The specific heat of hexane is 2.26 J/gºC. If 11,500 J of heat is added to

Light of wavelength 660 nm passes through two narrow slits 0.65 mm apart. The screen is 2.55 m away. A second source of unknown

stiv31 [10]

Answer:

Explanation:

Distance of 2 nd order fringe x₁ = 2 λ D/d [ λ is wave length , D is distance of screen , d is slit distance.

x₁ = 2 x 660 x10⁻⁹x 2.55/.65 x 10⁻³.= 5.17846 x 10⁻³ m.

Distance of fringe for 2nd radiation = 5.178 x 10⁻³ -1.17 x 10⁻³ = 4.008 x10⁻³

4.008 x 10⁻³ = 2 x λ x 2.55/.65 x 10⁻³

λ = 511 nm approx.

.

8 0

3 years ago

When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical powe

Grace [21]

Explanation:

the answer is in the above image

8 0

3 years ago

Two electrons travel towards each other at 0.2 c parallel to the laboratory x-axis. What is the relative velocity of one electro

Leya [2.2K]

1) In the reference frame of one electron: 0.38c

To find the relative velocity of one electron with respect to the other, we must use the following formula:

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

where

u is the velocity of one electron

v is the velocity of the second electron

c is the speed of light

In this problem:

u = 0.2c

v = -0.2c (since the second electron is moving towards the first one, so in the opposite direction)

Substituting, we find:

$u'=\frac{0.2c+0.2c}{1+\frac{(0.2c)(0.2c)}{c^2}}=\frac{0.4c}{1+0.04}=0.38c$

2) In the reference frame of the laboratory: -0.2c and +0.2c

In this case, there is no calculation to be done. In fact, we are already given the speed of the two electrons; we are also told that they travel in opposite direction, so their velocities are

+0.2c

-0.2c

5 0

3 years ago

What is the order of magnitude of the distance of Sun to nearest star in meters?

neonofarm [45]

Answer:

Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.

Explanation:

Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

Let's model the Milky Way galaxy as a disk.

The volume of a disk is:

V

=

π

⋅

r

2

⋅

h

Plugging in our numbers (and assuming that

π

≈

3

)

V

=

π

⋅

(

10

21

m

)

2

⋅

(

10

19

m

)

V

=

3

×

10

61

m

3

Is the approximate volume of the Milky Way.

Now, all we need to do is find how many stars per cubic meter (

ρ

) are in the Milky Way and we can find the total number of stars.

Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of

4

×

10

16

m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.

ρ

=

n

V

Using the volume of a sphere

V

=

4

3

π

r

3

ρ

=

1

4

3

π

(

4

×

10

16

m

)

3

ρ

=

1

256

10

−

48

stars /

m

3

Going back to the density equation:

ρ

=

n

V

n

=

ρ

V

Plugging in the density of the solar neighborhood and the volume of the Milky Way:

n

=

(

1

256

10

−

48

m

−

3

)

⋅

(

3

×

10

61

m

3

)

n

=

3

256

10

13

n

=

1

×

10

11

stars (or 100 billion stars)

Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.

4 0

2 years ago

Read 2 more answers

List 3 examples in which friction helps us or makes things easier in our daily life. Explain the effect of friction for each.

Kisachek [45]

<span>Friction can help us play tennis, knock down bowling pins, and cook dinner.</span>

8 0

3 years ago

Read 2 more answers