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siniylev [52]
3 years ago
11

Plz helppppppppp meeeeeeeeeeeeeeee

Physics
2 answers:
katen-ka-za [31]3 years ago
8 0

Answer: The Melted Crayon Represent In The Model Is Lava Because When Lava Hardens It Can Harden Into Rocks Called Igneous Rock, Sedimentary Rock, Magma, And Metamorphic Rock

Explanation:

Gwar [14]3 years ago
3 0
The melted crayon represents lava
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if a weight of 100 lbs is moved through a displacement of 1.2m, how much work was done to move the weight
OLEGan [10]

Answer:

534 j

Explanation:

100 lbs = 45.359 kg

work = F * d

         = 45.359 * 9.81  * 1.2 = ~ 534 j

6 0
3 years ago
n electric motor rotating a workshop grinding wheel at 1.10 102 rev/min is switched off. Assume the wheel has a constant negativ
Schach [20]

Answer:

Time taken to stop = 6 sec

Explanation:

Given:

Rotation of wheel = 1.10 × 10² rev/min

Final velocity (v) = 0

Angular acceleration (a) = - 1.92 rad/s²

Find:

Time taken to stop

Computation:

Initial velocity (u) = (1.10 × 10² × 2π rad) / 60 sec

Initial velocity (u) = 11.52 rad / sec

We know that,

V = U +at

t = (v-u)a

t = (0 - 11.52) / (-1.92)

Time taken to stop = 6 sec

3 0
3 years ago
If you see a rainbow near the time of sunset, where in the sky will the rainbow be?
Igoryamba
Closest to the sun f thats a choice or closest to the atmoshpere
8 0
4 years ago
Two identical space probes are orbiting Jupiter. Scientists determine that one of the space probes has a larger gravitational fo
galina1969 [7]

If the probes are identical, then the one that feels a larger gravitational
force is orbiting closer to Jupiter than the other one is.

If they're not identical, then the one with greater mass will feel more
gravitational force than the one with less mass, even if they're both
the same distance from Jupiter.  (We know this from the experimental
observation that fatter people weigh more, even on Earth.)
6 0
3 years ago
Read 2 more answers
A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk
Murrr4er [49]

Answer:

\omega_2=5.1rad/s

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is L=I\omega. If we call 1 the situation when the student is at the rim and 2 the situation when the student is at r_2=1.39m from the center, then we have:

L_1=L_2

Or:

I_1\omega_1=I_2\omega_2

And we want to calculate:

\omega_2=\frac{I_1\omega_1}{I_2}

The total moment of inertia will be the sum of the moment of intertia of the disk of mass m_D=119.1 kg and radius r_D=3.23m, which is I_D=\frac{m_Dr_D^2}{2}, and the moment of intertia of the student of mass m_S=54.3kg at position r (which will be r_1=r=3.23m or r_2=1.39m) will be I_{S}=m_Sr_S^2, so we will have:

\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}

or:

\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}

which for our values is:

\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s

6 0
3 years ago
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