Enable contrast version

Tutor profile: Stacy H.

Inactive
Stacy H.
Experienced Physics and Math Tutor
Tutor Satisfaction Guarantee

Questions

Subject: Physics (Newtonian Mechanics)

TutorMe
Question:

A (very large) balloon is filled with a lighter-than-air gas. The total buoyancy force acting on the balloon is 3.02 newtons. The mass of the balloon is 0.1 kg. Is the balloon accelerating? If so, how quickly is the balloon accelerating, and is it accelerating up or down?

Inactive
Stacy H.
Answer:

Lets assume that the upward direction is positive, and the downward direction is negative. First we should draw a free body diagram. We would draw the balloon with an arrow pointing up representing the buoyancy force F(buoyancy) = 1.98N, then we would add an arrow pointing down representing the force due to gravity F(gravity) which is equal to mass times gravitational acceleration, so F(gravity) = m x a(gravity) = (0.1kg x 9.8m/s^2). Next, lets sum the forces in the up-down direction. The sum of the forces will tell us if the balloon has a net force acting on it. Net force in the up-down direction = Buoyancy force - force due to gravity F(net) = F(buoyancy) - F(gravity) F(net) = 1.98N - (0.1kg x 9.8m/s^2) F(net) = 1.98N - 0.98N F(net) = 1N Because the net force acting on the balloon is non-zero and positive, we know the balloon is accelerating in the upward direction. Now, let's use Newton's second law to find out how quickly the balloon is accelerating in the upward direction: F(net) = m x a a = F(net) / m a = 1N / 0.1kg a = 10m/s^2 So the balloon is accelerating upwards at 10m/s^2 (That's fast!) A note: Remember, one newton actually equals one kg*m/s^2, which is why the units work out okay at the end of this problem.

Subject: Physics

TutorMe
Question:

Two masses (m1 and m2) are moving toward each other at two velocities (v1 and v2 respectively). Mass 1 is moving horizontally from left to right, and mass 2 is moving horizontally from right to left. When they collide they stick together. In which direction and at what speed are they moving after they collide and stick together? m1 = 2kg v1 = 12m/s m2 = 5kg v2 = 2m/s

Inactive
Stacy H.
Answer:

Conservation of momentum! Momentum must be conserved through the collision, so the momentum of the final stuck-together mass must equal the sum of the momentums of the two original masses. p(final) = p1 + p2 momentum = mass x velocity p = M x V So: p(final) = p1 + p2 M(final) x V(final) = (m1 x v1) + (m2 x v2) (m1 + m2) x V(final) = (m1 x v1) + (m2 x v2) Now plug in values and solve for V(final) Lets assume that any motion left to right is represented as positive velocity, and any motion right to left is represented as negative velocity. (m1 + m2) x V(final) = m1 x v1 + m2 x v2 (2kg + 5kg) x V(final) = (2kg x 12m/s) + (5kg x -2m/s) 7kg x V(final) = 24kgm/s - 10kgm/s 7kg x V(final) = 14kgm/s V(final) = 2m/s Because the value of V(final) is positive, we know that the final combined mass is moving from left to right. So the final motion of the combined mass is 2 m/s horizontal from left to right.

Subject: Algebra

TutorMe
Question:

7x + 19 = -2x + 4 + 14x Solve for x

Inactive
Stacy H.
Answer:

First, group like terms on each side of the equation: 7x + 19 = -2x + 4 + 14x - 2x + 14x = 12x 7x + 19 = 12x + 4 Now gather like terms all to one side, by adding or subtracting equally from both sides: 7x + 19 = 12x + 4 -7x -7x 19 = 5x + 4 - 4 - 4 15 = 5x Now, isolate the variable in question (x) by dividing both sides of the equation by the same amount: 15 = 5x /5 /5 3 = x

Contact tutor

Send a message explaining your
needs and Stacy will reply soon.
Contact Stacy

Request lesson

Ready now? Request a lesson.
Start Lesson

FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage
Made in California by Zovio
© 2020 TutorMe, LLC
High Contrast Mode
On
Off