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dexar [7]
3 years ago
8

b) Assume the rod is 0.60 m long and has a mass of 0.50 kg, and the clay blob has a mass of 0.20 kg and moves at an initial velo

city of 8.0 m/s. Calculate the final angular velocity of the rod. Be sure to put units in your calculation and show the resulting units in your answer.
Physics
1 answer:
SOVA2 [1]3 years ago
4 0

Answer:

The correct answer is "6.96 rad/s".

Explanation:

The given values are:

Length,

L = 0.6 m

Mass,

m₁ = 0.5 kg

m₂ = 0.2 kg

Initial velocity,

V = 8 m/s

Now,

The final angular velocity will be:

⇒ \omega =\frac{6m_1V}{(4m_1+3m_2)L}

By substituting the values, we get

⇒     =\frac{6\times 0.2\times 8}{(4\times 0.2+3\times 0.5)0.6}

⇒     =\frac{9.6}{1.38}

⇒     =6.96 \ rad/s

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Answer:

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Explanation:

7 0
3 years ago
A student team is to design a human powered submarine for a design competition. The overall length of the prototype submarine is
nikklg [1K]

Answer:

a) The speed is 61.42 m/s

b) The drag force is 10.32 N

Explanation:

a) The Reynold´s number for the model and prototype is:

Re_{m} =\frac{p_{m}V_{m}L_{m}   }{u_{m} }

Re_{p} =\frac{p_{p}V_{p}L_{p}   }{u_{p} }

Equaling both Reynold's number:

\frac{p_{p}V_{p}L_{p}   }{u_{p} }=\frac{p_{m}V_{m}L_{m}   }{u_{m} }

Clearing Vm:

V_{m} =\frac{p_{p}V_{p}L_{p} u_{m}   }{u_{p} p_{m} L_{m} }=\frac{999.1*0.56*8*1.849x10^{-5} }{1.138x10^{-3}*1.184*1 } =61.42m/s

b) The drag force is:

\frac{F_{Dm} }{p_{m}V_{m}^{2}L_{m}^{2}     } =\frac{F_{Dp} }{p_{p}V_{p}^{2}L_{p}^{2}     } \\F_{Dp} =\frac{F_{Dp}p_{p}V_{p}^{2}L_{p}^{2} }{p_{m}V_{m}^{2}L_{m}^{2}     } \\F_{Dp}=\frac{2.3*999.1*0.56^{2} *8^{2} }{1.184*61.42^{2}*1^{2}  } =10.32N

8 0
3 years ago
Read 2 more answers
radar wave is transmitted and later reflected off an aircraft and recieved 1.4×10^3 sec after being sent out. how far is the air
lutik1710 [3]

I think there's a typo because the answer I'm getting is very large.

This is what I'm getting

--------------------------------------

c = speed of light

c = 3.0 x 10^8 m/sec approximately

This is roughly 300 million meters per second

The time it takes the signal to reach the aircraft and come back is 1.4 x 10^3 seconds. Half of this time period is going one direction (say from the radar station to the aircraft), so (1.4 x 10^3)/2 = 7.0 x 10^2 seconds is spent going in this one direction.

distance = rate*time

d = r*t

d = (3.0 x 10^8) * (7.0 x 10^2)

d = (3.0*7.0) x (10^8*10^2)

d = 21.0 x 10^(8+2)

d = 21.0 x 10^10

d = (2.1 x 10^1) * 10^10

d = 2.1 x (10^1*10^10)

d = 2.1 x 10^11 meters

d = 210,000,000,000 meters (this is 210 billion meters; equivalent to roughly 130,487,950 miles)

3 0
3 years ago
. A 2.0-m wire carries a current of 15 A directed along the positive x axis in a region where the magnetic field is uniform and
Lady bird [3.3K]

Answer:

The resulting magnetic force on the wire is -1.2kN

Explanation:

The magnetic force on a current carrying wire of length 'L' with current 'I' in a magnetic field B is

F = I (L*B)

Finding (L * B) , where L = (2, 0, 0)m , B = (30, -40, 0)

L x B = \left[\begin{array}{ccc}i&j&k\\2&0&0\\30&-40&0\end{array}\right] = (0, 0, -80)

we can now solve

F = I (L x B) = I (-80)

F = -1200 kmN

F = -1200 kN * 10⁻³

F = -1.2kN

6 0
3 years ago
On a sky coaster (human pendulum) that reaches 10 meters from it's equilibrium position, a man of 120 kg is able to reach a maxi
dimaraw [331]

Answer:

14 m/s

Explanation:

Using the principle of conservation of energy, the potential energy is converted to kinetic energy, assuming any losses.

Kinetic energy is given by ½mv²

Potential energy is given by mgh

Where m is the mass, v is the velocity, g is acceleration due to gravity and h is the height.

Equating kinetic energy to be equal to potential energy then

½mv²=mgh

V

Making v the subject of the formula

v=√(2gh)

Substituting 9.81 m/s² for g and 10 m for h then

v=√(2*9.81*10)=14.0071410359145 m/s

Rounding off, v is approximately 14 m/s

6 0
3 years ago
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