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gtnhenbr [62]
3 years ago
8

NEED HELP RIGHT NOW PLSSSS!!

Physics
1 answer:
atroni [7]3 years ago
5 0
Do it in your own words
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Alex777 [14]

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If it doesn't do two complete cycles in 24 hours 50 minutes, then it's not tide.

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Two wheels are identical but wheel b is spinning with twice the angular speed of wheel
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<span>If two wheels are exactly the same but spin at different speeds, wheel b is twice te speed of wheel a, it is possible to find the ratio of the magnitude of radial acceleration at a singular point of the rim on wheel be to the spot is four.</span>
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3 years ago
A car traveled north 25 km for 1 hour. The car’s SPEED is 25 km/hr north.
kramer

Answer:

hgk

Explanation:

6 0
3 years ago
A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. A
Ghella [55]

The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is

<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J

That is,

• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point

• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium

so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.

By the work-energy theorem,

<em>W</em> = ∆<em>K</em> = <em>K</em>

where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So

<em>W</em> = 1/2 <em>mv</em> ²

where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get

<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s

8 0
3 years ago
Which temperature is the hottest? 98 F or 39 C or 303K?<br> F= 1.8C + 32<br> C= (F-32)/1.8
sergejj [24]

Answer:

The hottest temperature is  T_2 = 39^o C

Explanation:

From the question we are given

    T_1 =  98 F

  T_2 =  39^oC

  T_3 =  303 \  K

Generally converting T_3 to  Fahrenheit

    T_3' =  (T_3 -273 ) * \frac{9}{5}  + 32

=> T_3' =  (303 -273 ) * \frac{9}{5}  + 32

=> T_3' = 86 F

Converting  T_2 to  Fahrenheit

      T_2' =  T_2 * \frac{9}{5}  + 32

=> T_2' =  39 * \frac{9}{5}  + 32

=> T_2' =102.2 F  

Now comparing  the temperature  in Fahrenheit we see that T_2  is the hottest

3 0
3 years ago
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