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jek_recluse [69]
4 years ago
11

Eddie tosses his phone up to his friends window, which is 4 m above the ground. what energy conversion takes place with respect

to the phone?
A kinetic energy to elastic potential energy
B elastic potential energy to kinetic energy
C gravitational potential energy to kinetic energy
D kinetic energy to gravitational potential energy
Physics
2 answers:
lora16 [44]4 years ago
7 0

Answer:right answer is D apex

Explanation:

yan [13]4 years ago
3 0
C the anwer is c hope i help
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A 7.00 kg bowling ball is held 2.00 m above the ground. Using g- 9.8 m/s^2, how much energy does the bowling ball have due to it
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It should be B.137 just took the test.
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Why might the term red hot be misleading? (relating to stars)
Allushta [10]

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3 0
3 years ago
Hanna tosses a ball straight up with enough speed to rermain in the air for several seconds
Ipatiy [6.2K]
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7 0
3 years ago
Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19
Alisiya [41]

Answer:

The workdone is  W = 9.28 * 10^{3} J

Explanation:

From the question we are told that

   The height of the cylinder is  h = 0.588\ m

   The face Area is  A = 4.19 \ m^2

    The density of the cylinder is \rho  =  0.346 * \rho_w

     Where \rho_w is the density of freshwater which has a constant value

              \rho_w = 1000 kg/m^3

     

Now  

     Let the final height of the device under the water be  =  h_f

      Let  the initial volume underwater be = V_n

     Let the initial height under water be  = h_i

      Let the final volume under water be  = V_f

According to the rule of floatation

        The weight of the cylinder =  Upward thrust

This is mathematically represented as

          \rho_c g V_n = \rho_w gV_f

         \rho_c A h = \rho A h_f

So      \frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}

   =>     \frac{h_f}{h_c}  = 0.346

Now the work done is mathematically represented as  

          W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh

               =   \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.

              = \frac{g A \rho}{2}  [h^2 - h_f^2]

              = \frac{g A \rho}{2} (h^2)  [1  - \frac{h_f^2}{h^2} ]

Substituting values

        W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)

        W = 9.28 * 10^{3} J

4 0
3 years ago
An electron moves in a circular path perpendicular to a constant magnetic field with a magnitude of 2.50 mT. The angular momentu
liberstina [14]

Answer:

Explanation:

The angular momentum of electron mvR = 6 x 10⁻²⁵  Js

Magnetic field B = 2.5 x 10⁻³ T

radius of circular path R = mv / Bq

where m is mass , v is velocity and q is charge on electron

R² = mvR / Bq

R² = 6 x 10⁻²⁵ / 2.5 x 10⁻³ x 1.6 x 10⁻¹⁹

= 1.5 x 10⁻³

R = 3.87 x 10⁻² m

mvR = 6 x 10⁻²⁵

v = 6 x 10⁻²⁵  / mR

= 6 x 10⁻²⁵  / 9.1 x 10⁻³¹ x 3.87 x 10⁻²

= .17 x 10⁸

= 17 x 10⁶ m/s

6 0
3 years ago
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