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Marta_Voda [28]
4 years ago
12

13. You throw a ball vertically upward, and as it leaves your hand, its speed is 37.0 m/s. How long (in s) does the ball take to

return to the level where it left your hand after it reaches its highest point? (A) 1.38 seconds (B) 2.28 seconds (C) 3.78 seconds (D) 4.38 seconds (E) 5.18 seconds
Physics
1 answer:
katovenus [111]4 years ago
5 0

Answer:

(C) 3.78 seconds

Explanation:

At the highest point, the velocity is equal to 0m/s

v_{f}=v_{o}-gt

t=\frac{v_{o}}{g}  ; t is the time to reach the highest point

The  time the ball takes to return to its starting point after the ball  reach its maximum height is the same:

T_{descent}=t=\frac{v_{o}}{g}=\frac{37}{9.81}=3.78s

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A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodie
vivado [14]

Answer:

The length of the rod for the condition on the question to be met is L  =  1.5077 \ m

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is M  =  3.41 \ kg

           The mass of each small bodies is  m =  0.249 \ kg

           The moment of inertia of the three-body system with respect to the described axis is   I  =  0.929 \ kg \cdot  m^2

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

        I =  I_r + 2 I_m

Where  I_r is the moment of inertia of the rod about the describe axis which is mathematically represented as  

        I_r  =  \frac{ML^2 }{12}

And   I_m the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

           I_m  =   m * [\frac{L} {2} ]^2 =  m*  \frac{L^2}{4}

Thus  2 *  I_m  =  2 *  m  \frac{L^2}{4}  = m  *  \frac{L^2}{2}

Hence

       I  =  M  *   \frac{L^2}{12}  +  m  * \frac{L^2}{2}

=>   I  =    [\frac{M}{12}  + \frac{m}{2}] L^2

substituting vales  we have  

        0.929   =    [\frac{3.41}{12}  + \frac{0.249}{2}] L^2

       L  =  \sqrt{\frac{0.929}{0.40867} }

      L  =  1.5077 \ m

     

6 0
3 years ago
Compare the de Broglie wavelength of an electron moving at 1.30×107 miles per hour (5.81×106 m/s) to that of a bullet moving at
lesya [120]

Answer:

electron  λ = 12.5 nm , bullet  λ = 1.11 10⁻³³ m  and golf ball  λ = 4.7 10⁻³⁴ m

Explanation:

The Broglie wave duality principle states that all matter has wave and particle properties, it is expressed by the equation

      p = h / λ

Where lam is called broglie wavelength

Let's use the definition of momentum

       p = mv

Let's calculate the wavelengths

-Electron

     mv = h /λ

     λ = h / mv

     λ = 6.63 10⁻³⁴ / (9.1 10⁻³¹ 5.81 10⁶)  

     λ = 1.25 10⁻¹⁰ m

     λ = 12.5 nm

This is the X-ray region

-bullet

     λ = 6.63 10⁻³⁴ / (1.90 10⁻³  313)

     λ = 1.11 10⁻³³ m

It is too small, only particle characteristics are observed

-Golf ball

     λ = 6.63 10⁻³⁴ / (4.50 10⁻² 31.3)

    λ = 4.7 10⁻³⁴ m

Too small, only particle characteristics are visible

7 0
3 years ago
A 5.58 kg object with a speed of 35 m/s strikes a steel plate at an angle of 45.0 degrees with the normal to the plate, and rebo
MAVERICK [17]

Answer:

magnitude of vector is 276.19 kg m/s

Explanation:

The initial momentum is vector of magnitude

5.58 \times 35 = 195.3 (kg m/s)  And driven in a coherent manner with initial vector.

same magnitude is momentum after the impact, but it is oriented perpendicularly to initial momentum vector.

So, you have 2 momentum vector of  specified magnitude perpendicular to one another.

 The contrast between such two vectors is a right angle triangle hypotenuse of 195.3 sides

magnitude of vector is  \sqrt{ 195.3^2 + 195.3^2} = 276.19

8 0
3 years ago
Which factors are used to calculate the kinetic energy of an object? Check all that apply.
Verizon [17]
Gravity and velocity
8 0
3 years ago
Why is gasoline so flammable
Karo-lina-s [1.5K]
Flammable and combustible liquids themselves do not burn. It is the mixture of their vapours and air that burns. Gasoline, with a flashpoint of -40°C (-40°F), is a flammable liquid. Even at temperatures as low as -40°C (-40°F), it gives off enough vapour to form a burnable mixture in air.




Hope this helps




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8 0
4 years ago
Read 2 more answers
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