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labwork [276]
3 years ago
12

If the work required to stretch a spring 1 ft beyond its natural length is 12 ft-lb, how much work is needed to stretch the same

spring from its natural length to 3 in? Hint: 1 ft = 12 in
Physics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:W=\frac{3}{4} ft-lb

Explanation:

Given

Work required to stretch 1 ft is 12 ft-lb

and we have to find work required to stretch 3 in.

i.e. \frac{1}{4} ft

12=\frac{1}{2}K\left ( 1\right )^2 ------(1)

W=\frac{1}{2}k\left ( \frac{1}{4}\right )^2-----(2)

divide (1)&(2)

\frac{12}{W}=\left ( \frac{4}{1}\right )^2

W=\frac{12}{4\times 4}

W=\frac{3}{4} ft-lb

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Which ion channels mediate the falling phase of an action potontial?
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Answer:

Voltage-gated K+ channels

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4 years ago
A light ray incident from medium 1 to medium 2, where n1>n2. When the incident angle exceed the critical angle ac, the refrac
vovikov84 [41]

Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}

Where,  

{\theta_i}  is the angle of incidence

{\theta_r} is the angle of refraction = 90°

{n_2} is the refractive index of the refraction medium

{n_1} is the refractive index of the incidence medium

Thus,

n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}

The formula for the calculation of critical angle is:

{sin\theta_{critical}}=\frac {n_2}{n_1}

Where,  

{\theta_{critical}} is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

7 0
4 years ago
PLEASE HEELP!!!
Ann [662]

Answer:

The "pressure" of the electricity is electric potential. Electric potential is the amount of energy available to push each unit of charge through an electric circuit. The unit of electric potential is the volt. ... A volt is the force needed to move one amp through a conductor that has 1 ohm of resistance

8 0
3 years ago
Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.04 m away from a waterfall 0.585 m in heigh
S_A_V [24]

Answer:

V₀ = 5.47 m/s

Explanation:

The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 3.04 m

θ = Launch Angle = 41.7°

V₀ = Minimum Launch Speed = ?

g = 9.81 m/s²

Therefore,

3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)

V₀² = 3.04 m/(0.10126 s²/m)

V₀ = √30.02 m²/s²

<u>V₀ = 5.47 m/s</u>

6 0
3 years ago
Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram
Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

  • Force 1 = 500 Newton
  • Distance 1 = 1.2 meter
  • Force 2 = 500 Newton
  • Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

Work\;done = Force \times distance

<u>For </u><u>student 1</u><u>:</u>

Work\;done = 500 \times 1.2

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

Work\;done = 500 \times 5

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0
3 years ago
Read 2 more answers
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