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jeyben [28]
2 years ago
7

A comet is cruising through the solar system at a speed of 50,000 kilometers per hour foe 4 hours time. What is the total distan

ce traveled by the comet during that time
Physics
1 answer:
Triss [41]2 years ago
5 0

Answer:

<em>The distance covered by comet is </em>200,000 km

Explanation:

Speed is defined as the rate of change of distance with time. It is given by the equation speed= \bold{\frac{distance}{time}}

Thus distance= speed*time

In this problem it is given that speed of comet= \frac{50,000km}{hr}

time travelled by the comet= 4 hours

Thus distance= speed*time

                            = 500000*4

                            = \bold{200,000km}

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If a closed series circuit has a voltage of 12 V and R1 = 4 Ω and R2 = 2 Ω, what is the current running through R1 and R2?
Phantasy [73]
Total resistance=R1+ R2= 6Ω
Voltage=12v
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Current= 2A
In a series circuit, equal current passes through every resistance.
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2 years ago
What is the force holding you down?
Snezhnost [94]

Answer:

The force holding you down is gravity.

Explanation:

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3 0
3 years ago
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Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
2 years ago
How does Earth's surface and the structures on the surface change as a result of an earthquake? Help me pls and I will give Brai
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<span>Earthquakes often cause dramatic changes at Earth's surface. In addition to the ground movements, other surface effects include changes in the flow of groundwater, landslides, and mudflows. Earthquakes can do significant damage to buildings, bridges, pipelines, railways, embankments, dams, and other <span>structures</span></span>
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3 years ago
The lanthanides are all radioactive true or false
Oksi-84 [34.3K]

Answer:

False

Explanation:

All the lanthanides are not radioactive in nature. Just one of the lanthanides are radioactive.

  • The lanthanides belong to the f-block on the periodic table
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