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3 years ago

How are doctors both accurate and precise when preforming surgery or other things?

Physics

1 answer:

mr Goodwill [35]3 years ago

4 0

Answer:

learning about the different functions of the body and therefore they have enough practice in order to have a clean and safe operation

Explanation:

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CAN ANYONE HELP PLEASE I NEED THIS DONE BY TODAY!! I WILL MARK YOU BRIANLYIST.

bulgar [2K]

I have no idea my dude sorry

7 0

3 years ago

Imagine you re in an entrepreneur and have been provided a loan by a bank to set up a small electro plating unit. What would you

Lena [83]

Answer:

Electroplating involves deposition of a thin layer of metal onto the surface of a substance(compound) through electric current. This helps preserve the substance(compound) from rust and oxidative damage.

As an entrepreneur which has been provided a loan by a bank to set up a small electro plating unit. I would venture into electroplating electronic/electric appliances parts.This is because these types of appliances are in vogue now and used my millions of people for their day to day activities. This means that more people will come for various degrees of production and repairs which would include electroplating . This translates to more customers and more money for the repayment of the loan.

5 0

3 years ago

A deuteron particle consists of one proton and one neutron and has a mass of 3.34x10^-27 kg. A deuteron particle moving horizont

nikklg [1K]

_dThe radius of curvature of a subatomic particle under a magnetic field is given by the following formula:

$r=\frac{mv}{qB}$

Where:

$\begin{gathered} r=\text{ radius} \\ v=\text{ velocity} \\ q=\text{ charge} \\ B=\text{ magnetic field} \end{gathered}$

We can determine the quotient between the velocity and the charge of the deuteron particle from the formula. First, we divide both sides by the mass:

$\frac{r_d}{m_d}=\frac{v}{q_B_}$

Now, we multiply both sides by the magnetic field "B":

$\frac{Br_d}{m_d}=\frac{v}{q}$

Since the charge of the deuterion is the same as the charge of the proton and the velocity we are considering are the same this means that the quotient between velocity and charge is the same for both particles. Therefore, we can apply the formula for the radius again, this time for the proton:

$r_p=\frac{m_pv}{qB}$

And substitute the quotient between velocity and charge:

$r_p=\frac{m_p}{B}(\frac{Br_d}{m_d})$

Now, we cancel out the magnetic field:

$r_p=\frac{m_pr_d}{m_d}$

Now, we substitute the values:

$r_p=\frac{(1.67\times10^{-27}kg)(0.385m)}{(3.34\times10^{-27}kg)}$

Solving the operations:

$r_p=0.193m=19.3cm$

Therefore, the radius is 19.3 cm.

3 0

2 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

what is the mass of a cannon ball if a force 2500 N gives the cannon ball an acceleration of 200 m/s squared?

nignag [31]

Using Newton's second law of motion:

F=ma ; [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)

Given: Find: Formula: Solve for m:

F: 2500N mass:? F=ma Eq.1 m=F/a Eq. 2

a= 200m/s^2

Solution:

Using Eq.2

m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg

8 0

3 years ago