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Alecsey [184]
3 years ago
14

A trampoline spring has a force constant k = 800 N/m and is stretched exactly 17.5cm. What is the energy required to do this?

Physics
1 answer:
Artist 52 [7]3 years ago
6 0

Answer:

the energy required for the extension is 12.25 J

Explanation:

Given;

force constant of trampoline spring, k = 800 N/m

extension of trampoline spring, x = 17.5 cm = 0.175 m

The energy required for the extension is calculated as;

E = ¹/₂kx²

E = 0.5 x 800 x 0.175²

E = 12.25 J

Therefore, the energy required for the extension is 12.25 J

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Find the velocity. 10 points. Will give brainliest!
BARSIC [14]

Answer:

6.060606...

Explanation:

To figure out velocity, you divide the distance by the time it takes to travel that same distance, then you add your direction to it. So the distance would be 1000m and the time would be 2 minutes and 45 seconds and if you convert the minutes into fractions you would get 165 seconds than you would divide 1000m by 165 seconds and you would get 6.060606... seconds as her average velocity

4 0
3 years ago
2 pts
const2013 [10]

Answer:

I think the answer 1

Explanation:

im probably wrong too i dont know

5 0
2 years ago
Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau
____ [38]

The charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Similar charges repel each other, whereas charges that are opposed attract each other.

Given data;

Electric force,F = 9 × 10 ⁻⁹ N

Distance between charges,d = 7 × 10⁻⁴ m

Chrge,q₁ = q₂ =q C

From Columb's law;

\rm F = K \frac{q_1q_2}{d^2} \\\\ 9 \times 10^{-9}  = 9 \times 10^9 \frac{q^2}{(7 \times 10^{-4})^2} \\\\ q^2 = 4.9 \times 10^{-25} \\\\  q = 7 \times 10^{-13} \ C

Hence the charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

To learn more about Columb's law refer to the link;

brainly.com/question/1616890

#SPJ1

7 0
1 year ago
When all else remains the same, what effect would decreasing the focal length have on a convex lens?
aniked [119]
<h3>Answer;</h3>

<u>It would make the lens stronger. </u>

<h3>Explanation;</h3>
  • The focal length is the distance between the optical center or the center of the lens to the focal point of a convex or concave lens.
  • The power of the convex lens is lens ability to undertake refraction or bend light. It is given as the reciprocal of focal length.
  • Power of the lens = 1/ f; therefore the smaller the focal length the higher the power and the larger the focal length the lower the power.
  • Thus; decreasing the focal length of a convex lens makes the lens stronger.

4 0
3 years ago
Read 2 more answers
Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.
Fofino [41]

Answer:

Part a)

\alpha = 782.6 rad/s^2

Part B)

\omega = 587 rad/s

Part c)

a_t = 24.3 m/s^2

Explanation:

Part a)

As we know that

a = R \alpha

so we will have

a = 1.80 m/s^2

R = 0.230 cm

\alpha = \frac{a}{R}

\alpha = \frac{1.80}{0.230 \times 10^{-2}}

\alpha = 782.6 rad/s^2

Part B)

Angular speed of the yo-yo

\omega = \alpha t

so we have

\omega = 782.6 \times 0.750

\omega = 587 rad/s

Part c)

Tangential acceleration is given as

a_t = R \alpha

a_t = (3.10 \times 10^{-2})(782.6)

a_t = 24.3 m/s^2

6 0
3 years ago
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