1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kenny6666 [7]
3 years ago
7

A wheel that was initially spinning is accelerated at a constant angular acceleration of 5.0 rad/s^2. After 8.0 s, the wheel is

found to have made an angular displacement of 400 radians. (a) How fast was the wheel spinning initially? (b) What is the final angular velocity of the wheel?
Physics
1 answer:
notka56 [123]3 years ago
8 0

Answer:

a)  Initial angular speed = 30 rad/s

b) Final angular speed = 70 rad/s        

Explanation:

a) We have equation of motion s = ut + 0.5at²

    Here s = 400 radians

              t = 8 s

              a = 5 rad/s²

    Substituting

             400 = u x 8 + 0.5 x 5 x 8²

              u = 30 rad/s

   Initial angular speed = 30 rad/s

b) We have equation of motion v = u + at

     Here u = 30 rad/s

               t = 8 s

              a = 5 rad/s²  

    Substituting

             v = 30 + 5 x 8 = 70 rad/s    

   Final angular speed = 70 rad/s        

You might be interested in
Summarize ocean acidification in one sentence.
Snowcat [4.5K]

Answer:

The ocean absorbs a significant portion of carbon dioxide (CO2) emissions from human activities, equivalent to about one-third of the total emissions for the past 200 years from fossil fuel combustion, cement production and land-use change (Sabine et al., 2004). Uptake of CO2 by the ocean benefits society by moderating the rate of climate change but also causes unprecedented changes to ocean chemistry, decreasing the pH of the water and leading to a suite of chemical changes collectively known as ocean acidification. Like climate change, ocean acidification is a growing global problem that will intensify with continued CO2 emissions and has the potential to change marine ecosystems and affect benefits to society.

The average pH of ocean surface waters has decreased by about 0.1 unit—from about 8.2 to 8.1—since the beginning of the industrial revolution, with model projections showing an additional 0.2-0.3 drop by the end of the century, even under optimistic scenarios (Caldeira and Wickett, 2005).1 Perhaps more important is that the rate of this change exceeds any known change in ocean chemistry for at least 800,000 years (Ridgewell and Zeebe, 2005). The major changes in ocean chemistry caused by increasing atmospheric CO2 are well understood and can be precisely calculated, despite some uncertainty resulting from biological feedback processes. However, the direct biological effects of ocean acidification are less certain

image

1 “Acidification” does not mean that the ocean has a pH below neutrality. The average pH of the ocean is still basic (8.1), but because the pH is decreasing, it is described as undergoing acidification.

Page 2

Suggested Citation:"Summary." National Research Council. 2010. Ocean Acidification: A National Strategy to Meet the Challenges of a Changing Ocean. Washington, DC: The National Academies Press. doi: 10.17226/12904. ×

Save

Cancel

and will vary among organisms, with some coping well and others not at all. The long-term consequences of ocean acidification for marine biota are unknown, but changes in many ecosystems and the services they provide to society appear likely based on current understanding (Raven et al., 2005).

In response to these concerns, Congress requested that the National Research Council conduct a study on ocean acidification in the Magnuson-Stevens Fishery Conservation and Management Reauthorization Act of 2006. The Committee on the Development of an Integrated Science Strategy for Ocean Acidification Monitoring, Research, and Impacts Assessment is charged with reviewing the current state of knowledge and identifying key gaps in information to help federal agencies develop a program to improve understanding and address the consequences of ocean acidification (see Box S.1 for full statement of task). Shortly after the study was underway, Congress passed another law—the Federal Ocean Acidification Research and Monitoring (FOARAM) Act of 2009—which calls for, among other things, the establishment of a federal ocean acidification program; this report is directed to the ongoing strategic planning process for such a program.

Although ocean acidification research is in its infancy, there is already growing evidence of changes in ocean chemistry and ensuing biological impacts. Time-series measurements and other field data have documented the decrease in ocean pH and other related changes in seawater chemistry (Dore et al., 2009). The absorption of anthropogenic CO2 by the oceans increases the concentration of hydrogen ions in seawater (quanti-

Explanation:

3 0
3 years ago
A 1250 kg car starts from rest and speeds up to 14.0 m/s in 7.25s. How much force acted on the car ?
densk [106]
M = 1250 kg
V = 14.0 m/s
t = 7.25 s
F = ?

