g A ring and solid sphere are rolling without slipping so that both have a kinetic energy of 126 J. What is the rotation kinetic energy of the ring

Answer:

The answer is "[tex]5.06 \times 10^{-6} \ kg \ m^2[/tex]"

Explanation:

[tex]\to E_1=0..............(i)\\\\\to E_2= \frac{mV^2}{2} +\frac{Iw^2}{2} - mgh.............(ii)\\\\ \Delta E=0\\\\\to mgh= \frac{mV^2}{2} +\frac{Iw^2}{2} \\\\ \to 2 \ mgh= mV^2 +Iw^2\\\\ \to 2 \ mgh- mV^2 =Iw^2\\\\ \to m(2gh- V^2) =Iw^2\\\\ \to I= \frac{m(2gh- V^2)}{w^2}[/tex]

       [tex]= 5.06 \times 10^{-6} \ kg \ m^2[/tex]

Answer: The common difference between surface EMG and intramuscular EMG is that that former is non-invasive while the later is an invasive method

Explanation:

Electromyography (EMG) is used clinically for the examination of muscle excitations (muscle electrical activity) in both normal or abnormal conditions. There are two forms of EMG includes:

--> Surface EMT and

--> Intramuscular EMT

Surface EMT is a non invasive method of examination of muscle excitations for superficial and easily accessible muscles.

Intramuscular EMT is the invasive method of examination of muscle excitations usually for deep muscles.

The difference between the two forms of EMT includes:

- surface EMT is non- invasive while intramuscular EMT is invasive

- surface EMT is used to access superficial muscle while intramuscular EMT is used to access deep muscles.

- surface EMT requires less skill and time to carry out while intramuscular EMT requires special skills and takes more time while carrying out the procedure.

The difference between surface EMG and intramuscular EMG is that surface EMG is non-invasive while intramuscular EMG recording is an invasive method.

Electromyography is simply used for the examination of muscle excitations in both normal and abnormal conditions. It is the diagnostic procedure that is used in assessing the health of muscles.

It should be noted that surface EMT is a non-invasive method of examination of muscle excitations while intramuscular EMT is an invasive method of examination of muscle.

Also, the surface EMT is used to access superficial muscle while on the other hand, intramuscular EMT is used to access deep muscles.

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Answer:

density of the sediment, size of the sediment, and velocity of the agent of erosion.

Explanation:

Have a nice day!

Answer: b., c., e.

Explanation:

Answer:

Shape and position

Explanation:

Hope this helps! :)

Answer:

The value is  [tex]H = 0.036 \ kg/m^3[/tex]

Explanation:

From the question we are told that

   The saturation density of air is [tex]\rho = 0.04 \ kg/m^3[/tex]

   The temperature for this saturation density to occur  [tex]T_s = 96^o F[/tex]

    The relative humidity at this temperature is  [tex]R_h = 90\% = 0.90[/tex]

Generally the relative humidity is mathematically represented as

      [tex]R_H = \frac{H}{\rho}[/tex]

Here H represents humidity

=> [tex]0.90 = \frac{H}{0.04}[/tex]

=>  [tex]H = 0.90 * 0.04[/tex]

=>  [tex]H = 0.036 \ kg/m^3[/tex]

Answer:

.....

Explanation:

From the graph, it is clear that, the velocity is at a time of 1 s is highest. The velocity at 1 second corresponds to 1250 km/hr. Then it decreases with time.

The velocity - time graph shows the change in velocity with respect to time. The velocity is placed in y -axis and time is given in x - axis. The slope of the curve in velocity - time graph gives the acceleration of the object.

Similarly, the position of the object in meter after a t seconds can be determined from the velocity - time graph. It is the rate of change in velocity of the object.

From the graph, it is clear that, the curve has its peak at 1 second. After that the peak descends down. Hence, the maximum velocity of the car is at a time of 1 second at which the velocity is 1250 km/hr.

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Answer:

a) letting in-flowing and/or out-flowing tide flow through turbines in a dam.

Explanation:

Tidal power makes use of the energy from the tidal force and wave action in order to generate electricity. it is a predictable source of energy.

Tidal Barrages

The system allows tides to enter, seawater flows via the dam and is trapped into the basin when the tides subside and system’s gates close.

When the tides start to move out, the gates in the dam are opened up this consist of turbines and water begins to flow out, hitting the turbines this eventually turn to produce energy.

