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Bezzdna [24]
4 years ago
12

Which BMI standard for teens is considered under weight?

Physics
2 answers:
Leviafan [203]4 years ago
5 0

A: less than 5th percentile

Is the right answer...Merry Christmas!!!!!!!

Lena [83]4 years ago
3 0
First let’s pick one at random

Less than 5 percentile

That means they lined 100 kids up and less than 5% of them are that certain weight.

So if you apply this to every option you will see the answer is
A). Less than 5 percentile
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A certain string can withstand a maximum tension of 39. N without breaking. A child ties a 0.43 kg stone to one end and, holding
frez [133]

Answer:

Bottom of the circle.

Explanation:

At the top of the circle the tension and the weight contribute on being the centripetal force, at the middle of the circle only the tension contributes on being the centripetal force (the weight being perpendicular to it), while <u>at the bottom</u> of the circle the tension contributes on being the centripetal force (as always) <em>but the weight against to it</em>, so here is where the tension must be greater to allow the same centripetal force as the other cases, thus here is where the string will break.

6 0
3 years ago
Which statement describes what most likely occurs when a compass is placed next to a simple circuit made from a battery, a light
aleksley [76]
D. A magnetic field created by the electric current causes the compass needle to move.
6 0
3 years ago
Read 2 more answers
Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a
____ [38]

Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, v=2.79\times 10^8\ m/s

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s

Time period of oscillation measured by the observer is :

t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s

So, the time period of oscillation measured by the observer is 5.79 seconds.

3 0
3 years ago
which has a higher acceleration:a 10kg object acted upon with a net force of 20N or an 18kg object acted on by a net force of 20
MA_775_DIABLO [31]
<span>Answer: The acceleration of 10 kg object is greater than that of 18 kg object.

Explanation:
According to Newton's Second law:
F = ma --- (A)

Let's find the acceleration for both 10 kg and 18 kg objects!
The net force on both of these masses = F = 20N

(1) Acceleration of 10 kg object
Mass = m = 10 kg
Plug in the values in equation (A):
20 = 10 * a
Acceleration = a = 2 m/s^2

(2) Acceleration of 18 kg object
Mass = m = 18 kg
Plug in the values in equation (A):

20 = 18 * a
Acceleration = a = 1.11 m/s^2


2 > 1.11; therefore, 10 kg object has the higher acceleration compared to the acceleration of the 18 kg object.</span>
7 0
4 years ago
Read 2 more answers
What force must act on a 50.0-kg mass to give it an ancceleration of 0.30 m/s^2?
ratelena [41]

Answer:

15.0 N

Explanation:

see pic

3 0
3 years ago
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