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umka21 [38]
3 years ago
15

How would the elastic potential energy change if the displacement is multiplied by five?

Physics
1 answer:
nikdorinn [45]3 years ago
3 0

Answer: Work is done when a spring is extended or compressed . Elastic potential energy is stored in the spring. The elastic potential energy stored can be calculated using the equation: elastic potential energy = 0.5 × spring constant × (extension) 2.

Explanation: can you please mark me as brainliest

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4. Choose the velocity quantity from the following.
Tresset [83]

Answer:

a.8m/s is my ans it may help you

3 0
3 years ago
g If the x-component of a force vector is 5.69 newtons and its y-component is 8.00 newtons, then what is its magnitude?
beks73 [17]

Answer:

F = 9.82 N

Explanation:

given,

Force x-component = 5.69 N

Force y-component = 8 N

magnitude of force = ?

Resultant of force

F = \sqrt{F_x^2 + F_y^2}

F = \sqrt{5.69^2 + 8^2}

F = \sqrt{32.3761 + 64}

F = \sqrt{96.3761}

F = 9.82 N

Hence, the magnitude of force is equal to 9.82 N

4 0
2 years ago
Read 2 more answers
A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressin
BARSIC [14]

Answer:

U_k = 0.113

Explanation:

using the law of the conservation of energy:

E_i -E_f=W_f

\frac{1}{2}Kx^2=NU_kd

where K is the spring constant, x is the spring compression, N is the normal force of the block, U_k is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

N = mg

N = 3.35 kg * 9.81 m/s

N = 32.8635

So, Replacing values on the first equation, we get:

\frac{1}{2}(138)(0.123)^2= (32.8635)U_k(0.281m)

solving for U_k:

U_k = 0.113

8 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
Marizza181 [45]

Answer:

c) 2.02 x 10^16 nuclei

Explanation:

The isotope decay of an atom follows the equation:

ln[A] = -kt + ln[A]₀

<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>

[A] = Our incognite

k is constant decay:

k = ln 2 / Half-life

k = ln 2 / 4.96 x 10^3 s

k = 1.40x10⁻⁴s⁻¹

t is time = 1.98 x 10^4 s

[A]₀ = 3.21 x 10^17 nuclei

ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]

ln[A] = 37.538

[A] = 2.01x10¹⁶ nuclei remain ≈

<h3>c) 2.02 x 10^16 nuclei</h3>
7 0
2 years ago
C
Artemon [7]

Answer:

constant volicty of the pumper when they hit ground 7.03-/s

8 0
3 years ago
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