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4 years ago

12

the system that functions in the storage of minerals, such as calcium, is called the ____ system a. skeletal b.respiratory c. ca

rdiovascular d. lymphatic

2 answers:

-BARSIC- [3]4 years ago

6 0

<span> c. cardiovascular system</span>

Lemur [1.5K]4 years ago

4 0

The correct option is SKELETAL SYSTEM.
Calcium is a mineral that is found in many food and the body needs it to maintain strong bones and teeth and to carry out some biochemical reactions in the body. Calcium in the body is stored in bones and teeth; calcium supports the structure and hardness of these body parts.

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A solenoid having an inductance of 5.41 μH is connected in series with a 0.949 kΩ resistor. (a) If a 16.0 V battery is connected

vekshin1

Answer:

(A) $9.14\times 10^{-9}sec$

(B) $6.20\times 10^{-3}A$

Explanation:

We have given inductance $L=5.41\mu H=5.41\times 10^{-6}H$

Resistance $R=0.949kohm=0.949\times 10^3ohm$

Time constant of RL circuit is equal to $\tau =\frac{L}{R}$

$\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec$

Battery voltage e = 16 volt

(a) It is given current becomes 79.9% of its final value

Current in RL circuit is given by

$i=i_0(1-e^{\frac{-t}{\tau }})$

According to question

$0.799i_0=i_0(1-e^{\frac{-t}{\tau }})$

$e^{\frac{-t}{\tau }}=0.201$

${\frac{-t}{\tau }}=ln0.201$

${\frac{-t}{5.7\times 10^{-9} }}=-1.6044$

$t=9.14\times 10^{-9}sec$

(b) Current at $t=\tau sec$

$i=i_0(1-e^{\frac{-t}{\tau }})$

$i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})$

$i=6.20\times 10^{-3}A$

3 0

3 years ago

A 2.2 kg ball is attached to a ceiling by a string. The distance from the ceiling to the center of the ball is 1.06 m, and the h

Veronika [31]

Answer:

$U=-22.85J$

Explanation:

The potential energy associated with a body (m) located at a given height (h) above a reference point is defined as:

$U=mgh$

At the reference point the potential energy is zero. In this case the body is below the reference point (the ceiling), therefore, the value of h is negative:

$U=2.2kg*9.8\frac{m}{s^2}*(-1.06m)\\U=-22.85J$

3 0

3 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.032.03 times a second. A tack is stuck in the ti

Luda [366]

Answer:

4.55 m/s

Explanation:

The frequency is defined as the number of rotations in one second

So, f = 2.03 Hz

r = 0.357 m

The tangential speed is

v = r ω = r x 2 x 3.14 x f = 0.357 x 2 x 3.14 x 2.03 = 4.55 m/s

6 0

4 years ago

A solid sphere is released from the top of a ramp that is at a height

Alexus [3.1K]

Hi there!

We can use the work-energy theorem and apply it to this situation.

At the top of the ramp, the ball only has gravitational potential energy, and at the bottom of the ramp, the ball has BOTH translational and rotational kinetic energy.

We must use the following equations:
$GPE = mgh \\KE_T = \frac{1}{2}mv^2\\\\KE_R = \frac{1}{2}I \omega^2$

m = mass of sphere (kg)
g = acceleration due to gravity (m/s²)

h = height of ramp (m)

v = final velocity (m/s)
I = Moment of Inertia (kgm²)

ω = angular velocity (rad/sec)

Since:
$E_i = E_f\\\\mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega ^2$

In order to make things easier, since the ball is not slipping, we can relate angular velocity to translational velocity:
$\omega = \frac{v}{r}$

Also, recall the equation for the moment of inertia for a solid sphere:
$I = \frac{2}{5}mr^2$

We can use these to simply our equation:
$KE_R = \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{5}mv^2$

Now, we can rewrite the equation and solve for 'v'.

$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2\\\\mgh = \frac{7}{10}mv^2\\\\v = \sqrt{\frac{10gh}{7}} = \sqrt{\frac{10(9.8)(2.2 - 1.87)}{7}} = 2.149\frac{m}{s}$

a)

We can begin by solving for the time taken for the ball to land on the ground. The ball only has a horizontal velocity, so this is essentially a free-fall situation. Use the rearranged kinematic equation:
$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(1.87)}{9.8}} = .6178 s$

Now, use the following equation to solve for horizontal distance given horizontal velocity and time:
$d_x = v_x t\\\\d_x = 2.149 * .6178 = \boxed{1.328 m}$

b)
We can use the previously-stated relationship between translational and angular velocity to solve for the angular velocity.

$\omega = \frac{v}{r}$
It is given that the diameter is 0.14 m, so the radius is 1/2th the diameter, or 0.07 m.

Solve for the angular velocity:
$\omega = \frac{2.149}{0.07} = 30.706 \frac{rad}{sec}}$

Using the above fall time and dimensional analysis to convert from rad/sec to revolutions, we can solve for the # of revolutions made by the ball:
$\frac{30.706rad}{sec} * .6178 sec * \frac{1 rev}{2\pi rad} = \boxed{3.019 rev}$

5 0

3 years ago

For general projectile motion, the horizontal component

Aleksandr [31]

Answer:

Zero.

Explanation:

In the general projectile motion there is only one component of acceleration which is in the vertical direction.There is no any acceleration in the horizontal direction.As we know that horizontal component of the velocity is remains constant in the projectile motion.

Therefore we cans say that in the projectile motion ,the acceleration in the horizontal direction is zero.

Therefore the answer will be zero.

4 0

4 years ago