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Mariana [72]
3 years ago
9

HELP ME ASAP!!!!! This is 7th grade science! Will give brainliest, five starts, and a heart!

Physics
1 answer:
stira [4]3 years ago
6 0

Answer:

A- Carries urine to the bladder

B- excretes urine out of the body

C- Stores urine temporarily

D- Produces urine

You might be interested in
Write the formula of mechanical advantage​
Zanzabum

Answer:

the formula of mechanical advantage is

MA = load / effort

VR = effort distance / load distance

hope it is helpful to you

7 0
3 years ago
Please help me
qaws [65]

-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.

-- We know that the y-component of velocity is the derivative of the
y-component of position.

-- We're given the y-component of position as a function of time.

So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.

Now, the position function may look big and ugly in the picture.  But with the
exception of  't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation.  The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.

From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶

First derivative . . . y' (t) = (a₀ - g) t  -  6 (a₀ / 30t₀⁴ ) t⁵  =  (a₀ - g) t  -  (a₀ / 5t₀⁴ ) t⁵

There's your velocity . . . /\ .

Second derivative . . . y'' (t) = (a₀ - g) -  5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) -  (a₀ /t₀⁴ ) t⁴

and there's your acceleration . . . /\ .
That's the one you're supposed to graph.

a₀ is the acceleration due to the model rocket engine thrust
     combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀  is how long the model rocket engine burns

Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.

The big name in model rocketry is Estes.  Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.


6 0
3 years ago
A computer hard disk starts from rest, then speeds up with an angular acceleration of 190 rad/s2rad/s2 until it reaches its fina
Otrada [13]

Answer:

962 rpm.

Explanation:

given,

angular acceleration = 190 rad/s²

initial angular speed = 0 rad/s

final angular speed = 7200 rpm

                                 =7200\times\dfrac{2\pi}{60}

                                 =754\ rad/s

we need to calculate the revolution of disk after 10 s.

time taken to reach the final angular velocity

    using equation of angular motion

 \omega_f - \omega_i = \alpha t

 754 - 0 =190\times t

    t = 4 s

rotation of wheel in 4 s

\theta =\omega_i t+  \dfrac{1}{2}\alpha t

\theta = \dfrac{1}{2}\alpha t^2

\theta = \dfrac{1}{2}\times 190 \times 4^2

 θ = 1520 rad

 \theta = \dfrac{1520}{2\pi}

 \theta =242\ rev

now, revolution of the disk in next 6 s

angular velocity is constant

\omega_f = \dfrac{\theta_f-\theta_i}{t_f-t_i}

754 = \dfrac{\theta_f-1520}{10-4}

θ_f = 6044 rad

θ_f = \dfrac{6044}{2\pi}

revolution of the computer hard disk

θ_f =  962 rpm.

total revolution of the computer disk after 10 s is equal to 962 rpm.

3 0
3 years ago
A 41.0 kg child swings in a swing supported by two chains, each 2.98 m long. (a) If the tension in each chain at the lowest poin
Alex777 [14]

Answer:695.5 N

Explanation:

mass of child m=41 kg

Length of chain L=2.98 m

Tension in each chain T=348 N

(a)Tension at bottom point T=348 N

At lowest Point

T+T-mg=\frac{mv^2}{L}

2T-mg=\frac{mv^2}{L}

2\times 348-41\times 9.8=\frac{41\times v^2}{2.98}

v^2=\frac{294.2\times 2.98}{41}

v=\sqrt{21.38}=4.62 m/s

(b)Force exerted by Seat will be Equal to Normal reaction

N-mg=\frac{mv^2}{L}

N=mg+\frac{mv^2}{L}

N=41\times 9.8+\frac{41\times 21.38}{2.98}

N=695.95 N

8 0
3 years ago
In uniform circular motion, how does the acceleration change when the speed is increased by a factor of 3? When the radius is de
crimeas [40]

Answer:

The acceleration will become 9/2 times.

a' =9/2 a

Explanation:

We know that acceleration of a particle when it is moving in the circular path is given as

a=\omega^2\ r

r=radius

ω= angular speed

If the speed ω '= 3 ω

If the radius ,r'=\dfrac{r}{2}

The final acceleration =a'

a'=\omega^2'\ r'

a'=(3\omega)^2\times \dfrac{r}{2}

a'=9\omega^2\times \dfrac{r}{2}

a'=\omega^2\times \dfrac{9r}{2}

a'= \dfrac{9r}{2}\times \omega^2\times r

a'=\dfrac{9}{2}a

Therefore the acceleration will become 9/2 times.

5 0
3 years ago
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