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BlackZzzverrR [31]
3 years ago
8

I NEED HELP BADLY

Physics
1 answer:
Marrrta [24]3 years ago
7 0

Answer:

a) Andrea's initial momentum, 200 kg m/s

b) Andrea's final momentum, 0

c) Impulse,  = - 200 Ns

d) The force that the seat belt exerts on Andrea, - 400 N

Explanation:

Given data,

The initial velocity of the car is, u = 40 m/s

The mass of Andrea, m = 50 kg

The time period of deceleration, a = 0.5 s

The final velocity of the car, v = 0

a) Andrea's initial momentum,

               p = mu

                  = 50 x 40

                  = 200 kg m/s

b) Andrea's final momentum

                  P = mv

                     = 50 x 0

                    = 0 kg m/s

c) Impulse

                   I = mv - mu

                     = 0 - 200

                    = - 200 Ns

The negative sign indicated that the momentum is decreased.

d) The force that the seat belt exerts on Andrea

                   F = (mv - mu)t

                     = (0 - 200) / 0.5

                     = - 400 Ns

Hence,the force that the seat belt exerts on Andrea is, - 400 N

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What measurement do we use to determine the amount of force used to move an object by a simple machine
alexandr1967 [171]

Answer:

Newtons.

Explanation:

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

F = ma

Where;

F represents force measured in Newton.

m represents the mass of an object measured in kilograms.

a represents acceleration measured in meter per seconds square.

Newtons is a measurement we use to determine the amount of force used to move an object by a simple machine. It is the International System of Units (SI) used to measure force and has a symbol of N.

Basically, it was named after Sir Isaac Newton based on his fundamental works in the field of mechanics (motions).

8 0
3 years ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
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True or false the potential energy of a freely object increases as it begins to fall
Eva8 [605]
False, as an object falls its potential energy turns into kinetic energy thus decreasing the potential energy.
3 0
3 years ago
A thin rod (length = 1.09 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.
Murljashka [212]

Answer:

a) w = 4.24 rad / s , b) α  = 8.99 rad / s²

Explanation:

a) For this exercise we use the conservation of kinetic energy,

Initial. Vertical bar

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Final. Just before touching the floor

       Emf = K = ½ I w2

As there is no friction the mechanical energy is conserved

       Emo = emf

       mgh = ½ m w²

The moment of inertial of a point mass is

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       w = √ 2gh / L²

The initial height h when the bar is vertical is equal to the length of the bar

         h = L

         w = √ 2g / L

Let's calculate

       w = RA (2 9.8 / 1.09)

       w = 4.24 rad / s

b) Let's use Newton's equation for rotational motion

         τ = I α

         F L = (m L²) α

The force applied is the weight of the object, which is at a distance L from the point of gro

         mg L = m L² α  

          α  = g / L

          α  = 9.8 / 1.09

          α  = 8.99 rad / s²

5 0
3 years ago
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