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victus00 [196]
2 years ago
10

14. Which of the following is not an example of work being done?

Physics
1 answer:
Natali5045456 [20]2 years ago
7 0

Answer:

B. holding a coffee mug

Explanation:

Something must move a distance for work to be done.

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Need help ASAP
VARVARA [1.3K]

Answer:

the correct one is D,  

Ultraviolet, x-ray, gamma ray

Explanation:

Electromagnetism radiation are waves of energy that is expressed by the Planck relationship

          E = h f

where h is the plank constant and f the frequency of the radiation.

Also the speed of light is

          c = λ f

         

we substitute

          E = h c /λ

therefore to damage the cells of the body radiation of appreciable energy is needed

microwave radiation has an energy of 10⁻⁵ eV

infrared radiation                E = 10⁻² eV

visible radiation                   E = 1 to 3 eV

radiation Uv                         E = 3 to 6 eV

X-ray                                    E = 10 eV

 

gamma rays                         E = 10 5 eV

therefore we see that the high energy radiation is gamma rays, x-rays and ultraviolet light.

When checking the answers, the correct one is D

6 0
2 years ago
Unlike mechanical waves such as sound or earthquake waves, EM waves can travel through empty space. True or false
Alex777 [14]

Answer:

its true

Explanation:

5 0
2 years ago
The ball rolling along a straight and level path. The ball is rolling at a constant
Aleksandr-060686 [28]

Answer:6 joules

Explanation:

Mass(m)=3kg

Velocity(v)=2m/s

Kinetic energy=0.5 x m x v^2

Kinetic energy=0.5 x 3 x 2^2

Kinetic energy=0.5 x 3 x 2 x 2

Kinetic energy=6

6 0
2 years ago
In a typically constructed research study, the experimental group is selected:
blondinia [14]

Answer:

D. from a separate pool than is the control group.

Explanation:

in the picture the person answers is backwards but...

hope this helps have a nice day

7 0
1 year ago
6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres
sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

2ah=v_f^2-v_0^2

Where:

\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}

If we assume the initial velocity to be 0 we get:

2ah=v_f^2

The acceleration is the acceleration due to gravity:

2gh=v_f^2

Now, we take the square root to both sides:

\sqrt{2gh}=v_f

Now, we substitute the values:

\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f

solving the operations:

280\frac{m}{s}=v

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

F_d=\frac{1}{2}C\rho_{air}Av^2

Where:

\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

F_d=mg

Now, we determine the mass of the raindrop using the following formula:

m=\rho_{water}V

Where:

\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}

The volume is the volume of a sphere, therefore:

m=\rho_{water}(\frac{4}{3}\pi r^3)

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)

Solving the operations:

m=1.39\times10^{-5}kg

Now, we substitute the values in the formula for the drag force:

F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})

Solving the operations:

F_d=1.36\times10^{-4}N

Now, we substitute in the formula:

1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2

Now, we solve for the velocity:

\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2

Now, we substitute the values. We will use the area of a circle:

\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2

Substituting the radius:

\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2

Solving the operations:

70.67\frac{m^2}{s^2}=v^2

Now, we take the square root to both sides:

\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\  \\ 8.4\frac{m}{s}=v \\  \end{gathered}

Therefore, the velocity is 8.4 m/s

7 0
11 months ago
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