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koban [17]
4 years ago
15

Two astronauts in space with a baseball decide to play catch to pass the time. In the language of conservation of momentum, desc

ribe what happens to each astronaut as they start to toss the ball back and forth
Physics
2 answers:
IceJOKER [234]4 years ago
7 0

Answer:  Suppose that the first astronaut is still in place, then the full momentum of that astronaut is zero.

Now, when the astronaut throws the ball, now the ball has momentum, so the astronaut moves in the opposite direction to conserve the momentum (the movement of the arm also creates a response in the body of the astronaut)

(all of this can be explained also by the third Newton's law, for example, the astronaut that accelerates the baseball also experiences a force that the baseball does in him)

Usually, in the earth, the force of gravity keeps the players in place, but in the space, this is not the case, so the tiny force that the ball does in the astronaut is enough to accelerate the astronaut.

It is the same for the other one, the ball comes with a little bit of momentum, so when he catches the ball, the momentum must be conserved, so the astronaut will move in the same direction that the ball was moving.

Anna35 [415]4 years ago
4 0
As the first astronaut throws the ball, lets assume it goes with v velocity and the mass of the ball be m
the momentum comes out be mv, thus to conserve that momentum the astronaut will move opposite to the direction of the ball's motion with the velocity mv/M (where M is the mass of the astronaut).
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The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water
sashaice [31]

Answer: the constant angular velocity of the arms is 86.1883 rad/sec

Explanation:

First we calculate the linear velocity of the single sprinkler;

Area of the nozzle = π/4 × d²

given that d = 8mm = 8 × 10⁻³

Area of the nozzle = π/4 × (8 × 10⁻³)²

A = 5.024 × 10⁻⁵ m²

Now total discharge is dived into 4 jets so discharge for single jet will be;

Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec

So using continuity equation ;

Q_single = A × V_single

V_single = Q_single/A

we substitute

V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)

V_single = 29.8566 m/s

Now resolving the forces as shown in the second image,

Vt = Vcos30°

Vt = 29.8566 × cos30°

Vt = 25.8565 m/s

Finally we calculate the angular velocity;

Vt = rω

ω_single = Vt / r

from the given diagram, radius is 300mm = 0.3m

so we substitute

ω_single = 25.8565 / 0.3

ω_single = 86.1883 rad/sec

Therefore the constant angular velocity of the arms is 86.1883 rad/sec

7 0
3 years ago
Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the
Hoochie [10]

You've told us:

-- 130°x  =  212°F

and

-- 10°x  =  32°F

Thank you.  Those are two points on a graph of °x vs °F .  With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are  (130, 212)  and  (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation.  I'll use the second point  (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are !  Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad.  Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for  x, and the solution will be the same temperature in  F  too.

or

We can write [ F = 1.5F + 17 ], solve it for  F, and the solution will be the same temperature in  x  too.

F = 1.5F + 17

Subtract  F  from each side:  0.5F + 17 = 0

Subtract 17 from each side:   0.5F = -17

Multiply each side by 2 :  F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for  x:  

F = 1.5(-34) + 17

F = -51 + 17

<em>F = -34</em>

It works !  -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

<em>yay !</em>

6 0
3 years ago
Question 15 plz picture above
Pie
15) C. The amount of each element that begins....
8 0
3 years ago
In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an
VashaNatasha [74]

Answer:

V=3.475ft^3/s=3.48ft^3/s

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})

P= 73.7*10^{3}ft.lbf/s

we have other values such h=400ft and  \gamma= 62.42lb/ft^3 (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

V=\frac{P}{\gamma h \eta_0}

Where \eta_0 is the turbine efficiency, at which is,

\eta_0 = \frac{P}{\gamma Vh}

Replacing,

V=\frac{73.7*10^{3}}{62.42*400*0.85}

V=3.475ft^3/s

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

m=\rho V\\m= 3.475*1.94 \\m=6.7415 slugs/s

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

m= 217lbm/s

6 0
3 years ago
A pendulum at position A is released and swings through position B to position Con the other side.
zaharov [31]

Explanation:

Given the conditions A,B and C when the pendulum is released, at point A the initial velocity of the pendulum is zero(0), the potential energy stored is maximum(P.E= max),

the conditions can be summarized bellow

point A

initial velocity= 0

final velocity=0

P.E= Max

K.E= 0

point B

initial velocity= maximum

final velocity=maximum

P.E=K.E

point C

initial velocity= min

final velocity=min

P.E= 0

K.E= max

3 0
3 years ago
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