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irga5000 [103]
3 years ago
9

An adult with a BMI of 25 to 30 is considered

Physics
2 answers:
Romashka-Z-Leto [24]3 years ago
6 0
Well, the average adult (BMI) would be <span>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.

</span><span>A. overweight.
B. underweight.
C. healthy.
D. obese.</span>
soldier1979 [14.2K]3 years ago
4 0
Option A overweight

HOPE IT HELPS!!
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Consider an electric dipole in a uniform electric field. In which orientation does the dipole-field system have the greatest pot
Gekata [30.6K]

Answer: when the dipole moment and electric field are parallel

Explanation: The formulae that relates the potential energy of a dipole, dipole moment and strength of electric field is given as

u = p * E cosθ

Where, u= dipole potential energy, p = dipole moment, E = strength of electric field.

The expression is at maximum when θ = 0 (cos 0° = 1)

Hence the function for potential energy will be greatest when θ = 0° which implies that the dipole moment and strength of electric field are parallel to each other.

5 0
3 years ago
Isla made ice by putting water in the freezer. Freezing is an example of (2 points)
SSSSS [86.1K]

Answer:

a physical change

6 0
3 years ago
Read 2 more answers
What is the frequency of a wave if the speed is 24 m/s and the wave is 2 meters
vekshin1
Velocity=frequency(wavelength)
24m/s=f(2m)
24/2=f(2)/2
12Hz=f
4 0
3 years ago
A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horiz
Rufina [12.5K]

Answer:

0.88752 kgm²

0.02236 Nm

Explanation:

m = Mass of ball = 1.2 kg

L = Length of rod = 0.86 m

\theta = Angle = 90°

Rotational inertia is given by

I=mL^2\\\Rightarrow I=1.2\times 0.86^2\\\Rightarrow I=0.88752\ kgm^2

The rotational inertia is 0.88752 kgm²

Torque is given by

\tau=FLsin\theta\\\Rightarrow \tau=2.6\times 10^{-2}\times 0.86sin90\\\Rightarrow \tau=0.02236\ Nm

The torque is 0.02236 Nm

5 0
3 years ago
a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u
n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

    a_{c}  = \frac{v^{2} }{r}

   a_{c}  = 23.04/10.3

    a_{c}  = 2.23 m/s²

It continues to fly but now with some tangential acceleration

a_{t} = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

a_{net}  =  \sqrt{a_{c} ^{2} +a_{t} ^{2}   }

a_{net}  =  \sqrt{4.97 + 0.396}

a_{net}  =  2.316 m/s²

So the magnitude of  net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

#SPJ4

8 0
1 year ago
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