Answer:
α = -9.67 rad/s²
Explanation:
the magnitude of the angular acceleration of the wheel can be calculated using the expression below;
ω² = ω₀² + 2α(θ- θ₀)..............eqn(1)
ω² = ω₀² + 2αΔθ...............eqn(2)
Δθ= θ- θ₀= 15.6 rad
ω = 13.5 rad/s
ω₀ = 22.0 rad/s
where ω₀= initial angular velocity
ω= final angular velocity
We can now input all the parameters into the equation(2) above
(13.5)² = (22.0)² + 2αΔθ
2αΔθ= (13.5)² - (22.0)²
2αΔθ= 182.25 - 484
2αΔθ=301.75
But our Δθ= 15.6 rad
2α *15.6 rad = 301.75
2α=301.75/15.6
2α=19.343
α=9.67
the magnitude of the angular acceleration of the wheel is 9.67 rad/s²
HELP PLEASE!!!! :( Does anyone have Lab: Conservation of Linear Momentum for online Physics class?? WILL GIVE YOU 80 PTS
Answer:
this is it
Explanation:
the control and experiemental groups i didnt know
Which of the following best describes the relationship between the speed of light and the type of medium through which it travels?
A.
Light travels fastest in solids
B.
Light travels fastest in liquids
C.
The less dense the medium, the slower a light wave travel
D.
The more dense the medium, the slower a light wave will travel
Answer:
D
Explanation:
When light makes contact with any object it naturally slows down, and the more dense the object the slower it travels.
Which of the following is how the substance in a longitudinal wave flows?
A. Up and Down
B. At Random
C. Perpendicular to the direction of the wave
D. In the direction of the wave
Please help I need this turned in tonight.
Answer:
The answer is D. In the direction of the wave
Explanation:
I took the test, hope this helps! :)
At 96° F the saturation density of air is 0.04 kg/m3. If the weather report says the relative humidity is 90 % when the temperature is 96°F, the humidity is
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]
21.Which of the following contain the molecules with the highest average speed?
C
L 나
G G
ico
cube
at -10°C
water
at roon
temperature
wood
at room
temperature
steam
at 110°C
Need help ASAP please help
Answer:
.....
Explanation:
What are the 6 basic machines, and what are they and what can the be, and look like.
Answer:
Pulleys, levers, inclined planes, wedges, screws, and wheel and axle.
Explanation:
describe the motion of the fireball in the air
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:
Which factors affect the rate of deposition? Check all that apply.
color of the sediment
density of the sediment
size of the sediment
texture of the sediment
velocity of the agent of erosion
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:
What question did use of a telescope by hubble answer A) Is earth the center of the universe B) Is the sun the center of the universe C) How do the planets move in the solar system D) Are there more galaxies beyond the milky way
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
Learn more here: brainly.com/question/16319804
Consider the accuracy of recording muscle electrical activity using electrodes placed on the skin surface, compared to directly in the muscle. What are the differences between surface and intramuscular EMG recordings
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.
Learn more about EMG on:
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which property of potential energy distinguishes it from kinetic energy
Answer:
Shape and position
Explanation:
Hope this helps! :)
As the rate of radioactive decay becomes smaller half life become...
Which form of energy has the most potential globally
a) Solar energy
b) Bio energy
c) Wind energy
d) Geothermal energy
Answer:
i think the answer is solar,
Explanation:
because it is a natural resource im sorry if im wrong
A lead fishing weight of mass 0.20 kg is tied to a fishing line that is 0.50 m long. The weight is then whirled in a vertical circle. The finishing line will break if its tension exceeds 100.0 N. (a) If the weight is whirled at higher and higher speeds, at what point in the vertical circle will the string break (top, bottom, or random position)
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]
Maybe I’ll make one more question. What type of electricity is made when you spin something really fast?
Answer:
i didnt know ther was differnt types
Explanation:
You tie the loose end of a 0.1 kg yo-yo string to your finger and then release the yo-yo so that it spins down toward the ground (the yo-yo is released from rest and the end of the string tied to your finger remains motionless). After the yo-yo falls a distance of 0.9 m, it has a translational speed of 4 m/s and an angular speed of 180 rad/s. What is the moment of inertia of the yo-yo
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]
Harnessing tidal power would involve:_______.a) letting in-flowing and/or out-flowing tide flow through turbines in a dam.b) huge floats mechanically hooked to generators.c) electrodes in the water.d) bioluminescence of marine organisms.e) all of the above
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.
what can you infer from the statement. velocity is an object is zero
Why is struggling an important part of the process of learning?
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.
The graph below shows the velocity of a car as it attempts to set a speed
record.
Velocity vs. Time
1400
1300
1200
1100
1000
4 45
3
1 (s)
At what point is the car the fastest?
A. t = 1.0 s
B. t = 4.2 s
C. t = 3.0 s
D. t = 4.5 s
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.
What is velocity - time graph ?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.
Find more on velocity - time graph :
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a. A remote-control car with a constant velocity drives off the top of a wall that is 10.0 m high and lands 4.60 m from the base of the wall. Draw a diagram of the problem. Label the known and unknown quantities. b. What is the car's speed before it drives off the top of the wall? c. What is the car's vertical speed just before impact? d. How far from the base of the wall would the remote-control car have landed if it had been subject to the gravitational field on the Moon (1.7 N/kg) ?
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.
Bicyclist travels at an average velocity of 11.2 km/h [W]. How far will the bicyclist travel in 175 minutes?
Answer:
32.67km
Explanation:
~= round off
11.2km/h~3.11m/s
175minutes=10500s
10500*3.11~32, 666.67m
=32.66667km
~ 32.67km
What is the average emf between the ends of the wings of a plane flying at a speed of 500 km/hr when the vertical component?
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
If a second planet were of the same radius R and made of the same material but had a hollow center of radius 0.50 R , what would be the acceleration of gravity at its surface
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.
How much tension must a rope withstand if it is used to accelerate a 1200-kg car vertically upwards at 0.8m/s2?
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:
asheed and Sofia are riding a merry-go-round that is spinning steadily (uniform circular motion). Sofia is twice as far from the axis as is Rasheed. Sophia's acceleration is _________ that of Rasheed. *
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
Moderate injuries are generally __________.
A.
life-threatening
B.
soft tissue injuries
C.
bumps and bruises
D.
too serious for first aid
Please select the best answer from the choices provided.
A
B
C
D
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 :)
Which action would decrease the strength of attraction between a magnet and an iron object
A. Moving the magnet farther away from the iron object
B. Moving the iron object closer to the magnet
C. Decreasing the number of electrons on the magnet
D. Increasing the number of electrons on the magnet
Answer:
a. moving the magnet farther away and an iron object.
Explanation:
Answer:A
Explanation:
A p e x
What is the momentum of 100 kg running at 4m/s north?