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3 years ago

Which BMI standard for teens is considered under weight?

Physics

2 answers:

Leviafan [203]3 years ago

5 0

A: less than 5th percentile

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

Lena [83]3 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 ball is thrown upward from an initial height of 1.5m the ball reaches a height of 5m then falls to the ground . What Is the di

jeka94

Answer:

The distance traveled by the ball is 8.5 m

Explanation:

Initial height of the ball, h₁ = 1.5 m above the ground

final height of the ball, h₂ = 5m

Upward distance = distance traveled by the ball from a height of 1.5m to 5m = 5m - 1.5m = 3.5 m

Downward distance = distance traveled by the ball from 5m height to the ground =5m - 0 = 5m

Total distance traveled = upward distance + downward distance

Total distance traveled = 3.5 m + 5m = 8.5 m

Therefore, the distance traveled by the ball is 8.5 m

4 0

3 years ago

Mercury was named after the roman god of speed why is it an appropriate name for the planet

elena-14-01-66 [18.8K]

Because it's the planet in our solar system with the shortest,
fastest orbit around the sun ... only 88 Earth days.

The people who named it didn't know that ... they still thought that
the sun and all the planets revolve around the Earth. But they did
see it zip from one side of the sun to the other, faster than any other
planet ... the result of having the shortest, fastest orbit of any planet.

5 0

3 years ago

A sprinter accelerates from rest to 10.0 m/s in 1.28 s . Part A Part complete What is her acceleration in m/s2? a a = 7.81 m/s2

Mashutka [201]

Explanation:

It is given that,

Initial speed of sprinter, u = 0

Final speed of sprinter, v = 10 m/s

Time taken, t = 1.28 s

a. We need to find the acceleration of sprinter. It can be calculated using first equation of motion as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{10\ m/s}{1.28\ s}$

$a=7.81\ m/s^2$

b. Final speed of the sprinter, v = 36 km/h

Time, t = 0.000355 h

Acceleration, $a=\dfrac{36}{0.000355}$

$a=101408.45\ km/h^2$

Hence, this is the required solution.

3 0

3 years ago

3. A football is kicked with a speed of 35 m/s at an angle of 40°.

jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

$u_y = u sin \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_y = (35)(sin 40)=22.5 m/s$

b) 26.8 m/s

The initial horizontal velocity is given by:

$u_x = u cos \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_x = (35)(cos 40)=26.8 m/s$

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

$v_y = u_y + at$

where

$v_y$ is the vertical velocity at time t

$u_y = 22.5 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, $v_y =0$ ; substituting, we find the time t at which this happens:

$0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s$

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}gt^2$

where

s is the vertical displacement at time t

$u_y = 22.5 m/s$

$g=-9.8 m/s^2$

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

$s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m$

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

$t=2(2.30 s)=4.60 s$

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

$v_x = 26.8 m/s$

Therefore, the horizontal distance travelled during the whole motion is

$d=v_x t = (26.8)(4.60)=123.3 m$

So, the ball lands 123.3 m far from the initial point.

4 0

3 years ago

(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W

Kitty [74]

Answer:

The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

$D\sin\theta=n\lambda$

$D=\dfrac{n\lambda}{\sin\theta}$

Where, D = width of slit

$\lambda$ = wavelength

Put the value into the formula

$D=\dfrac{n\lambda}{\sin\theta}$

Here, $\sin\theta$ should be maximum.

So. maximum value of $\sin\theta$ is 1

Put the value into the formula

$D=\dfrac{1\times\lambda}{1}$

$D=\lambda$

(b). If the minimum number is 50

Then, the width is

$D=\dfrac{50\times\lambda}{1}$

$D=50\lambda$

(c). If the minimum number is 1000

Then, the width is

$D=\dfrac{1000\times\lambda}{1}$

$D=1000\lambda$

Hence, The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

4 0

3 years ago