A body moves due north with velocity 40 m/s. A force is applied

on it and the body continues to move due north with velocity 35 m/s. W. .What is the direction of rate of change of momentum,if it takes

some time for that change and what is the direction of applied

external force?​

Answers

Answer 1

Answer:

the direction of rate of change of the momentum is against the motion of the body, that is, downward.

The applied force is also against the direction of motion of the body, downward.

Explanation:

The change in the momentum of a body, if the mass of the body is constant, is given by the following formula:

[tex]\Delta p=\Delta (mv)\\\\\Delta p=m\Delta v[/tex]

p: momentum

m: mass

[tex]\Delta v[/tex]:  change in the velocity

The sign of the change in the velocity determines the direction of rate of change. Then you have:

[tex]\Delta v=v_2-v_1[/tex]

v2: final velocity = 35m/s

v1: initial velocity = 40m/s

[tex]\Delta v =35m/s-40m/s=-5m/s[/tex]

Hence, the direction of rate of change of the momentum is against the motion of the body, that is, downward.

The applied force is also against the direction of motion of the body, downward.


Related Questions

Motion maps for two objects, Y and Z, are shown.

A motion map. The position line is a long black arrow pointing right with x as the reference point at left. Above the line are three dots, each with a vector pointed away from x back to back in a line labeled B. Above B, there are four dots, each with a shorter vector pointing away from x in a line labeled A starting closer to x .

Object Z passes object Y after how many seconds?

2
3
4
5

Answers

Answer: it takes 3 seconds (b)

Explanation:

Answer: B. 3

Explanation:

Each black point on the map represents one second. There are three black points with vectors representing Z's movement before Y begins to move.

From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum upward displacement from the launch point is 150 m. What are the (a) horizontal and (b) vertical components of its launch velocity

Answers

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

v₀ₓ = 63.5 m/s

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

v₀y = 54.2 m/s

What is a major criticism of Maslow's hierarchy of needs? Select one: a. It is subjective. b. It does not take gender differences into account. c. It is humanistic. d. It only accounts for the objective world.

Answers

The correct answer is A. It is subjective

Explanation:

In 1943, the recognized psychologist Abraham Maslow proposed a theory to understand and classify human needs. The work of Maslow included five different categories to classify all basic needs, psychological needs, and self-esteem needs; additionally, in this, Maslow proposed individuals need to satisfy the needs of previous levels to satisfy more complex needs. For example, the first level includes physiological needs such as hunger and these are necessary to get to more complex needs such as the need for safety or self-satisfaction.

This hierarchy is still used all around the world to understand human needs; however, it was been widely criticized because the classification itself is related to Maslow's perspective as this was mainly based on Maslow's ideas about needs, which makes the hierarchy subjective. Also, due to its subjectivity,  the hierarchy may apply only in some individuals or societies.

Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that F1x and F2x represent the components, of the corresponding forces. Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that and represent the components, of the corresponding forces. F1x=−F2x F1x=F2x m1=m2 m1≪m2

Answers

Answer:

a) m₁ = m₂  F₁ₓ = F₂ₓ

b) m₁ << m₂   F₂ₓ =0

Explanation:

This interesting exercise is unclear your statement, so that in a center of mass system has an acceleration of zero it is necessary that the sum of the forces on each axis is zero, to see this we write Newton's second law

     ∑ F = m a

for acceleration to be zero implies that the net force is zero.

we must write the expression for the center of mass

        [tex]x_{cm}[/tex] = 1 / M (m₁ x₁ + m₂ x₂)

now let's use the derivatives

      [tex]a_{cm}[/tex] = d² x_{cm}/dt² = 1 / M (m₁ a₁ + m₂a₂)

where M is the total mass M = m₁ + m₂

     so that the acceleration of the center of mass is zero

               0 = 1 / M (m₁ a₁ + m₂a₂)

               m₁ a₁ = - m₂ a₂

In the case that we have components on the x axis, the modulus of the two forces are equal and their direction is opposite, therefore

   F₁ₓ = -F₂ₓ

b)r when the two masses are equal , in the case of a mass greater than the other m₁ << m₂

      acm = d2 xcm / dt2 = 1 / M (m1 a1 + m2a2)

so that the acceleration of the center of mass is zero

               0 = 1 / M (m1 a1 + m2a2)

               m1 a1 = - m 2 a2

with the initial condition, we can despise m₁, therefore

                0 = m₂a₂

 if we use Newton's second law

              F₂ = 0

       

I tell you that in this case with a very high mass difference the force on the largest mass must be almost zero

How many times can a three-dimensional object that has a radius of 1,000 units fit something with a radius of 10 units inside of it? How many times can something with a radius of 2,000 units fit something with a radius of 1 unit?

Answers

Answer:

# _units = 1000

Explanation:

This exercise we can use a direct proportion rule.

If a volume of radius r = 1 is one unit, how many units can fit in a volume of radius 10?

