 m²/s -70 m·v +25 s ·v² + 25 m v

56 m²/s = 25 s · v²

56 m²/s / 25 s = v²

v = 1.50 m/s

The speed of the moving walkway is 1.50 m/s

7 0
3 years ago
Part A
jok3333 [9.3K]

Answer:

Using 3 periods to get an accurate reading:

3T = (6.40 S - 0.90 s) = 5.50 S So, T = 1.83 s

m = 0.250 kg

Using algebra:

k= \frac{4\pi ^{2}m }{T^{2} }=\frac{4\pi ^{2}0.250 }{1.90^{2} }=2.95kg/sec

7 0
2 years ago
Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe
Mice21 [21]

As per the question the distance of venus from sun is given as 0.723 AU

We have been asked to calculate the time period of the planet venus.

As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically

                                        T^{2} \alpha R^{3}

                                         ⇒ T^{2} = KR^{3} where is k is the proportionality  constant

We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days

Hence T_{1} =365.5 days

The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun

Hence R_{1} =1 AU

The distance of venus from sun is 0.723 AU i.eR_{2} =0.723

From keplers law we know that-\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{3} }

                            ⇒T_{2} ^{2} =T_{1} ^{2} *\frac{R_{2} ^{3} }{R_{1} ^{3} }

Putting the values mentioned above we get-

                                      T_{2} ^{2} =50,350.132851075

                                         ⇒ T_{2} =\sqrt{50,350.132851075}

                                        ⇒T_{2} = 224.388352752710 days.

Hence the time period of venus is 224.388352752710 days

                                         

                     






                           

7 0
4 years ago
Read 2 more answers
Can someone please help me with this question... thank u ❤️​
hjlf

Answer:

Which of the following... : constant velocity horizontal motion

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8 0
3 years ago
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