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DedPeter [7]
3 years ago
7

Solve for the BMI weight 58kg Height 1.61​

Physics
1 answer:
vaieri [72.5K]3 years ago
7 0

Answer:

Explanation:

BMI= weight/(height × height)          ; weight in kilogram and height in metter

     = 58kg / (1.61m  × 1.61m )

     = (58/ 2.5921) kg/m^{2}

     = 22.375  kg/m^{2}

     ≈ 22.4 kg/m^{2}

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If the distance to a point source of sound is doubled, by what multiplicative factor does the intensity change?
alina1380 [7]

If the distance to a point source of sound is doubled, by a multiplicative factor of 4, the intensity changes.

Intensity of sound is the sound which is perpendicular to sound wave propogation per unit area. It is dependent on the Surface of source sound.

Intensity is the Power per unit area. Its SI unit is Watt/m².

As we move away from a source of sound, the sound starts to diminish. This is due to the decreasing sound intensity with distance.

It can also be understood by the fact that on increasing distance, the Power radiated by the source spreads over a larger area. Hence, the Intensity decreases gradually.

Since, Intensity is proportional to the square of the distance.

Hence, on doubling the distance, Intensity reduces to one fourth of the initial intensity or reduces by a multiplicative factor of 4.

Learn more about Intensity here, brainly.com/question/17583145

#SPJ4

8 0
2 years ago
Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted
galina1969 [7]

Answer:

1) When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}

Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}

Where:

f = Focal length of the mirror = R/2

d_{i} = Image distance from the mirror

d_{o} = Object distance from the mirror

h_{i} = Image height

h_{o} = Object height

d_{o} is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

d_{i} is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0
3 years ago
What is the average velocity of atoms in 2.00 mol of neon (a monatomic gas)
Dimas [21]

Answer:

v = 876 m/s

Explanation:

It is given that,

Number of mol of Neon is 2 mol

Temperature, T = 308 K

Mass, m = 0.02 kg

Value of R - 8.31 J/mol-K

We need to find the average velocity of atoms in 2.00 mol of neon. Neon is a monoatomic gas. Let v is the velocity. So,

\dfrac{1}{2}mv^2=\dfrac{3}{2}nRT\\\\v=\sqrt{\dfrac{3nRT}{m}} \\\\v=\sqrt{\dfrac{3\times 2\times 8.314\times 308}{0.02}} \\\\v=876.47\ m/s

So, the correct option is (B).

8 0
3 years ago
Read 2 more answers
Evaluate the gravitational potential energy between two 5.00-kg spherical steel balls separated by a center-tocenter distance of
navik [9.2K]

Answer:

U = 8.30×10-⁹J

Explanation:

m1 = m2 = 5.00kg masses of the spheres

d = 15.0cm = 15×10-²m

r = 5.10cm = 5.10×10-²m

R = d + r = 15×10-² + 5.10×10-²

R = 20.10 ×10-²m = 0.201m

G = 6.67×10-¹¹Nm²/kg²

U = Gm1×m2/R = potential energybetween the spheres

U = 6.67×10-¹¹×5.00×5.00/0.201

U = 8.30×10-⁹J

7 0
3 years ago
Calculate the pressure on the ground from an 80 kg woman leaning on the back of one of her shoes with a 1cm diameter heel, and c
rodikova [14]

Answer:

Pressure of woman will be 99.87\times 10^5N/m^2

Pressure of the elephant will be 1716560.50N/m^2

Explanation:

We have given that mass of the woman m = 80 kg

Acceleration due to gravity g=9.8m/sec^2

Diameter of shoes = 1 cm =0.01 m

So radius r=\frac{d}{2}=\frac{0.01}{2}=0.005m

So area A=\pi r^2=3.14\times 0.005^2=7.85\times 10^{-5}m^2

We know that force is given  F = mg

So F=80\times 9.8=784N

Now we know that pressure is given by P=\frac{F}{A}=\frac{784}{7.85\times 10^{-5}}=99.87\times 10^5N/m^2

Now mass of elephant m = 5500 kg

So force of elephant = 5500×9.8 = 53900 N

Diameter = 20 cm

So radius r = 10 cm

So area will be A=3.14\times 0.1^2=0.0314m^2

So pressure will be P=\frac{53900}{0.0314}=1716560.50N/m^2

3 0
3 years ago
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