Power is produced when the tidal range, which is the difference between low and high tide, has to be more than 5 meters.

Answer:

a. Please see the attached graph and drawing combined created with Microsoft Excel

b. Approximately 3.2217 m/s

c. Approximately 14.0 m/s  

d. Approximately 11.05 meters

Explanation:

a. The given parameters are;

The height of the wall, h = 10.0 m

The distance from the base of the wall the car lands = 4.60 m

The time, t, it takes the car to land is given by the equation for free fall as follows;

h = 1/2·g·t²

Where;

g = The acceleration due to gravity = 9.81 m/s²

From the equation for free fall, we have;

h/(1/2·g) = t²

∴ t = √(h/(1/2·g) ) = √(10/(1/2·9.81) ) ≈ 1.4278

The time it takes the car to land, t ≈ 1.4278 seconds

b. The horizontal speed of the car = Horizontal distance/(Time) = 4.6/1.4278 ≈ 3.2217 m/s

The horizontal speed of the car before it drives off the wall  ≈ 3.2217 m/s

c. The car's vertical speed just before impact is given by the following equation;

v = u + gt

Where;

u = The initial vertical speed = 0

t = The time it takes before impact ≈ 1.4278 seconds

∴ v = 9.81 × 1.4278 ≈ 14.0 m/s  

The car's vertical speed just before impact, v ≈ 14.0 m/s  

d. Whereby the car is subject to the gravitational field of the moon, we have;

Gravitational force per kilogram = 1.7 N/kg

∴ Gravitational acceleration = 1.7 m/s².

The time it takes the car to land whilst subject to the gravitational field of the moon is therefore;

[tex]t_{Moon}[/tex] = √(h/(1/2·g)) = √(10/(1/2 × 1.7)) ≈ 3.43

[tex]t_{Moon}[/tex] ≈ 3.43 seconds

The horizontal distance covered, at the car's horizontal speed in the time of free fall ≈ 3.2217 m/s ×  3.43 seconds ≈ 11.05 meters

The horizontal distance covered, at the car's horizontal speed in the time of free fall  = The distance the car will land from the base of the wall ≈ 11.05 meters

The distance the car will land from the base of the wall ≈ 11.05 meters.

Answer:

The answer is D. In the direction of the wave

Explanation:

I took the test, hope this helps! :)

Answer:

Following are the solution to this question

Explanation:

please find the complete question in the attached file.

In point a:

The answer is "bottom".

In point b:

Using formula:

[tex]T= mg + \frac{m V^2}{2}[/tex]

[tex]\to 100= 0.2 \times 9.81 + \frac{0.2 \times V^2}{0.5}[/tex]

[tex]\to 100= 1.962+ \frac{0.2 + V^2}{0.5}\\\\\to 100- 1.962= \frac{0.2 + V^2}{0.5}\\\\\to 98.038= \frac{0.2 + V^2}{0.5}\\\\\to 49.019=0.2+V^2\\\\\to 48.819=V^2\\\\ \to 6.987 \ \frac{m}{s}[/tex]

Answer:

Pulleys, levers, inclined planes, wedges, screws, and wheel and axle.

Explanation:

Answer:

Students need a safe environment to take risks and struggle. It’s uncomfortable to struggle, but struggling—falling down and getting back up—is an important facet to learning. Productive struggle is not about being in pain or becoming frustrated.

Please mark branliest!

Explanation: it shows that your really trying your best and when you mess up its telling you that your actually learning from your mistakes.

Answer:

The same as

Explanation:

First of all, they are in a merry go round, which is a circular one. And thus, we would be dealing with circular motions.

The angular velocity of a merry go round, in this instance, is given by

w = Δ/Δt

The formula has nothing whatsoever to do with distance, and as such, both Rasheed and Sofia would have the same angular velocity.

Transferring this further, the angular acceleration is given as

α = Δw/Δt

Remember I said the velocity has nothing to do with distance, well, so does the acceleration as we can see from the formula stated. And therefore, both Rasheed and Sofia would have the same angular acceleration

Explanation:

doesn't get the question and the answer to

Answer:

(this is the sample answer:After the launch, the fireball moves upward, than it falls back down to EarthEartEarthEarthEarthEarth

Explanation:

Answer:

i didnt know ther was differnt types

Explanation:

Answer:

D

Explanation:

When light makes contact with any object it naturally slows down, and the more dense the object the slower it travels.