Impulse equation
F . t = m . V

:. F = mv/t

F = (1250 kg x 14.0 m/s) / 7.25s

F = 2414 N
4 0
4 years ago
A driver sets out on a journey. for the first half of the distance she drives at the leisurely pace of 30 mi/h; during the secon
Gemiola [76]
The average speed of a moving object is the rate of change of a certain distance with respect with time. It is equal to the total distance that was traveled by the object over the total time it takes to travel that distance. For this problem we need to assume that the total distance that was traveled would be equal to 120 miles. So, for the first half of the distance or 60 miles at a speed of 30 miles per hour, the time taken would be two hours. For the remaining 60 miles at a speed of 60 miles per hour, 1 hour is total time traveled. So, we calculate the average speed as follows:

Average speed = total distance / total time
Average speed = 120 miles / 2 hr + 1 hr
Average speed = 40 mi / hr
5 0
4 years ago
If the magnitude of the electric field in air exceeds roughly 3 ✕ 106 N/C, the air breaks down and a spark forms. For a two-disk
Westkost [7]

Answer:

1.843 x 10^-5 C  

Explanation:

<u><em>Givens:   </em></u>

It is given that the air starts ionizing when the electric field in the air exceeds a magnitude of 3 x 10^6 N/C, which means that the max electric field can stand without forming a spark is 3 x 10^6 N/C.  

Also it is given that the radius of the disk is 50 cm, it is required to find out the max amount of charge that the disk can hold without forming spark, which means the charge that would produce the max magnitude of the electric field that air can stand without forming spark, and since we know that the electric field in between 2 disk "Capacitor" is given by the following equation  

E = (Q/A)/∈o                                (1)

Where,

Q: total charge on the disk.

A: the area of the disk.  

<u><em>Calculations:  </em></u>

We want to find the quantity of charge on the disk that would produce an electric field of 3 x 10^6 N/C, knowing the radius of the disk we can find the cross-section of the disk, thus substituting in equation (1) we find the maximum quantity of charge the disk can hold  

Q = EA∈o

   = (3 x 10^6) x (π*0.50) x (8.85 x 10^-12)  

  = 1.843 x 10^-5 C  

note:

calculations maybe wrong but method is correct

8 0
3 years ago
The voltage between the cathode and the screen of a television set is 22 kV. If we assume a speed of zero for an electron as it
il63 [147K]

Answer:

v= 8.8*10⁷ m/s

Explanation:

  • Assuming no friction present, the change in electrical potential energy, must be equal in magnitude, to the change in kinetic energy of the electron.
  • The change in the electrical potential energy, can be expressed as follows:

       \Delta U = (-e)*\Delta V

  • The change in kinetic energy, assuming that the electron started from rest, can be written as follows:

       \Delta K = \frac{1}{2} *m*v^{2}

       ⇒\Delta K = -\Delta U

  • From the equation above, replacing by ΔK and ΔU, we have:

       -\Delta U =- (-e)*\Delta V =\Delta K = \frac{1}{2} *m*v^{2}

  • Solving for v:

        v=\sqrt{\frac{2*e*\Delta V}{m_{e}} } =\sqrt{\frac{2*1.6e-19C*22e3V}{9.1e-31kg}} }\\ v= 8.8e7 m/s

4 0
3 years ago
Other questions:
  • A spring in a pogo-stick is compressed 12 cm when a 40. kg girl stands on the stick. what is the force constant for the pogo-sti
    12·1 answer
  • What force is used to make maglev trains run along their tracks?
    7·1 answer
  • Why is the main reason that attitudes are more often revealed in spoken rather than written language
    9·1 answer
  • A 3.0-kg cart rolls down a hill. Assume that the height of the cart changes from a maximum of 4.0 meters at the top of the hill
    8·1 answer
  • The strengths of the fields in the velocity selector of a Bainbridge mass spectrometer are B=0.500 T and E=1.2x105 V/m. The stre
    10·1 answer
  • Suppose that a star has a spectrum that includes red, blue, and violet lines spaced in the pattern of the lines from hydrogen bu
    13·1 answer
  • A 438 gram ball is traveling east at 39 m/s when it is hit by a 2.4 kg softball bat. After being in contact with the bat for 302
    11·1 answer
  • A 50kg car is moving at a velocity of 4 m/s. After a collision, the car is moving at a velocity of 5m/s.
    6·1 answer
  • a 250kg box is being push with a force of 100N to the left and 225n to the right determine the acceleration
    8·1 answer
  • If each wire carries a current of 67 a in the same direction, what is the magnitude of the magnetic force per meter (in nm) exer
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!