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Light with a wavelength of 700 nm (7×〖10〗^(-7) m) is incident upon a double slit with a separation of 0.30 mm (3 x 10-4 m). A screen is located 1.5 m from the double slit. At what distance from the screen will the first bright fringe beyond the center fringe appear?

Answers

Answer 1

Answer:

Answer:

$0.0035\ \text{m}$

Explanation:

y = Distance from the center point

d = Separation between slits = 0.3 mm

D = Distance between slit and screen = 1.5 m

$\lambda$ = Wavelength = 700 nm

m = Order = 1

We have the relation

$d(y)/(D)=m\lambda\n\Rightarrow y=(Dm\lambda)/(d)\n\Rightarrow y=(1.5* 1* 700* 10^(-9))/(0.3* 10^(-3))\n\Rightarrow y=0.0035\ \text{m}$

The distance from the screen at which the first bright fringe beyond the center fringe appear is $0.0035\ \text{m}$ .

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5. (Serway 9th ed., 7-3) In 1990, Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much work was done on the object by Arfeuille in this lift, assuming the object was lifted at constant speed? (b) What total force was exerted on Arfeuille’s teeth during the lift? (Ans. (a) 472 J; (b) 2.76 kN)

Answers

Para resolver este problema es necesario aplicar los conceptos de Fuerza, dados en la segunda Ley de Newton y el concepto de Trabajo, como expresión de la fuerza necesaria para realizar una actividad en una distancia determinada.

El trabajo se define como

W = F*d

Where,

F = Force

d = Distance

At the same time we have that the Force by second's Newton law is equal to

F = mg

Where,

m = mass

g = Gravitational acceleration

PART A) Using our values and replacing we have that

$W = F*d\nW = mg*d\nW=281.5*9.8(17.1*10^(-2)\nW = 471.738 J\approx 472J$

PART B) Using Newton's Second law we have that,

$F = mg \nF= 281.5*9.8\nF= 2758.7 N \approx 2.76kN$

_ is the name given to the heat energy received from the sun

Answers

Answer:

The think the answer is solar radiation.

Explanation:

here, we gain the heat from the sun through a radiation. When it travels from the sun the harmful radiation are absorbed by ozone layer and heat enegry is provided to the surface of the Earth.

hopeit helps..

Find T1, the magnitude of the force of teTo practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.6 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.7 m from the other end. A monkey of mass 1.3 kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.nsion in string 1, at the moment that the monkey is halfway between the ends of the bar. Express your answer in newtons using three significant figures. View Available Hint(s)

Answers

Answer:

$T_2 = 24.95 N$

$T_1 = 13.3 N$

Explanation:

As we know that total torque on the rod must be zero when monkey is at mid point of the rod

So we have

torque due to Tension at other end = torque due to weight of monkey + rod

so we will have

$T_2 (3 - 0.7) = (Mg + mg)(1.5)$

here we know that

M = 2.6 kg

m = 1.3 kg

$T_2(2.3) = (2.6 + 1.3)(9.81)(1.5)$

$T_2 = 24.95 N$

Now similarly we can say that

$T_1 + T_2 = (m + M)g$

$T_1 + 24.95 = (2.6 + 1.3)(9.81)$

$T_1 = 13.3 N$

If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.(a) What is the heat capacity of this object?
(b) What is the object's specific heat?

Answers

When an object gets heated by a temperature ΔT energy needed, E = mcΔT

Here energy is given E = 2050 J

Mass of object = 150 g

Change in temperature ΔT = 15 $^0C$ = 15 K

a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.

So heat capacity = E/ΔT = 2050/15 = 136.67 J/K

b) We have E = mcΔT

c = $(2050)/(150*10^(-3)*15) = 911.11 J/kgK$

So object's specific heat = 911.11 J/kgK

If you rub a balloon on your hair, youcan hang the balloon on the wall.
Why does the balloon stick to the
wall?

Answers

Answer:

The balloon is electrostatically charged

Explanation:

After rubbing it on the hair, the balloon is electrically charged, and as such, when approaching the wall it draws opposite charges from the wall creating locally on the wall's surface an accumulation of the charges opposite to the balloon . and repelling deeper into the wall those charges of the same sign.

A train of 150 m length is going toward north direction at a speed of 10 ms–1 and a parrot is flying towards south direction parallel to the railway track with a speed of 5 ms–1. The time taken by the parrot to cross the train is equal to.?

Answers

Answer:

10ms

Explanation:

The bird must travel the length of the train (150m), with a combined speed of 15m/s this means it will take 10s to cross an accumulated 150ms.

Answer:

The time taken by the parrot to cross the train is = 10 m/s

Explanation:

given:

A train of 150 m length is going toward north direction at a speed of 10 m/s

and a parrot is flying towards south direction parallel to the railway track with a speed of 5 m/s

find;

The time taken by the parrot to cross the train

using the distance over speed relation formula t = d / v

where:

v = velocity

d = distance

t = time

v = 10 m/s + 5 m/s = 15 m/s (combine velocity)

d = 150 m

t = ?

plugin values into the formula

t = 150

t = 10 m/s

therefore,

The time taken by the parrot to cross the train is = 10 m/s

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