Answer:

B. soft tissue injuries

Explanation:

i took the test on edge and its really the only one that would make sense if you payed attention to the lesson  :)

Answer:

this is it

Explanation:

the control and experiemental groups i didnt know

Answer:

32.67km

Explanation:

~= round off

11.2km/h~3.11m/s

175minutes=10500s

10500*3.11~32, 666.67m

=32.66667km

~ 32.67km

Answer:

i think the answer is solar,

Explanation:

because it is a natural resource im sorry if im wrong

Answer:

This question is incomplete

Explanation:

This question is incomplete because there are some parameters that are missing. However, when calculating the average emf and speed/velocity is provided, then

average emf = Brvl

where Br is the vertical component of the earth's magnetic field (in weber per square meter, Wb/m²)

v is velocity (in m/s)

l is the length (in meter) of the wing span

Answer:

[tex]g_2=\dfrac{7}{8}g[/tex]

Explanation:

G = Gravitational constant

M = Mass of planet

R = Radius of planet

Acceleration due to gravity on first planet

[tex]g=\dfrac{GM}{R}[/tex]

Assuming that the planets have the same mass density [tex]\rho[/tex]

Density of first planet

[tex]\rho=\dfrac{M}{V}\\\Rightarrow \rho=\dfrac{M}{\dfrac{4}{3}\pi R^3}\\\Rightarrow M=\rho \dfrac{4}{3}\pi R^3[/tex]

Density of second planet

[tex]\rho=\dfrac{M_2}{V_2}=\dfrac{M_2}{\dfrac{4}{3}\pi R^3-\dfrac{4}{3}\pi (0.5R)^3}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(1-\dfrac{1}{2^3})}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(1-\dfrac{1}{8})}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(\dfrac{7}{8})}\\\Rightarrow M_2=\rho\dfrac{4}{3}\pi R^3(\dfrac{7}{8})\\\Rightarrow M_2=M\dfrac{7}{8}[/tex]

Acceleration due to gravity on the second planet

[tex]g_2=\dfrac{GM_2}{R}\\\Rightarrow g_2=\dfrac{GM\dfrac{7}{8}}{R}\\\Rightarrow g_2=\dfrac{7}{8}g[/tex]

The acceleration due to gravity of the planet would be [tex]\dfrac{7}{8}[/tex] times the acceleration due to gravity on the first planet.

Answer:

T = 12732 [N]

Explanation:

To solve this problem we must use newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

By means of the attached free body diagram, we can better understand the solution to this problem.

As we see in the diagram the tension Force T lifts the car upwards so the movement is upwards. Therefore we have two forces one upward (positive) of the force T and the other negative downward due to the weight of the vehicle.

As the movement is up the acceleration is also up (with a positive sign).

ΣF = m*a

where:

F = forces [N] (units of Newtons)

m = mass = 1200 [kg]

a = acceleration = 0.8 [m/s²]

T - 1200*9.81 = 1200*(0.8)

T = 12732 [N]

Answer:

12,720

Explanation:

Answer:

a. moving the magnet farther away and an iron object.

Explanation:

Answer:A

Explanation:

A p e x

Answer: C

Explanation:

The characteristics and problems of telescopes we can find the correct answer to the question that the Hubble telescope helps to answer is:

           D) Are there more galaxies beyond the milky way

Telescopes are instruments that can perceive very small amounts of light and have very large magnifications, in general to be able to see objects increasingly weak in intensity, telescopes become larger and the observation times of the object increase.

When we increase the observation times, the problem is that atmospheric movement changes the observation conditions and makes it difficult to obtain good images. One way to solve the problem was with telescopes in near space, the environmental disturbance is eliminated.

When the Hubble telescope was placed in space, there were already many answers but they continue to wonder what the limit of the universe is, this telescope allowed us to observe a body much farther away than ground-based telescopes, which is why it answered the question how many and to where there are galaxies out of the milky way.

Let's review the statements given to see the correct ones;

A) False. Copernicus clarified that the Sun is the center of the solar system

B) False. The center of the universe is not known, but the sun is a star in one of the arms of the milky way that is not the center of the universe either.

C) False. It has been known for a century what the movement of the planets in the solar system

D) True. The telescope in space allowed us to expand the limit of vision of the universe and respond to where there are galaxies outside the Milky Way.

In conclusion with the characteristics and problems of telescopes we can find the correct answer to the question that the Hubble telescope helped to answer is:

      D) Are there more galaxies beyond the milky way

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