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Alik [6]
3 years ago
14

A thin rod of length 0.79 m and mass 130 g is suspended freely from one end. It is pulled to one side and then allowed to swing

like a pendulum, passing through its lowest position with angular speed 3.61 rad/s. Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.

Physics
1 answer:
-Dominant- [34]3 years ago
3 0

Answer:

a) 0.3965 j

b) 0.3112 m

Explanation:

The picture attached explains it all. Thank you

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The sky is blue because: Select one: a. the index of refraction for air is slightly larger for blue than for red b. atomic hydro
lapo4ka [179]

Answer:

a. the index of refraction for air is slightly larger for blue than for red

5 0
3 years ago
Dr. Aloysius found four homogeneous mixtures, but he thinks that only three are alloys. He tests each one and determines their e
Lina20 [59]

Answer:

A. carbon and boron

Explanation:

Carbon and boron is not an alloy.

An allow forms between metals and metals using their huge electron could.

Carbon is a non-metal, boron is a also a non-metal

Two non-metals combining together does not make an alloy.

Iron, nickel, aluminum  are all metals.

6 0
3 years ago
Read 2 more answers
A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

4 0
3 years ago
A swimmer, capable of swimming at a speed of 1.0 m/s in still water (i.e., the swimmer can swim with a speed of 1.0 m/s relative
Vesnalui [34]
If the current takes him downstream we must find the resultant vector of the velocities: V res= \sqrt{1^{2}+0.91^{2}  } = \sqrt{1.8281}= 1.3520747 Then if the river is 3000 m-wide the swimmer will have to pass:
 1.3520747 · 300 = 4056.14 m                t = 4056.14 m : 1 m/s
a ) It takes 4056.15 seconds ( 1 hour 7 minutes and 36 seconds ) to cross the river.  
b ) 0.91 · 3000 = 2730 m
He will be 2730 m downstream.
5 0
3 years ago
A hot-air balloon is rising upward with a constant speed of 2.03 m/s. When the balloon is 8.13 m above the ground, the balloonis
HACTEHA [7]
The answer is 4.2s...
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3 years ago
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