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Ierofanga
5 days ago
14

Kendra separately pushes two boxes in a straight line at constant velocity over the same distance on level ground. Each box begi

ns and ends at rest. Kendra exerts no force to stop either box; friction brings them both to rest. The contact surfaces and areas between each box and the floor are the same for both, but one box is heavy and the other is light.
Which of the statements is correct?

A) Kendra does the same amount of work on both boxes, and the same amount of net work is also done on both boxes.

B) Kendra does more work on the heavy box, and more net work is also done on the heavy box.

C) Kendra does more work on the heavy box, but the net work done on both boxes is equal.

D) Kendra does the same amount of work on both boxes, but more net work is done on the heavy box.

E) Kendra does more work on the light box, but more net work is done on the heavy box.
Physics
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Laser beam with wavelength 632.8 nm is aimed perpendicularly at opaque screen with two identical slits on it, positioned horizon
Softa [3030]

Response:

1 slit width = 0.158 mm, slit separation = 0.633 mm, distance between diffraction maxima = 12.7 mm

Explanation:

This involves defining several terms:

λ, the wavelength of the beam

D, the distance from the plane to the slit

x, the distance between minima in the diffraction pattern (in single slit setups)

w, the fringe width for double slit setups

1. In a double slit experiment, the fringe width is also recognized as the distance between Maxima.

Thus, w=λD/d, leading to d=λD/w when rearranged.

So, d= (632.8 x 10^-9 x 1 x 10^3)/1

which equals d= 0.633mm.

For single slit diffraction, minima are defined by a*sinΘ= m*x

where a is the slit width and m is an integer.

For small angles, it simplifies to Θ= (x/D) = (λ/a), allowing us to solve for a:

a = λD/a

a = (632.8x10^-9 x 1)/4

yielding a= 0.158mm.

2. A grating with 20 slits/mm gives d= 1/20mm= 0.05x10^-3.

To find y (the distance between maxima), apply y= λL/d:

Finally, y= (632.8 x 10^-9 x 1)/0.05x10^-3, which results in y= 12.7mm.

6 0
2 months ago
A metal object with a mass of 19g is heated to 96c, then transferred to a calorimeter containg 75g of water at 18c. the water an
Yuliya22 [3333]

Let Cp represent the specific heat of the metal object. To find this, we can set up a heat balance equation (heat lost by metal = heat gained by water):

- 19g * Cp * (22degC – 96degC) = 75g * 4.184J/g degC * (22degC - 18degC)

<span>Cp = 0.893 J/g degC</span>

6 0
2 months ago
Read 2 more answers
Help asap please!! An aluminum block of mass 12.00 kg is heated from 20 C to 118 C. If the specific heat of aluminum is 913 J-1
Sav [3153]
Q = mCΔT, in which Q = energy required, m = mass of the block, C = specific heat, ΔT = temperature change.

Utilizing the values provided;

Q = 12*913*(118-20) = 1073688 J = 1073.688 kJ.

The correct option is B.
7 0
3 months ago
Two friends, Burt and Ernie, are standing at opposite ends of a uniform log that is floating in a lake. The log is 3.2m long and
Softa [3030]

Answer:

x = 0.29 m

Explanation:

It is known that the total external force acting on the mass system equals ZERO,

so the center of mass of the entire system will stay stationary.

We find that

m_1\Delta x_1 + m_2\Delta x_2 + m_3\Delta x_3 = 0

Since Ernie approaches Burt's position, we have:

33 x + 290 x + 32(-3.2 + x) = 0

355 x = 102.4

therefore, we conclude that

x = \frac{102.4}{355}

x = 0.29 m

8 0
2 months ago
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