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-Dominant- [34]
3 years ago
11

A woman weighs 65 kg which is _____lbs.

Physics
1 answer:
pogonyaev3 years ago
7 0
She would be 143.3 lbs.
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A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A
Kruka [31]

Answer:

\omega_f = 585.37 \ rev/s

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)=  800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

I \omega_i = I' \omega_f

(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f

M\times 800 = ( M + m )\omega_f

3\times 800 = (3+1.1)\times \omega_f

2400 = (4.1)\times \omega_f

\omega_f = 585.37 \ rev/s

3 0
3 years ago
A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t
andrew11 [14]

Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = \frac{r}{2}

f = \frac{0.40}{2} = 0.20 m

Now,

m = - \frac{v}{u}

- 2 = -\frac{v}{u}

\frac{v}{u} = 2                  (2)

Now, by lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

v = \frac{uf}{u - f}            (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = \frac{uf}{u - f}

2 = \frac{f}{u - f}

2(u - 0.20) = 0.20

u = 0.30 m

6 0
3 years ago
the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg​
lukranit [14]

Answer:

Mass of the steel cube =  7800 kg

Volume of the steel  = 1.025 cubic centimetre

Explanation:

Given:

The density of the steel   =  7.8

Side of the cube = 12 cm

<u>(1)The mass of steel cube :</u>

We know that,

Density  = \frac{Mass}{Volume}

We are given with density and sides of the cube

then volume of the cube

= (side)^3

= 10^3

= 1000 cubic centimetre

Now

7.8 = \frac{mass}{1000}

mass = 7.8 \times 1000

mass =  7800 kg

<u>(2)volume of steel:</u>

Given the mass  = 8 kg

Density  = \frac{Mass}{Volume}

Substituting the values

7.8 = \frac{8}{volume}

volume = \frac{8}{7.8}

volume = 1.025 cubic centimetre

3 0
3 years ago
An RL circuit contains a resistor with R = 6800 Ω and an inductor with L = 2300 µH. If the impedance of this circuit is 160,000
Rainbow [258]

| Impedance | = √ [R² +(ωL)²]

R² = 6800² = 4.624 x 10⁷
 
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²

          = 2.0884 x 10⁻⁴  f²

| Z | =  √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ]  =  1.6 x 10⁵

     (1.6 x 10⁵)²  =  (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)

     (2.56 x 10¹⁰) - (4.624 x 10⁷)  =  2.0884 x 10⁻⁴ f²


Frequency² =   (2.56 x 10¹⁰ - 4.624 x 10⁷)  /  2.0884 x 10⁻⁴

                    =       2.555 x 10¹⁰ / 2.0884 x 10⁻⁴

                    =          1.224 x 10¹⁴ 

                    =          122,400 GHz          <== my calculation

                                      11.1 MHz           <== online impedance calculator

Obviously, I must have picked up some rounding errors
in the course of my calculation. 
  











7 0
3 years ago
Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine
TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because,  the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by tanθ = y/D

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ =  λD/d and y₋₁ =  -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ =  3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ =  λD/d and y₂ =  2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

7 0
3 years ago
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