    # _units = V₁₀ / V₁

The volume of a body of radius 1 is

       V₁ = 4/3 π r₁³

        V₁ = 4/3π

the volume of a body of radius r = 10

        V₁₀ = 4/3 π r₂³

        V10 = 4/3 π 10³

     

the number of times this content is

         #_units = 4/3 π 1000 / (4/3 π 1)

        # _units = 1000

How far does a roller coaster travel if it accelerates at 2.83 m/s2 from an initial
velocity of 3.19 m/s for 12.0 s?

Answers

Answer:

b

Explanation:

(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on the street, (a) what was its initial speed, (b) what altitude does it reach, (c) when was it thrown, and (d) when does it reach the street again? Giancoli, Douglas C.. Physics (p. 45). Pearson Education. Kindle Edition.

Answers

Answer:

Assuming that the vertical speed of the ball is 14 m/s we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

 

Explanation:

a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:  

[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]

Where:

[tex]V_{f}[/tex]: is the final speed = 14 m/s

[tex]V_{0}[/tex]: is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

[tex] V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s [/tex]  

b) The maximum height is:

[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]

[tex] h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m [/tex]

c) The time can be found using the following equation:

[tex] V_{f} = V_{0} - gt [/tex]

[tex] t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s [/tex]

d) The flight time is given by:

[tex] t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s [/tex]

         

I hope it helps you!    

A 60-kg skier is stationary at the top of a hill. She then pushes off and heads down the hill with an initial speed of 4.0 m/s. Air resistance and the friction between the skis and the snow are both negligible. How fast will she be moving after she is at the bottom of the hill, which is 10 m in elevation lower than the hilltop

Answers

Answer:

The velocity is  [tex]v = 8.85 m/s[/tex]

Explanation:

From the question we are told that

    The mass of the skier is [tex]m_s = 60 \ kg[/tex]

      The initial speed is [tex]u = 4.0 \ m/s[/tex]

       The height is  [tex]h = 10 \ m[/tex]

According to the law of energy conservation

     [tex]PE_t + KE_t = KE_b + PE_b[/tex]

Where [tex]PE_t[/tex] is the potential energy at the top which is mathematically evaluated as

       [tex]PE_t = mg h[/tex]

substituting values

       [tex]PE_t = 60 * 4*9.8[/tex]

      [tex]PE_t = 2352 \ J[/tex]

And  [tex]KE_t[/tex] is the kinetic energy at the top which equal to zero due to the fact that velocity is zero at the top of the hill

And  [tex]KE_b[/tex] is the kinetic energy at the bottom of the hill which is mathematically represented as

         [tex]KE_b = 0.5 * m * v^2[/tex]

  substituting  values

         [tex]KE_b = 0.5 * 60 * v^2[/tex]

=>     [tex]KE_b = 30 v^2[/tex]

Where v is the velocity at the bottom

   And [tex]PE_b[/tex] is the potential  energy at the bottom which equal to zero due to the fact that height  is zero at the bottom of the hill

So  

        [tex]30 v^2 = 2352[/tex]

=>      [tex]v^2 = \frac{2352}{30}[/tex]

=>       [tex]v = \sqrt{ \frac{2352}{30}}[/tex]

        [tex]v = 8.85 m/s[/tex]

         

Answer:

The Skier's velocity at the bottom of the hill will be 18m/s

Explanation:

This is simply the case of energy conversion between potential and kinetic energy. Her potential energy at the top of the hill gets converted to the kinetic energy she experiences at the bottom.

That is

[tex]mgh = 0.5 mv^{2}[/tex]

solving for velocity, we will have

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

hence her velocity will be

[tex]v=\sqrt{2 \times 9.81 \times 10}=14.00m/s[/tex]

This is the velocity she gains from the slope.

Recall that she already has an initial velocity of 4m/s. It is important to note that since velocities are vector quantities, they can easily be added algebraically. Hence, her velocity at the bottom of the hill is 4 + 14 = 18m/s

The Skier's velocity at the bottom of the hill will be 18m/s

HELP ASAP!
There is a lever with 5 m long. The fulcrum is 2 m from the right end. Each end hangs a box. The whole system is in balance. If the box hung to the right end is 12 kg, then what is the mass of the box hung to the left end?

Answers

Answer:

8kg

Explanation:

For the box to be in equilibrium. The clockwise moment ensued by the box on the right should be same as that ensued by the one on the right. Hence :

M ×3 = 12 ×2

M = 24/3 = 8kg

Note mass is used because trying to compute the weight by multiplying by the acceleration of free fall due to gravity on both sides will cancel out.

A spaceship travels toward the Earth at a speed of 0.97c. The occupants of the ship are standing with their torsos parallel to the direction of travel. According to Earth observers, they are about 0.50 m tall and 0.50 m wide. Calculate what the occupants’ height and width according to the others on the spaceship?

Answers

Answer:

Explanation:

We shall apply length contraction einstein's relativistic formula to calculate the length observed by observer on the earth . For the observer , increased length will be observed for an observer on the earth

[tex]L=\frac{.5}{\sqrt{1-(\frac{.97c}{c})^2 } }[/tex]

[tex]L=\frac{.5}{.24}[/tex]

L= 2.05

The length will appear to be 2.05 m . and width will appear to be .5 m  to the observer on the spaceship. . It is so because it is length which is moving parallel to the direction of travel. Width will remain unchanged.  

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