An object is sitting on the floor. A 22.4 N force is pulling the object to the right and an 11 N force is pulling the object to the left. The gravitational force on the object is 70 N. What is the net force?

Answers

Answer 1

answer = 33.4 net force.


Related Questions

Ball X has a mass of 8kg and is moving toward ball Y (which is sitting still) at 2m/s. After they collide, ball X is
sitting still. How fast is ball Y moving after the collision if it has a mass of 4kg?

Answers

Answer:

v = 4 m/s

Explanation:

Given :

Ball X :

Mass, m1 = 8kg ;

Initial Velocity, u1 = 2 m/s

Final velocity, v1 = 0

Ball Y:

Mass, m2 = 4kg ;

Initial Velocity, u2 = 0 m/s

Final velocity, v2 = v

(m1u1 + m2u2) = (m2v2 + m1v1)

(8*2 + 4*0) = (4*v + 4*0)

16 + 0 = 4v + 0

16 = 4v

v = 16 / 4

v = 4 m/s

Answer:

Explanation:

This is the Law of Momentum Conservation which for us looks like this:

[tex][m_xv_x+m_yv_y]_b=[m_xv_x+m_yv_y]_a[/tex] and that should look familiar to you if this is what you are doing in physics. Filling in our particular info:

[(8.0 × 2.0)+ (4.0 × 0.0)] = [(8.0 × 0.0) + (4.0v)] and

16 + 0 = 0 + 4.0v and

16 = 4.0v so

v = 4.0 in the direction of ball X

with respect to air the refractive index of ice is 1.31 and that of rock salt is 1.54 calculate the refractive index of rock salt with respect to ice​

Answers

Answer:

1.17

Explanation:

Given that,

The refractive index of ice wrt air = 1.31

The refractive index of rock salt wrt air = 1.54

We need to find the refractive index of rock salt with respect to ice.

We know that,

refractive index = (speed of light in air or vaccum)/( speed of light in that medium)

So,

The speed of light in ice = c/(1.31)

The speed of light in rock salt = c/(1.54)

So, the refractive index of rock salt with respect to ice​ is :

[tex]\mu=\dfrac{1.54}{1.31}\\\\=1.17[/tex]

So, the required refractive index of rock salt wrt ice is 1.17.

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