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zlopas [31]
3 years ago
6

An electron is carried from the positive terminal to the negative terminal of a 9 v battery. How much work is required in carryi

ng this electron?
Physics
1 answer:
shutvik [7]3 years ago
5 0

Answer:

The work required for the electron is 1.44x10^{-18} J

Explanation:

The work of the electron is calculated in the equation:

W= e * V

e= 1.6x10^{-19} C

C= Coulomb = s*A

V=  Volts = \frac{W}{A}

W= \frac{J}{s}

W= 1.6 x10^{-19}C *9 V\\W=1.44x10^{-18} C*V\\ W=1.44x10^{-18} s*A*\frac{W}{A} \\W=1.44x10^{-18} W*s\\W=1.44x10^{-18} \frac{J}{s}*s \\W=1.44x10^{-18} J\\

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A shopper is pushing a cart down a grocery store aisle. Starting from rest, the shopper applies a constant force to the cart for
inysia [295]

A shopping cart that starts from rest, is accelerated for 4 s, moves at constant velocity for 4 s, and is decelerated for 4s until returning to rest, has an average acceleration of 0 m/s².

A shopper is pushing a cart down a grocery store aisle. The movement of the cart is:

  • It starts from rest.
  • From t = 0 s to t = 4.0 s it is accelerated with a constant force.
  • From t = 4 s to t = 8.0 s it receives just enough force to balance the friction on the cart.
  • From t = 8 s to t = 12 s it is decelerated until it comes to rest.

All in all, at the initial time (t = 0 s), the velocity is 0 m/s (rest) and at the final time (t = 12 s) the velocity is 0 m/s as well (rest). The average acceleration in that period is:

a = \frac{v_{12}-v__o}{t_{12}-t_0} = \frac{0m/m-0m/s}{12s-0s}  = 0 m/s^{2}

A shopping cart that starts from rest, is accelerated for 4 s, moves at constant velocity for 4 s, and is decelerated for 4s until returning to rest, has an average acceleration of 0 m/s².

Learn more: brainly.com/question/16274121

3 0
2 years ago
Read 2 more answers
A 0.37-kg object connected to a light spring with a force constant of 23.2 N/m oscillates on a frictionless horizontal surface.
mars1129 [50]

Answer:

a) v = 31.67 cm / s , b)   v = -29.36 cm / s , c) v= 29.36 cm/s, d) x = 3.46 cm

Explanation:

The angular velocity in a simple harmonic movement is

       w = √ K / m

       w = √ 23.2 / 0.37

       w = 7,918 rad / s

a) the expression against the movement is

        x = A cos (wt + Ф)

Speed ​​is

        v = dx / dt = - A w sin (wt + Ф)

 The maximum speed occurs for cos = ± 1

        v = A w

        v = 4.0 7,918

        v = 31.67 cm / s

b) as the object is released from rest

        0 = -A w sin (0+ Фi)

        sin Ф = 0

         Ф = 0

The equation is

        x = 4.0 cos (7,918 t)

        v = -4.0 7,918 sin (7,918 t)

        v = - 31.67 sin (7.918t)

     

Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians

          7,918 t = cos⁻¹ 1.5 / 4.0

          t = 1,186 / 7,918

          t = 0.1498 s

We look for speed

         v = -31.67 sin (7,918 0.1498)

         v = -29.36 cm / s

c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign

         v = 29.36 cm / s

d) let's look for the time for the condition v = v_max / 2

         31.67 / 2 = 31.67 sin ( 7,918 t)

          7.918t = sin⁻¹ 0.5

         t = 0.5236 / 7.918

         t = 0.06613

With this time let's look for displacement

         x = 4.0 cos (7,918 0.06613)

        x = 3.46 cm

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3 years ago
What term best describes the geologic event taking place in the above illustration?
grigory [225]

u lying you made me get it wrong, for ya'll out there who want the real answer is sea floor spreading

8 0
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Answer:

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