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sammy [17]
3 years ago
14

Which chemical equation is balanced?

Physics
1 answer:
scoundrel [369]3 years ago
6 0
The 4na + 02 —> wna20
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Find an expression for the minimum frictional coefficient needed to keep a car with speed v on a banked turn of radius R designe
solniwko [45]
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr 
</span><span>Θ = arctan(v0² / gr) </span>

<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>

<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
4 0
3 years ago
A 6.4-N force pulls horizontally on a 1.5-kg block that slides on a smooth horizontal surface. This block is connected by a hori
Elena L [17]

-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless.  I guess that's what "smooth" means.

-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.

-- Your force is 6.4 N.
                                    Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
                                                                 </em>
That's about  <em>2.634 m/s²</em>  <em>

</em>
(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)

-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is

       Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = <em>2.45 N</em>.

That's the force that's accelerating the little block, so that must be the tension
in the string.


7 0
3 years ago
A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Iteru [2.4K]

Answer:

A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

                         

Explanation:

We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

d = v*t

Where:

v: is the tangential speed of the disk

t: is the time = 30 s  

The tangential speed can be found as follows:

v = \omega*r

Where:

ω: is the angular speed = 100 rpm

r: is the radius = 50 cm = 0.50 m

v = \omega*r = 100 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}*0.50 m = 5.24 m/s    

Now, the distance traveled by the disk is:

d = v*t = 5.24 m/s*30 s = 157.2 m

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

I hope it helps you!

3 0
3 years ago
PLEASE HELP ME TIMED TEST!!!
malfutka [58]

Answer:

56

Explanation:

I just want the points to be completely honest with you.  

7 0
3 years ago
Read 2 more answers
an engine has been design to work between source and the sink at temperature 177 degree Celsius and 27 degree Celsius respective
irina [24]

<u>Given data</u>

Source temperature (T₁) = 177°C = 177+273 = 450 K

Sink temperature (T₂) = 27°C = 27+273 = 300 K

Energy input (Q₁) = 3600 J ,

Work done = ?

                We know that, efficiency (η) = Net work done ÷ Heat supplied

                                                           η =   W ÷ Q₁  

                                                           W = η × Q₁

               First determine the efficiency ( η ) = ?

                                Also, we know that ( η ) = (T₁ - T₂) ÷ (T₁)

                                                                        = 33.3% = 0.333

               Now, Work done is W = η × Q₁

                                                    = 0.33 × 3600

                                                 <em>  W = 1188 J</em>

<em>Work done by the engine is 1188 J</em>

4 0
3 years ago
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