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Natasha_Volkova [10]

4 years ago

13

Which statement best describes the motion of an object that is being subjected to two forces of equal magnitude in opposite dire

ctions?

1 answer:

Ludmilka [50]4 years ago

8 0

The net force on the object is zero, so it has no acceleration. It moves in a straight line and at a constant speed (which may be zero).

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The largest single publication in the world is the 1112-volume set of British Parliamentary Papers for 1968 through 1972. The co

Marat540 [252]

Answer:

$m_l=550\ kg$ is the mass of librarian.

Explanation:

Given:

mass of the system, $m_s=3.3\times 10^{3}\ kg$

velocity of librarian relative to the ground, $v_l=2.5\ m.s^{-1}$

velocity of the cart relative to the ground, $v_c=0.5\ m.s^{-1}$

N<u>ow using the principle of elastic collision:</u>

Net momentum of the system is zero.

$m_l\times v_l=(3300-m_l)\times v_c$

$m_l\times 2.5=(3300-m_l)\times 0.5$

$m_l=550\ kg$ is the mass of librarian.

6 0

3 years ago

Read 2 more answers

Why does a moving object come to a stop on a frictional surface?

ohaa [14]

It stops cause of gravitaional pull

3 0

4 years ago

Mark creates a graphic organizer to review his notes about electrical force. Which labels belong in the regions marked X and Y?

sveta [45]

Answer:

The correct answer is A

Explanation:

The question requires as well the attached image, so please see that below.

Coulomb's Law.

The electrical force can be understood by remembering Coulomb's Law, that describes the electrostatic force between two charged particles. If the particles have charges $q_1$ and $q_2$ , are separated by a distance r and are at rest relative to each other, then its electrostatic force magnitude on particle 1 due particle 2 is given by:

$|F|=k \cfrac{q_1 q_2}{r^2}$

Thus if we decrease the distance by half we have

$r_1 =\cfrac r2$

So we get

$|F|=k \cfrac{q_1 q_2}{r_1^2}$

Replacing we get

$|F|=k \cfrac{q_1 q_2}{(r/2)^2}\\|F|=k \cfrac{q_1 q_2}{r^2/4}$

We can then multiply both numerator and denominator by 4 to get

$|F|=k \cfrac{4q_1 q_2}{r^2}$

So we have

$|F|=4 \left(k \cfrac{q_1 q_2}{r^2}\right)$

Thus if we decrease the distance by half we get four times the force.

Then we can replace the second condition

$q_{2new} =2q_2$

So we get

$|F|=k \cfrac{q_1 q_{2new}}{r_1^2}$

which give us

$|F|=k \cfrac{q_1 2q_2}{r_1^2}\\|F|=2\left(k \cfrac{q_1 q_2}{r_1^2}\right)$

Thus doubling one of the charges doubles the force.

So the answer is A.

8 0

3 years ago

Read 2 more answers

If velocity is positive, which woul

shusha [124]

Answer:

C. An inital volocity that is faster than the final volocity

Explanation:

.

5 0

3 years ago

The stoplight had just changed and a 2200 kg cadillac had entered the intersection, heading north at 2.8 m/s , when it was struc

tekilochka [14]

<span>3. The attempt at a solution So basically what I did was divided into components. x: (3)(2000) = (3000)*v_x y: (v_vw)*(10000) = (3000)*v_y v_x, v_y is the velocity (after collision) in the x and y direction, respectively, of both cars stuck together (since it is an inelastic collision). v_vw is the initial velocity of the Volkswagen. Now what I did was that the angle is 35 degrees north of east. So basically made a triangle and figured that tan(35) = (v_y)/(v_x). This means (v_x)*(tan35) = v_y. Then, I simplified the component equations to get: x: 2 = v_x y: v_vw = 3*v_y Then plugging in for v_y, I got: v_vw = 3(2)(tan35) = 4.2 m/s as the velocity of the volkswagen. However, the answer key says 8.6 m/s. Could someone please help me out? Thanks Phys.org - latest science and technology news stories on Phys.org â€˘ Game over? Computer beats human champ in ancient Chinese game â€˘ Simplifying solar cells with a new mix of materials â€˘ Imaged 'jets' reveal cerium's post-shock inner strength Oct 24, 2012 #2 ehild Homework Helper Gold Member What directions you call x and y? Reference https://www.physicsforums.com/threads/2d-momentum-problem.646613/</span>

7 0

3 years ago