Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

An object is moving with constant velocity. Which of the following statements is true? A) The net force on the object is zero. B) A constant force is being applied in the direction of motion. C) There are no forces acting on the object. D) There is no frictional force acting on the object.

OpenStudy (anonymous):

C) There are no forces acting on the object. That doesn't mean there aren't ANY forces acting on the object; it just means they are all canceling out. So there COULD be friction, but if there is, there's also some force that just matches the friction in the opposite direction.

OpenStudy (anonymous):

If an object is moving at a constant velocity, it must not be accelerating. If it is not accelerating, but is at a constant mass (which it is), that must mean that it has a net force of zero acting upon it. It is possible for there to be no forces on the object, but it is also possible for forces to be acting upon the object and just cancel out. I do agree with whitekitten_cat that any forces acting upon the object are canceling out, but since it is still possible for forces to be acting upon the object that cancel out, I think the answer is A) The net force on the object is zero.

OpenStudy (anonymous):

I had this question awhile ago but It is C. But good observations

OpenStudy (anonymous):

yeah i just typeed in a as my answer and the computer accepted it,so whitekitten is right

OpenStudy (anonymous):

thank you. anymore ill help out. hey you think you can answer mine

OpenStudy (anonymous):

what's urs

OpenStudy (anonymous):

Write an equation for the circle with center (-4, 4) and containing the point (-1, 16). Use the ^ key for the exponents. Write your answer as the example: (x+2)^2+(y-9)^2=100

OpenStudy (anonymous):

use the formula of the equation of a circle and solve,which is (x-h)^2 + (y-k)^2=r^2

OpenStudy (anonymous):

i tried that

OpenStudy (anonymous):

u substituted ur center into the equation?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

wat did u get as ur answer?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!