All categories

3 years ago

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

1 answer:

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

You might be interested in

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

Artemon [7]

Answer:

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

8 0

3 years ago