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-Dominant- [34]
3 years ago
5

A light spring with a force coefficient 11.85 N/m is compressed by 14 cm as it is held between a 0.27 kg block on the left and a

0.6 kg block on the right, both resting on a horizontal surface. The spring exerts a force on each of the blocks, tending to push them apart. The blocks are then released from rest.Find the acceleration of each block if the coefficient of friction between the blocks and the surface is:
[Note: This question may require you to perform your calculations to a few extra digits of precision to get the answer to within 1%.]

(a) μ = 0: Left______ Right _______

(b) μ = 0.158: Left______ Right _______
Physics
1 answer:
lilavasa [31]3 years ago
4 0

Answer:

A) Left = 6.14 m/s2 Right=2.765 m/s2   B) Left = 4.59 m/s2 Right= 1.215 m/s2

Explanation:

In the question we are given the spring constant which is 11.85 N/m and the compression of the spring is 14 cm. There are two blocks in front of the springs which are 0.27 kg and 0.6 kg respectively.

First, we need to calculate how much force the spring are going to exert on the masses when they are released. Since this force is not going to change for both cases, we only need to do it once.

The formula for calculation the force is F = k.x where k is the spring constant or force coefficient 11.85 N/m and the x is the compression which is 14 cm or 0.14 meters. If we put them in the equation we can find that F = 1.659 N

A) In the first case scenario, where the friction is equal to 0, we can use the formula F=m.a where F is the force applied by the release of the spring, m is the mass of the block and a is the acceleration.

For the first block, when we put 1.659 N and 0.27kg in the equation, a is calculated to be 6.14 m/s2.

For the second block, the same force of 1.659 N and 0.6 kg, a is calculated to be 2.765 m/s2.

B) In the second case scenario, where the friction is equal to 0.158, we first need to calculate its effect on each block. We need to use the formula      Fk = μ.N where μ is the friction constant and N is the normal force of the block which is m.g (where g = 9,81 m/s2).

The friction force for the first block is calculated to be 0.4184 N and the friction force for the second block is calculated to be 0.9299 N.

The total force for the first block is 1.659 N - 0.4184 N = 1.2405 N.

The total force for the second block is 1.659 N - 0.9299 N = 0.7290 N.

When we insert these numbers into the same equation we used to find the acceleration in the first case scenario, we can calculate the acceleration of the first block as 1.2405 = 0.27 x a  and a = 4.59 m/s2.

When we insert these numbers into the same equation we used to find the acceleration in the first case scenario, we can calculate the acceleration of the second block as 0.7290 = 0.6 x a  and a = 1.215 m/s2.

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Which two surfaces have the lowest coefficient of friction
jok3333 [9.3K]

Answer:

Explanation:

As an example, ice on steel has a low coefficient of friction – the two materials slide past each other easily – while rubber on pavement has a high coefficient of friction – the materials do not slide past each other easily. The coefficients of friction ranges from near 0 to greater than 1.

5 0
3 years ago
Two particles, each with charge 55.3 nC, are located on the y axis at y 24.9 cm and y -24.9cm (a) Find the vector electric field
ser-zykov [4K]

Answer:

Ex = kq 2x / ∛ (x² + y²)²  and  Ex = 2008 N / C

Explanation:

a)   The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,

The equation for the electric field produced by a point charge is

         E = k q / r²

With r the distance between the point charge and the positive test charge

We look for each electric field

Particle 1.  Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance

          r² = x² + y²

          E1 = k q / (x² + y²)

Particle 2.   located at x = -24.9 m

          r² = x² + y²

          E2 = k q / (x² + y²)

We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.

In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find

Let's measure the angle from axis X

     cos θ = CA / H = x / (x2 + y2) ½

     E1x = E1 cos θ

      E2x = = E1 cos θ

The resulting field

      Ey = 0

      Ex = E1x + E2x 2 E1x

      Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)

      Ex = kq 2x / ∛ (x² + y²)²

b) For this part we substitute the numerical values

      Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂

      Ex = 497.15   x / (x² + 0.062)  ³/₂  

Point where can the value of the electric field x = 38.1 cm = 0.381 m

       Ex = 497.15 0.381 / (0.381² + 0.062)  ³/₂  

       Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2

       Ex= 189.41 /0.0943

       Ex = 2008 N / C

c)  E = 1.00 kN / C = 1000 N / C

To solve this part we must find x in the equation

       Ex = 497.15 x / (x² + 0.062)  ³/₂  

Let's use some arithmetic

       Ex / 497.15 = x / (x² + 0.062)  ³/₂  

       [Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃

       ∛[Ex / 497.15]² = (∛x²) / (x² + 0.062)                 (1)

The roots of this equation are the solution to the problem,

     

For Ex = 1.00 kN / C = 1000 N / C

 

      [Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312

       1.312 = (∛x² ) / (x² + 0.062)

       1.312 (x² + 0.062) = ∛x²

       1.312 X² - ∛x² + 1.312 0.062 = 0

       1.312 X² - ∛x² + 0.0813 = 0

We need used computer

4 0
3 years ago
an object is 40M below the surface of sea water of density 1020 kg/m3 calculate the pressure difference
olga nikolaevna [1]

Answer:

P = 400248 [Pa]

Explanation:

We can calculate the hydrostatic pressure exerted by a column of liquid by means of the following equation.

P=Ro*g*h

where:

Ro = density = 1020 [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = depth = 40 [m]

Therefore:

P=1020*9.81*40\\P=400248 [Pa]

8 0
3 years ago
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Answer:

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Explanation:

3 0
3 years ago
A child is playing on a swing. As long as he does not swing too high the time it takes him to complete one full oscillation will
Aleks [24]

Answer:

We know that for a pendulum of length L, the period  (time for a complete swing) is defined as:

T = 2*pi*√(L/g)

where:

pi = 3.14

L = length of the pendulum

g = gravitational acceleration = 9.8 m/s^2

Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.

Then the period is independent of:

The mass of the child

The initial angle

Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.

7 0
3 years ago
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