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N76 [4]
4 years ago
13

A spring scale hung from the ceiling stretches by 6.1 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl

aced with a 2.5 kg mass.
Part A What is the stretch of the spring?
Physics
1 answer:
lapo4ka [179]4 years ago
8 0
To solve this problem, we need to use Hooke's Law. 

Hooke's Law: 

F = -kx

where F = mg
m = mass of the object
g = gravitational force
k = spring constant
x = length of stretch of the spring

Solve for the spring constant, k:

2.0 kg * 9.81 m/s^2 = -k*(6.1cm/100cm/m)

k = 321.64 

The stretch of the spring with mass of 2.5 kg is then determined: 

2.5 kg * 9.81 m/s^2 = -321.64 *(x)
x = 7.62 cm
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If you push any floating object down from equilibrium and release it, it bobs up and down. That looks like an oscillation, so le
GarryVolchara [31]

Answer:

  F_{y} = ( ρ_fluid g A) y

Explanation:

This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force

for the first part, let's write Newton's equilibrium equation

        B₀ - W = 0

        B₀ = W

        ρ_fluid g V_fluid = W

the volume of the fluid is the area of ​​the cube times the height it is submerged

      V_fluid = A y  

For the second part, the body introduces a quantity and below this equilibrium point, the equation is

        B - W = m a

        ρ_fluid g A (y₀ + y) - W = m a

        ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a

       ρ_fluid g A y + (B₀-W) = ma

the part in parentheses is zero since it is the force when it is in equilibrium

      ρ_fluid g A y = m a

      this equation the net force is

      F_{y} = ( ρ_fluid g A) y

we can see that this force varies linearly the distance and measured from the equilibrium position

8 0
3 years ago
A train pulls away from a station with a constant acceleration of 0.42 m/s2. A passenger arrives at a point next to the track 6.
Rina8888 [55]

Answer:

2.69 m/s

Explanation:

Hi!

First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:

x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m

So, the position as a function of time is:

xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m

Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:

xP(t)=V*t

In order for the passenger to catch the train

xP(t)=xT(t)

(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t

To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:

0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2

This equation give us the minimum velocity the passenger must have in order to catch the train:

V^2 - 7.22534(m/s)^2 = 0

V^2 = 7.22534(m/s)^2

V = 2.6879 m/s

4 0
3 years ago
If Fg=mg solve for g
Bas_tet [7]

Answer:Fg = mg however newtons second law states that the net force acting on an object is equal to it's mass times it's acceleration so what allows us to say that Fg = mg because certainly not for every single situation the net force is going to equal to the force of gravity please explain... what allows us to say Fg = mg

Source https://www.physicsforums.com/threads/fg-mg-questioned.336776/

Explanation:

6 0
3 years ago
7. Calculations.
kotykmax [81]

Answer:

5 ms-2

Explanation:

F = ma

F = 100N

m = 20kg ( you should make sure the unit is kg before you answer the question)

100 = 20a

a = 100÷ 20

a = 5 ms-2

4 0
4 years ago
What is the definition of speed and the formula for calculation?
pashok25 [27]
For a body moving at a uniform velocity you can calculate the speed by dividing the distance traveled by the amount of time it took, for example one mile in 1/2 hour would give you 2 miles per hour. If the velocity is non-uniform all you can say is what the average speed is. 

HOPE IT HELPS YOU
3 0
3 years ago
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