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Nataliya [291]
2 years ago
15

A popular ride at some playgrounds consists of two circular disks joined by rungs . The center of the unit is attached to a hori

zontal axle around which the unit rotates. The rotational inertia of the unit is 550 kg⋅m2 , and its radius is 1.00 m. A 30-kg child runs and grabs the bottom rung of the ride when it is initially at rest. Because the child can pull herself up to meet the rung at waist height, treat her as a particle located on the rung. The friction between unit and axle is minimal and can be ignored. What is the minimum speed v the child must have when she grabs the bottom rung in order to make the unit rotate about its axle and carry her over the top of the unit? Express your answer with the appropriate units.
Physics
1 answer:
weeeeeb [17]2 years ago
7 0

Answer:

v = 36.667 m/s

Explanation:

Knowing the rotational inertia as

Lₙ = 550 kg * m²

r = 1.0 m

m = 30.0 kg

To determine the minimum speed v must have when she grabs the bottom

Lₙ =  I * ω

I = ¹/₂ * m * r²

I = ¹/₂ * 30.0 kg * 1.0² m

I = 15 kg * m²

Lₙ =  I * ω  ⇒ ω =  Lₙ / I

ω = [ 550 kg * m² /s  ] / ( 15 kg * m² )

ω = 36.667 rad /s

v = ω * r

v = 36.667 m/s

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Consider a very small hole in the bottom of a tank 17.0 cm in diameter filled with water to a heightof 90.0 cm. Find the speed a
umka21 [38]

Answer:

Speed of water, v = 4.2 m/s

Explanation:

Given that,

Diameter of the tank, d = 17 cm

It is placed at a height of 90 cm, h = 0.9 m

We need to find the speed at which the water exits the tank through the hole. It can be calculated using the conservation of energy as :

\dfrac{1}{2}mv^2=mgh

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 0.9}

v = 4.2 m/s

So, the speed of water at which the water exits the tank through the hole is 4.2 m/s. Hence, this is the required solution.

5 0
2 years ago
You have negotiated with the Omicronians for a base on the planet Omicron Persei 7. The architects working with you to plan the
steposvetlana [31]

Answer:

5.724 meters / second^2

Explanation:

We are given two pieces of information, 5.24 flurg = 1 meter, 1 grom = 0.493 second. If that is so, we can say that there are two possible conversion units,  5.25 flurg / meter, and 0.493 second / grom.

_____

We want to convert 7.29 flurg / grom^2 ( I believe? ) to the units meters / second^2. But, let's break this down into bits. It would be convenient to first convert 7.29 flurg / grom^2 to the units meters / grom^2, by dividing the conversion factors as to cancel out the appropriate things, which we will go into detail on a bit later ( using the first conversion factor ). Respectively we can convert meters / grom^2 to meters / grom * s, canceling out the flurg ( through the second conversion factor ). And now we would need to get rid of the grom, dividing similarly.

_____

( 1 ) ( flurg / grom^2 ) / ( flurg / meters  ) - first conversion unit

= flurg / grom^2 * meters /flurg

= ( meters * flurg ) / ( grom^2 * flurg )  

= meters /grom^2,

7.29 flurg / grom^2 / 5.24 flurg / meter = ( About ) 1.39 meter / grom^2

( 2 ) ( meter / grom^2 ) / ( second / grom  ) - second conversion unit

= meter / grom^2 * grom / second

= ( meter * grom ) / ( grom^2 * second )

= meter / ( grom * second ),

( 1.39 meter / grom^2 ) / 0.493 second / grom = ( About ) 2.82195 meter /  grom * second

( 3 ) ( 2.82195 meter / ( grom * second ) ) / 0.493 second / grom = 5.724 meter / second^2

( And thus, the value of gOP7 in the units the architects will use should be about 5.724 meters / second^2 )

8 0
2 years ago
23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n
PSYCHO15rus [73]

Answer:

a)  (95.4 i^ + 282.6 j^) N , b) 298.27 N  71.3º and c)   F' = 298.27 N   θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

      sin θ = Fjy / Fj

      cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

          T = 90 -32 = 58º

         Fjy = Fj sin 58

         Fjx = FJ cos 58

         Fjx = 180 cos 58 = 95.4 N

         Fjy = 180 sin 58 = 152.6 N

Andrea's force is

         Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

       Fx = Fjx

       Fx = 95.4 N

Axis y

       Fy = Fjy + Fa

       Fy = 152.6 + 130

       Fy = 282.6 N

       F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

       F² = Fx² + Fy²

       F = √ (95.4² + 282.6²)

       F = √ (88963)

       F = 298.27 N

       tan θ = Fy / Fx

       θ = tan-1 (282.6 / 95.4)

       θ = tan-1 (2,962)

       θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

       F' = 298.27 N

       θ' = 180 + 71.3

       θ = 251.4º

4 0
2 years ago
Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50.
Nady [450]

Answer:

v = 38.73 m/s

Explanation:

Given

Extension of the bow, x = 50 cm = 0.5 m

Force of the arrow, F = 150 N

Mass of the arrow, m = 50 g = 0.05 kg

speed of arrow, v = ? m/s

We start by finding the spring constant

Remember, F = kx, so

k = F/x

k = 150 / 0.5

k = 300 N/m

the potential energy if the bow when pulled back is

E = 1/2kx²

E = 1/2 * 300 * 0.5²

E = 0.5 * 300 * 0.25

E = 37.5 J

The speed of the arrow will now be found by using the law of conservation of energy

1/2kx² = 1/2mv²

kx² = mv²

v² = kx²/m, on substituting, we have

v² = (300 * 0.5²) / 0.05

v² = 75 / 0.05

v² = 1500

v = √1500

v = 38.73 m/s

8 0
2 years ago
Rhea kicks a soccer ball at 13 km/h to Sean. After kicking the ball, the speed of the soccer ball from Rhea’s reference frame is
BartSMP [9]

Answer:  Sean is standing still, and Rhea is running toward Sean while   kicking the ball

Explanation: Your welcome :)

5 0
2 years ago
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