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Alchen [17]
3 years ago
7

Ana poured equal amounts of water into four bowls. Then she put the bowls in the sun for different amounts of time. She recorded

the temperature of the water in each bowl after its time was up. Which bowl should have the warmest water?
Physics
2 answers:
sattari [20]3 years ago
4 0
The bowl which was out for the longest amount of time.
fiasKO [112]3 years ago
4 0

Answer:

Bowl 4

Explanation:

me big smart :)

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Answer: A 2m/s^2

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(24 - 0) / 12 = 2
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What percent of all bolts fire within a cloud
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5 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
2 years ago
According to the picture, which is the least dense?
alexdok [17]

Answer:

a the chess peice

Explanation:

my head

4 0
2 years ago
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