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antiseptic1488 [7]
3 years ago
7

A block is attached to a spring hanging vertically. After being slowly lowered, it hangs at rest with the spring stretched by 15

.0 cm. If the block is raised back up and released from rest with the spring unstretched, what maximum distance does it fall
Physics
1 answer:
artcher [175]3 years ago
6 0

Answer:

The equilibrium position is 15 cm below the starting point. The motion is symmetric about the equilibrium position, so the two turning points are 30 cm apart.

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Mass of the milkyway galaxy :

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2 years ago
Calculate the tension on the rope.<br> A. 57.89 N<br> B. 32.73 N<br> C. 69.84 N<br> D. 12.55 N
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Hi there!

\large\boxed{\text{B. 32.73N}}

To calculate the tension, we must calculate the acceleration of the system.

Begin with a summation of forces:

∑F = -M₁gsinФ + T - T + M₂g

Simplify and solve for acceleration: (Tensions cancel out)

a = \frac{-M_1gsin\theta + T - T + M_2g}{M_1+M_2}

Plug in values. Let g = 10 m/s²

a = \frac{-3(10)(sin30)+8(10)}{3+8} = 5.91 m/s^{2}

Now, to find tension, let's sum up the forces acting on ONE block. For simplicity, we can look at the hanging block:

∑F = -T + W

ma = -T + W

Rearrange to solve for T:

T = W - ma

We know the acceleration, so plug in the values:

T = (8)(10) - (8)(5.91) = 32.73 N

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3 years ago
The rate constant of a reaction is 7.8 × 10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C? Enter yo
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Answer:

k_2 = 7.815 * 10^-3 s^-1

Explanation:

Given:

- rate constant of reaction k_1 = 7.8 * 10^-3 s^-1    @  T_1 = 25 C

- rate constant of reaction k_2 = ?    @  T_2 = 75 C

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- Gas constant R = 8.314472 KJ / mol . K

Find:

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Solution:

- we will use a combined form of Arrhenius equations that relates rate constants k as function of E_a and temperatures as follows:

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