According to the Periodic Table of Elements, Calcium (Ca) has:

Inessa05 [86]

Savings account is where she can have access to her money anytime in case of an emergency.

<h3>What is a Savings account?</h3>

This is a type of account run by a financial institution which yields interests on deposits made.

This type of account usually comes with features in which the money can be withdrawn via an ATM card or by visiting the bank to collect it. This therefore makes it the most appropriate choice for Analisa.

Read more about Savings account here brainly.com/question/25787382

7 0

3 years ago

A shipping pallet holds 10 boxes. Each box holds 300 parts of differenttypes. The part weight is normally distributed with a mea

Nat2105 [25]

Answer:

Explanation:

300

5 0

4 years ago

A CD has a mass of 17 g and a radius of 6.0 cm. When inserted into a player, the CD starts from rest and accelerates to an angul

Stells [14]

Answer:

τ =9.41 * 10⁻⁴ N*m

Explanation:

Kinematics of the CD

The CD rotates with constant angular acceleration and its angular acceleration is calculated as follows:

$\alpha = \frac{\omega_{f}- \omega_{i}}{t}$ Formula (1)

Where:

α : angular acceleration. (rad/s²)

ωf: final angular velocity (rad/s)

ωi : initial angular velocity (rad/s)

t = time interval (s)

Data

ωf= 20 rad/s

ωi =0

t = 0.65 s.

Calculating of the angular acceleration of the CD

We replace data in the formula (1)

$\alpha = \frac{20 -0}{0.65}$

α = 30.77 rad/s²

Newton's second law in rotation:

F = ma has the equivalent for rotation:

τ = I * α Formula (2)

where:

τ : It is the net torque applied to the body. (N*m)

I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Calculating of the moment of inertia of the CD

The moment of inertia of a disk with respect to an axis perpendicular to the plane and passing through its center is calculated by the following formula:

I = (1/2) M*R² Formula (3)

Data

M= 17 g = 17/1000 kg = 0.017 kg : CD mass

R= 6.0 cm= 6/100 m = 0.06 m : CD radius

We replace data in the formula (3) :

I = (1/2) ( 0.017 kg)*(0.06 m)² = 3.06 * 10⁻⁵ kg*m²

Calculating of the net torque acting on the CD

Data

α = 30.77 rad/s²

I = 3.06 * 10⁻⁵ kg*m²

We replace data in the formula (2) :

τ = I * α

τ =( 3.06 * 10⁻⁵ kg*m²) * (30.77 rad/s²)

τ =9.41 * 10⁻⁴ N*m

3 0

3 years ago

Help!!

mezya [45]

Answer:

Oil Spills.

Explanation:

3 0

3 years ago

Read 2 more answers

A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk

Murrr4er [49]

Answer:

$\omega_2=5.1rad/s$

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is $L=I\omega$ . If we call 1 the situation when the student is at the rim and 2 the situation when the student is at $r_2=1.39m$ from the center, then we have:

$L_1=L_2$

Or:

$I_1\omega_1=I_2\omega_2$

And we want to calculate:

$\omega_2=\frac{I_1\omega_1}{I_2}$

The total moment of inertia will be the sum of the moment of intertia of the disk of mass $m_D=119.1 kg$ and radius $r_D=3.23m$ , which is $I_D=\frac{m_Dr_D^2}{2}$ , and the moment of intertia of the student of mass $m_S=54.3kg$ at position $r$ (which will be $r_1=r=3.23m$ or $r_2=1.39m$ ) will be $I_{S}=m_Sr_S^2$ , so we will have: