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2 years ago

In the equation vx^2=v0x^2+2ax(x-x0) what does the terms vx, v0x, x, and x0 stand for respectively?

Physics

2 answers:

Jobisdone [24]2 years ago

7 0

Explanation:

The equation of motion of an object is given by :

$v_x^2=v_{ox}^2+2ax(x-x_o)$

Where

$v_x$ is velocity of a particle at position x

$v_{ox}$ is the velocity at position x = 0

x is the position of an object

$x_o$ is position at t = 0

So, the correct option is (b) "velocity at position x, velocity at position x=0, position x, and the original position". Hence, this is the required solution.

tatuchka [14]2 years ago

6 0

B. velocity at position x, velocity at position x=0, position x, and the original position

In the equation

$v_{x}^{2}$ = $v_{ox}^{2}$ +2 a x (x - x₀)

$v_{x}$ = velocity at position "x"

$v_{ox}$ = velocity at position "x = 0 "

x = final position

$x_{o}$ = initial position of the object at the start of the motion

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2 years ago

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1 year ago

A 5.0 kg cannonball is dropped from the top of a tower. It falls for 1.6 seconds before slamming into a sand pile at the base of

stepan [7]

Answer:

15.7 m/s

Explanation:

The motion of the cannonball is a accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground (gravitational acceleration). Therefore, the velocity of the ball at time t is given by:

$v(t)=u + gt$

where

u = 0 is the initial velocity

g = 9.8 m/s^2 is the acceleration

t is the time

If we substitute t=1.6 s into the equation, we find the final velocity of the cannonball:

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2 years ago

A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the

olga2289 [7]

Answer:

a) F = 30 N, b) I = 12 N s , c) I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

p₀ = m v

The speed can be found by kinematics

v = v₀ - g t

v = 0 - 10 0.4

v = -4.0 m / s

Final after division

pf = m₁ v₁f + m₂ v₂f

p₀ = pf

M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

F = mg

F = 3 10 = 30 N

b) The expression for momentum is

I = Ft

Before the explosion the only force that acts is the weight

I = mg t

I = 3 10 0.4

I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

M v = m₁ v₁f + m₂ v₂f

v₂f = (M v - m₁ v₁f) / m₂

v₂f = (3 (-4) - 1 6) / 2

v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

I = F t = Dp

F t = pf –po)

F t= [m₁ v₁f + m₂ v₂f]

I = [1 6 + 2 (-9) -0]

I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

ΔI =If – I₀

ΔI = - 12 – (-12)

ΔI = -0 N s

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2 years ago

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Answer:

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