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Deffense [45]
3 years ago
5

A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the

width of the central bright spot on the screen change if the slit width is doubled?
a. It will become four times as large.
b. It will double.
c. It will be cut in half.
d. It will become eight times as large.
e. It will be cut to one-quarter its original size.
Physics
1 answer:
garik1379 [7]3 years ago
7 0

Answer:

c : it wil be cut in half.

The pattern is formed on a distant screen so we can use the Fraunhofer difracction for a single slit. The formula of the width of the central bright spot is given by Δx = (2λz)/a, where λ is the wavelength and a is the width of the slit. So if the inicial width (a_1) is doubled (a_2= 2 x a_1),the width of the central spot will be cut in half Δx = (2λz)/a_2 = (2λz)/2xa_1 .

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A piece of hot copper of mass 4.00 kg has it's temperature decrease by 36.90 ºC when it is placed in a body of water of unknown
Ainat [17]

Answer:

1.48 kg

Explanation:

mass of copper = 4 kg

Decrease in temperature of copper = 36.9°C

increase in temperature of water = 9.20°C

Let the mass of water is m.

Specific heat of water = 4181 J/Kg°C

Specific heat of copper = 385 J/kg°C

According to the principle of caloriemetery

Heat lost by copper = heat gained by water

mass of copper x specific heat of copper x decrease in temperature = mass of water x specific heat of water x increase in temperature

4 x 385 x 36.9 = m x 4181 x 9.20

56826 = 38465.2 m

m = 1.48 kg

Thus, the mas of water is 1.48 kg.

mas of water

5 0
4 years ago
What is the mass of a 735-N weight on Earth?
Allisa [31]
W = mg,      Assuming g ≈ 9.8 m/s² on the earth surface.

735 N =  m* 9.8

735/9.8 =  m

75 = m

Mass , m = 75 kg.  B.
3 0
3 years ago
A 50kg block is at rest on the ground.
yKpoI14uk [10]

Answer:

(b) 490 N downwards

Explanation:

(b) Formula : w= mg

m is the mass of the object

g is a gravitational constant (9.8)

w= mg

= 50×9.8

= 490 N downwards

7 0
3 years ago
A playground merry-go-round of radius r = 1 m has a moment of inertia of i = 240 kg*m^2. and is rotating at a rate of ω = 8 rev/
Ghella [55]
When the child jumps onto the merry-go-around the moment of inertia of the system changes. If we consider the child to be point-like mass then its moment of inertia would be:
I_{ch}=mr^2
We get the new moment of inertia by simply adding the child's moment of inertia to the old moment of inertia.
I_{new}=I_{old}+I_{ch}=240+35(1)^2=275 $kgm^2
Since there is no force mention we must assume that angular momentum is conserved.
L=const.\\ L=I_{old}\omega_0=I_{new}\omega'\\ \omega'=\frac{I_{old}\omega_0}{I_{new}}
When we plug in all the numbers we get:
\omega'=\frac{240\cdot8}{275}=6.98 \ \frac{rev}{min}

3 0
3 years ago
A horizontal pipe of inner diameter 2.2 cm carries water with a density of 1000.0 kg/m3 flowing at a rate of 1.5 kg/s. If the pi
EleoNora [17]

The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is,\rm d_1 = 2.2 \ cm

initial radius,

r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm

The initial crossection area;

\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times  (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2

The final crossection area;

\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2

The initial flow rate is;

R = density ×velocity ×area

\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1  = 3.947 \ m/sec

The speed of the water in the wider part will be;

From the continuity equation;

\rm A_1 V_1 = A_2V_2  \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec

Hence, the speed of the water in the wider part will be 1.194 m/sec.

To learn more about the speed, refer to the link;

brainly.com/question/7359669

#SPJ1

7 0
2 years ago
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