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djyliett [7]
3 years ago
14

A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 me

ters per second. What is the magnitude of the total momentum of the two balls after collision?
Physics
1 answer:
enyata [817]3 years ago
6 0

Answer:

We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.  

Momentum is a VECTOR quantity having both magnitude and direction.  The first ball has momentum P =m*v = 2*4 = 8 at 90degrees.  The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees.  They sum to zero when you perform vector addition.

Explanation:

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Maru [420]

Whatever the statement is, I know one thing for sure:  It's NOT included on the list of choices that you've provided.

6 0
3 years ago
A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in
german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

\Sigma F = -f = m\cdot a (Eq. 1)

Where:

f - Kinetic friction force, measured in newtons.

m - Mass of the box, measured in kilograms.

a - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight (W = m\cdot g) and uniform accelerated motion (v = v_{o}+a\cdot t), we expand the previous expression:

-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)

And the magnitude of the friction force exerted on the box is calculated by this formula:

f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right) (Eq. 1b)

Where:

W - Weight, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

t - Time, measured in seconds.

If we know that W = 52.4\,N, g = 9.807\,\frac{m}{s^{2}}, v_{o} = 1.37\,\frac{m}{s}, v = 0\,\frac{m}{s} and t = 2.8\,s, the magnitud of the kinetic friction force exerted on the box is:

f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s}  }{2.8\,s} \right)

f = 2.614\,N

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0
3 years ago
A 6,600 kg train car moving at +2.0 m/s bumps into and locks together with one of mass 5,400 kg moving at -3.0 m/s.
Katarina [22]
Option B would be right one
according to momentum conservation
6600*2 = 13200kgm/s
5400*3 = 16200kgm/s
16200-13200 = 3000
now 6600-5400 = 1200 kg
thus 3000/1200 = 2.5 v
5 0
3 years ago
The following diagram represents a cart with an initial velocity of 1.0 m/s sliding along a frictionless track from point A
ki77a [65]

Answer:

point b

Explanation:

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This is the si unit of current
skelet666 [1.2K]

Answer:

ampere

and it is denoted by A

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