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2 years ago

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Ingrid is participating in a relay race. While jogging at 9 km/h, she tosses a relay stick at 16 km/h to her teammate, who is st

anding still. How fast is the relay stick moving relative to Ingrid?

2 answers:

shepuryov [24]2 years ago

9 0

C.16 km/h
Got it correct on edgenuity.
speed and veolocity quiz

Aleks04 [339]2 years ago

4 0

Answer: 16 km/h

Ingrid is jogging at a speed of 9 km/h. She passes the relay stick to her team and tosses the stick at 16 km/h. In Ingrid's frame of reference, speed of relay stick is 16 km/h because, she is at rest in her own reference frame. so, relative to Ingrid, speed of relay stick is 16 km/h.

If we were asked to find the speed of relay stick relative to her teammate, then the speed of Ingrid would have been added to the speed of tossing the stick.

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A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c) </em> $\frac{4492.18}{v_{car} ^{2} }$

<em></em>

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = <em>6738.27 J</em>

<em></em>

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = <em>61.908 J</em>

<em></em>

<em></em>

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

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Hope this is helpful <span>Weightlessness

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A 120-kg refrigerator that is 2.0 m tall and 85 cm wide has its center of mass at its geometrical center. You are attempting to

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To solve this problem it is necessary to apply the concepts related to Force of Friction and Torque given by the kinematic equations of motion.

The frictional force by definition is given by

$F= \mu mg$

Our values are here,

$\mu=0.3$

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$g=9.8m/s^2$

Replacing,

$F=0.30*120*9.8 = 352.8N$

Consider the center of mass of the body half its distance from the floor, that is d = 0.85 / 2 = 0.425m. The torque about the lower farther corner of the refrigerator should be zero to get the maximum distance, then

$F*x = mg*d$

Re-arrange for x,

$x= \frac{mg*d}{F}$

$x= \frac{mg*d}{\mu mg}$

$x= \frac{d}{\mu}$

$x= \frac{0.425}{0.3}$

$x = 1.42m$

Then we can conclude that 1.42m is the distance traveled before turning.

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Two ice skaters, Lilly and John, face each other while stationary and push against each others hands. John's mass is twice the m

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