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2 years ago

Calculate the workdone to stretch an elastic string by 40cm if a force of 10 newton produces an extension of 4cm in it

Physics

1 answer:

notka56 [123]2 years ago

5 0

Answer:

Workdone = 20 Joules

Explanation:

Given the following data;

Force = 10N

Extension, e = 4cm to meters = 4/100 = 0.04 meters

Workdone extension = 40cm to meters = 40/100 = 0.4 meters

To find the work done;

First of all, we would find the spring constant using the formula;

Force = spring constant * extension

10 = spring constant * 0.04

Spring constant = 10/0.04

Spring constant = 250 N/m

Next, we find the work done;

Workdone = ½ke²

Where;

k is the spring constant.

e is the extension.

Substituting into the formula, we have;

Workdone = ½ * 250 * 0.4²

Workdone = 125 * 0.16

Workdone = 20 Joules

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A xenon arc lamp is covered with an interference filter that only transmits light of 400-nm wavelength. When the transmitted lig

zepelin [54]

Answer:

3. both are true.

Explanation:

Energy increses with decrease in wavelenght.

For photoemission to occur, a threshold energy barrier must be broken.

Higher energy means more electrons will be emmited.

The electrons emmited will posses energy that is less than the incident energy by the value of the threshold energy.

So the higher the energy, the higher the energy possessed by the electrons.

4 0

3 years ago

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A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

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2 years ago

A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surf

Sati [7]

Answer:

a) The minimum thickness of the oil slick at the spot is 313 nm

b) the minimum thickness be now will be 125 nm

Explanation:

Given the data in the question;

a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

t $_{min$ = λ/2n

given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20

we substitute

t $_{min$ = 750 / 2(1.20)

t $_{min$ = 750 / 2.4

t $_{min$ = 312.5 ≈ 313 nm

Therefore, The minimum thickness of the oil slick at the spot is 313 nm

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

minimum thickness of the oil slick at the spot will be;

t $_{min$ = λ/4n

given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50

we substitute

t $_{min$ = 750 / 4(1.50)

t $_{min$ = 750 / 6

t $_{min$ = 125 nm

Therefore, the minimum thickness be now will be 125 nm

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2 years ago

In a given chemical reaction, the energy of the products is greater than the energy of the reactants. Which statement is true fo

Flura [38]

Energy is released in the reaction

Explanation:

In a given where the energy of the products is greater than that of the reactants, we can infer that energy is released in the reaction.

This indicates that the reaction is an exothermic or exergonic reaction.

These reaction types are accompanied by release of energy.

In an exothermic change energy is released to the surroundings.
The surrounding becomes hotter at the end of the change.
This applies in exergonic reaction which leaves a reaction having more energy than it originally started with.

Learn more:

Exothermic process brainly.com/question/10567109

#learnwithBrainly

3 0

3 years ago

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What is the maximum number of unpaired d electrons that an atom or ion can possess?

Sonbull [250]

Answer:

Explanation:

The d subshell has 5 orbitals, each capable of holding a maximum of two electrons. Hund's rule tells us that every orbital in a sub-level must first be singly occupied by electrons before any orbital is doubly occupied. Therefore five electrons will fill the five orbitals within the d subshell.

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3 years ago