Determine the momentum of a system that consists of two objects. One object, m1, has a mass of 6 kg and a velocity of 13 m/s towards the east and a second object, m2, has a mass of 14 kg and a velocity of 7 m/s in that same direction.

Answers

Answer 1

Momentum = (mass) x (velocity)

Momentum m₁  =  (6 kg) x (13 m/s east)  =  78 kg-m/s east

Momentum m₂  =  (14 kg) x (7 m/s east)  =  98 kg-m/s east

Since they're both in exactly the same direction you can simply addum up to get their total momentum.  It's 176 kg-m/s east .

Answer 2

The  momentum of a system that consists of two objects should be 176 kg m/s on east hand side.

Calculation of the momentum:

Since we know that

Momentum = (mass) x (velocity)

So,

Momentum m₁  =  (6 kg) x (13 m/s east)  =  78 kg-m/s east

Momentum m₂  =  (14 kg) x (7 m/s east)  =  98 kg-m/s east

Since they are on same direction so the total momentum should be

= 78 + 98

= 176 kg-m/s east

hence, The  momentum of a system that consists of two objects should be 176 kg m/s on east hand side.

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Related Questions

Two identical conducting spheres are charged with a net charge of +5.0 q on the first sphere and a net charge of −8.0 q on the second sphere. The spheres are brought together, allowed to touch, and then separated. What is the net charge on each sphere now?

Answers

Answer:

The net charge on each sphere is -1.5 q

Explanation:

Conductors are materials that allow the electrons which are the carriers of the charges to move between them, and when two conductors come in contact, the available charge is shared by the two conductors and the resultant like charges will spread on the surface of the conductor due to the repellent effect between similar charges such that if the conductors are identical, the resultant charge becomes evenly shared by the conductors when they become separated again

The given parameters of the conducting spheres meant to touch are;

The net charge on the first sphere, Q₁ = +5.0 q

The net charge on the second sphere, Q₂ = -8.0 q

The net charge on each sphere after touching and then separated, 'Q', is given as follows;

[tex]Q = \dfrac{Q_1 + Q_2}{2}[/tex]

Therefore, by substituting the known values of the variables, we have;

[tex]Q = \dfrac{5 \ q+ (-8 \ q)}{2} = -\dfrac{3 \ q}{2} = -1.5 \ q[/tex]

The net charge on each sphere, Q = -1.5 q.

The net charge on each sphere after spheres are brought together, allowed to touch is -1.5 q

What is charge?

Conductors are materials that allow the electrons which are the carriers of the charges to move between them, and when two conductors come in contact, the available charge is shared by the two conductors and the resultant like charges will spread on the surface of the conductor due to the repellent effect between similar charges such that if the conductors are identical, the resultant charge becomes evenly shared by the conductors when they become separated again

The given parameters of the conducting spheres meant to touch are;

The net charge on the first sphere, Q₁ = +5.0 q

The net charge on the second sphere, Q₂ = -8.0 q

The net charge on each sphere after touching and then separated, 'Q', is given as follows;

[tex]Q=\dfrac{Q_1+Q_2}{2}[/tex]

Therefore, by substituting the known values of the variables, we have;

[tex]\dfrac{5q+(-8q)}{2}=-1.5q[/tex]

Hence the net charge on each sphere after spheres are brought together, allowed to touch is -1.5 q

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There is a current of 2.95 A in one of the copper wires for 60 seconds. Calculate the charge flow through the wire.

Answers

Answer:

177 coulomb

Explanation:

Given data

I=2.95A

T=60

The relationship between current I and quantity of charge Q.

Q=IT

Substitute

Q=2.95*60

Q=177 coulomb

Suman with a mass of 45kg climbs in 3m heigh ladder in 10 seconds. Calculate her power. Plz help me​

Answers

Answer:

work done=fdcosø

f=mg=450N

d=+3m

ø=180

450×3×cos(180°) power=work done/

time taken

1350×-1 p=135watts

John slides down a hill at a 4 m height initially at rest. If all energy is conserved and friction is excluded, what is John's velocity at the bottom of the ramp?

Answers

The velocity at the bottom : v = 8.85 m/s

Further explanation

Given

height of the hill = 4 m

Required

The velocity at the bottom

Solution

The law of conservation energy :

ME₁=ME₂

(PE+KE)₁ = (PE+KE)₂

initially at rest⇒vo=0⇒KE₁=0

At the bottom⇒h=0⇒PE₂=0

So the equation becomes :

PE₁=KE₂

mgh=1/2.mv²

gh = 1/2v²

[tex]\tt v=\sqrt{2gh}[/tex]

v = √2x9.8x4

v = 8.85 m/s

Hell it’s due soon??

Answers

Explanation:

don't know don't know don't know

Outward.................

Sometimes, a crackling sound is heard while taking off a sweater during winters. Explain

Answers

Answer:

it is somewhat frozen

Explanation:

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