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3 years ago

An adult with a BMI of 25 to 30 is considered

2 answers:

6 0

Well, the average adult (BMI) would be 18.5 to 24.9. So, I wouldn't really call tis (obese), but I would consider this (overweight).If the question is stating (BMI of 25 to 30), then this would be over than 25.0. Then is would be overweigth.

A. overweight.
B. underweight.
C. healthy.
D. obese.

soldier1979 [14.2K]3 years ago

4 0

Option A overweight

HOPE IT HELPS!!

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1 watt power is defined as the 1 Joule of energy consumed per second..

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If you enjoyed teaching small or large classes and being active every day which career would be a good fit for you

Vlad [161]

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3 years ago

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kati45 [8]

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3 years ago

what is the wavelength λ2λ2lambda 2 of the second laser that would place its second maximum at the same location as the fourth m

nordsb [41]

The distance separating two wave crests is known as the wavelength. The wavelength of the second laser, where its second maximum would be located, is 0.109375 mm.

Double slit interference (constructive) can be described by the equation dsinθ=m λ

Here, the order is (m), the wavelength is (), and the slit distance is (d).

Information provided-

The laser's wavelength is d/8.

The slit separation is 0.500 mm in length.

Put the distance into consideration to determine the wavelength values as,

λ1=0.5 x 10⁻³/8

λ1=0.0625 x 10⁻³

Due to the second laser's wavelength, its second maximum would be at the same spot as the first laser's fourth minimum.

Therefore, we must first determine the fourth order minima as,

dsinθ= (4-1/2) λ1

dsinθ= (7/2) λ1

Using the second maxima, dsinθ= 2λ₂

Put the following value for in the equation above:

7/2λ1 = 2λ₂

7/2 x (0.0625 x 10⁻³) = 2λ₂

λ₂=0.000109375m.

Therefore, the wavelength of the second laser, where its second maximum would be located, is 0.109375 mm.

Learn more about wavelength here-

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4 0

1 year ago

A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origi

Liula [17]

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

a) calculating work done by the force over the journey of the train

F = mx + b ------ ( 1 )

m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8 ----- ( 2 )

hence to determine the work done by the force

W = $\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx$ Note: the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

b) calculate the speed of the train at the end of its journey

we will apply the work energy theorem

W = 1/2 m*v^2 - 1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )

V^2 = 12.11

hence V = 3.48 m/s

6 0

3 years ago