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3 years ago

5

What is the difference between a license and a certificate in the healthcare system?

1 answer:

Tems11 [23]3 years ago

7 0

Answer:

A

Explanation:

You might be interested in

How many units measure one wavelength?<br><br><br> a<br> 16<br> b<br> 8<br> c<br> 2<br> d<br> 4

pochemuha

B it makes more sense

8 0

4 years ago

An element had 82 protons, 83 neutrons, and 82 electrons. What is the symbol of the element

serg [7]

Name: Lead
Symbol: Pb
Atomic Number: 82
Atomic Mass: 207.2 amu
Melting Point: 327.5 °C (600.65 K, 621.5 °F)
Boiling Point: 1740.0 °C (2013.15 K, 3164.0 °F)
Number of Protons/Electrons: 82
Number of Neutrons: 125
Classification: Other Metals
Crystal Structure: Cubic
Density @ 293 K: 11.34 g/cm3
Color: bluish

3 0

3 years ago

A spring has a length of .200 m when a .300 kg hangs from it,and a length of .750 m when a 1.95 kg hangs from it.what is the for

musickatia [10]

1) 29.5 N/m

2) 0.100 m

Explanation:

1)

The force constant of the spring can be found by using the fact that the force on the spring is proportional to the extension of the spring (Hooke's Law). Therefore, we can write:

$\Delta F= k \Delta x$

where

$\Delta F = F_2 - F_1$ is the change in the force on the spring, where

$F_1 = m_1 g = (0.300)(9.8)=2.94 N$ is the force applied when the hanging mass is

$m_1 = 0.300 kg$

$F_2 = m_2 g = (1.95)(9.8)=19.1 N$ is the force applied when the hanging mass is

$m_2 = 1.95 kg$

$\Delta x=x_2 -x_1$ is the change in extension of the spring, where

$x_1=0.200 m$ is the extension of the spring when the hanging mass is 0.300 kg

$x_2=0.750 m$ is the extension of the spring when the hanging mass is 1.95 kg

Solving for k,

$k=\frac{F_2-F_1}{x_2-x_1}=\frac{19.1-2.94}{0.750-0.200}=29.5 N/m$

2)

When the first mass is hanging on the spring, we have

$F_1 = k (x_1 - x_0)$

where:

$F_1$ is the force applied on the spring (the weight of the hanging mass)

k is the spring constant

$x_1$ is the extension of the spring wrt its natural length

$x_0$ is the natural length of the spring (the unloaded length)

Here we have

$F_1=2.94 N$

k = 29.5 N/m

$x_1=0.200 m$

Solving for $x_0$ , we find:

$x_0 = x_1 - \frac{F_1}{k}=0.200 - \frac{2.94}{29.5}=0.100 m$

5 0

3 years ago

Sam and George are riding in separate cars on the freeway. The lanes they are driving in are adjacent. We choose our coordinate

butalik [34]

Answer:

(a) Relative speed of Sam = -7 mph

(b) Relative speed of George = 7 mph

Explanation:

We have given speed of Sam 65 mph in north direction.

And speed of George is 72 mph also in north direction

We have find the relative velocity of both Sam and George

(a) Speed of Sam relative to George = speed of Sam - speed of George = 65 -72 = -7 mph

(b) Speed of George with respect to Sam = speed of George - speed of Sam = 72-65 = 7 mph

3 0

3 years ago

A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc

Elanso [62]

Answer:

λ = 470.66 nm

Explanation:

for bright fringe $y_m = \frac{m\lambda D}{d}$

D= distance between slit and screen

d= distance between the slits

for first order bright fringe m = 1,

$y_1 = \frac{1\lambda D}{d}$

$y_1 = {706*D}{d}$

for dark fringe,we have

$y_m = {(m + 1/2)\lambda D}{d}$

Now to get the dark fringes at the same location we should have;

(706)D/d = (m + 1/2)λD/d

put m = 1

(1 + 1/2)λ = (706)

λ = 470.66 nm

6 0

4 years ago