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ale4655 [162]
3 years ago
7

Please Help!!

Physics
1 answer:
kotegsom [21]3 years ago
8 0

Answer:

Potential energy of spring = 24 Joules.

Explanation:

Given the following data;

Spring constant = 85N/m

Extension, e = 0.75m

Mass = 25kg

To find the potential energy of a spring

Potential energy of a spring is given by the formula;

P.E = ½ke²

Substituting into the equation, we have

P.E = ½*85*0.75²

P.E = 42.5 * 0.5625

P.E = 23.91 ≈ 24 Joules

P.E = 24 Joules

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A mass on a spring A oscillates at twice the frequency of the same mass on spring B. Which statement is correct?A.The spring con
Nataliya [291]

Answer:

A.The spring constant for B is one quarter of the spring constant for A.

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If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.

T_B = 2T_A

As T = 2\pi\sqrt{\frac{m}{k}}, and the 2 springs have the same mass

2\pi\sqrt{\frac{m}{k_B}} = 2\pi\sqrt{\frac{m}{k_A}}

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[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r
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Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)  it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity V_{f} for an object traveling distance h taking the initial vertical component of velocity as V_{i} the kinematics equation is written as

V_{f}^{2}=V_{i}^{2}+2ah where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

V_{f}^{2}=V_{i}^{2}+2gh

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}= 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

V_{f}^{2}=V_{i}^{2}+2gh

Since we have final velocity of 25 m/s

V_{i}^{2}=2gh-V_{f}^{2}

V_{i}=\sqrt{(V_{f}^{2}-2gh)}

V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}= 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0
3 years ago
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