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alina1380 [7]
4 years ago
14

An archer pulls a bow string 0.5 m. If the spring constant is 16,000 N/m, what is the energy stored in the bow string?

Physics
1 answer:
mars1129 [50]4 years ago
3 0

2000J

Explanation:

Given parameters:

Extension = 0.5m

Spring constant = 16000N/m

Unknown:

Energy stored in the bow string = ?

Solution:

The energy stored in a bow string is an elastic potential energy.

It can be calculated using the expression below;

     Elastic energy = \frac{1}{2} K e²

Where k is the spring constant

            e is the extension

Input the parameters;

  Elastic energy = \frac{1}{2} K e²

                          =\frac{1}{2} x 16000 x 0.5²

                          = 2000J

learn more:

Potential energy brainly.com/question/10770261

#learnwithBrainly

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ASHA 777 [7]

The charge would be in equilibrium so there would be no charge in the body of the conductor.


Answer:

(b) there cannot be any charge in the body of the conductor

5 0
3 years ago
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A uniform electric field exists in the region between two oppositely charged plane-parallel plates. An electron is released from
Zigmanuir [339]

Answer:

Explanation:

  • given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s
  • q = 1.6 x 10_19C
  • using S = at^2/2
  • acceleration = 0.032 X 2 /(1.30×10−8)^2

a = 3.79 x 10^14m/s^2

  • From F = ma
  • F = qE
  • ma = qE

E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19

E = magnitude of this electric field. = 2156.3N/C

b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as

= 2 X 3.79 x 10^14 X 0.032

= 4.92 X 10^6m/s

5 0
3 years ago
Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the
Natasha_Volkova [10]

The force per unit length between the two wires is 6.0\cdot 10^{-5} N/m

Explanation:

The magnitude of the force per unit length exerted between two current-carrying wires is given by

\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}

where

\mu_0 = 4\pi \cdot 10^{-7} Tm/A is the vacuum permeability

I_1, I_2 are the currents in the two wires

r is the separation between the two wires

For the wires in this problem, we have

I_1 = 1.75 A

I_2 = 3.45 A

r = 2.00 cm = 0.02 m

Substituting into the equation, we find

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m

Learn more about current and magnetic fields:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

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4 0
4 years ago
4. Determine the net force for the free-body
krek1111 [17]

Answer:

Fapp =44N is the correct answer

Explanation:

Please mark brainliest

3 0
3 years ago
8. A car travels at a constant velocity of 70 mph for one hour. By the end of the second hour, the car’s velocity was 60 mph. At
Mrac [35]

<u>Answer:</u>

  Positive acceleration is in third hour and negative acceleration is in second hour.

<u>Explanation:</u>

  Velocity of car in first hour =  70 mph

  Velocity of car in second hour = 60 mph

  Velocity of car in third hour = 80 mph

   Acceleration = Change in velocity / Time

   Acceleration in second hour = (60 - 70)/1 = -10 mph²

   Acceleration in third hour = (80 - 60)/1 = 20 mph²

   So positive acceleration is in third hour and negative acceleration is in second hour.

8 0
4 years ago
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