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lianna [129]
4 years ago
11

Two planes, which are 37403740 miles apart, fly toward each other. their speeds differ by 35mph35â¢mph. if they pass each other

in 44 hours, what is the speed of each?
Physics
1 answer:
Orlov [11]4 years ago
5 0
33750 miles because it in the sky
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A force of 500 N acts horizontally on a 10,000 g body. What is its horizontal acceleration
Paladinen [302]
200n because it's 2×5=10so maybe try solving the problem like that ok does that help
8 0
4 years ago
When a sound wave moves through a medium such as air, the motion of the molecules of the medium is in what direction (with respe
Alchen [17]

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7 0
3 years ago
The drawing shows a tire of radius R on a moving car
masha68 [24]

Answer:

 H / R = 2/3

Explanation:

Let's work this problem with the concepts of energy conservation. Let's start with point P, which we work as a particle.

Initial. Lowest point

          Em₀ = K = 1/2 m v²

Final. In the sought height

         Em_{f} = U = mg h

Energy is conserved

        Em₀ =  Em_{f}

        ½ m v² = m g h

        v² = 2 gh

Now let's work with the tire that is a cylinder with the axis of rotation in its center of mass

Initial. Lower

        Em₀ = K = ½ I w²

Final. Heights sought

        Emf = U = m g R

        Em₀ =  Em_{f}

       ½ I w² = m g R

The moment of inertial of a cylinder is

       I = I_{cm} + ½ m R²

      I= ½ I_{cm}  + ½ m R²

Linear and rotational speed are related

       v = w / R

       w = v / R

We replace

      ½ I_{cm}  w² + ½ m R² w²  = m g R

moment of inertia of the center of mass      

      I_{cm}  = ½ m R²

     ½ ½ m R² (v²/R²) + ½ m v² = m gR

    m v² ( ¼ + ½ ) = m g R

   

       v² = 4/3 g R

As they indicate that the linear velocity of the two points is equal, we equate the two equations

        2 g H = 4/3 g R

         H / R = 2/3

7 0
3 years ago
Kalyan ramji sain, of india, had a mustache that measured 3.39 m from end to end in 1993. suppose two charges, q and 3q, are pla
jeka94

Answer:

q=3.19*10^{-8}C

Explanation:

The expression for the electric force between two charges

F=k\frac{q_1q_2}{r^2}\\

where K is the Coulomb's constant (k=9*10^{9}Nm^2/C^2), and we have that

q1=q

q2=3q

r=3.39m

F=2.4*10^{-6}N

By replacing in the formula we have

F=2.4*10^{-6}N=(9*10^{9}\frac{Nm^2}{C^2})\frac{(q)(3q)}{(3.39m)^2}\\\\

and by taking apart q we have

q=3.19*10^{-8}C

hope this helps!!

4 0
3 years ago
Submit Quiz
kkurt [141]

Explanation:

the question is unanswerable

7 0
3 years ago
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