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alukav5142 [94]
1 year ago
11

If a 20 N block experiences a force of kinetic friction of 8.0 N

Physics
1 answer:
shutvik [7]1 year ago
6 0

The coefficient of kinetic friction (μ) between the block and the table is 0.4.

<h3>What is kinetic friction?</h3>

This sis the frictional force between an object in motion with the surface in contact.

μN = ff

where;

  • N is normal reaction due to weight of the block
  • ff is frictional force
  • μ is coefficient of friction

μ = ff/N

μ = 8/20

μ = 0.4

Thus, the coefficient of kinetic friction (μ) between the block and the table is 0.4.

Learn more about coefficient of friction here: brainly.com/question/20241845

#SPJ1

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A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee
Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

  • Initial Velocity; u = 0
  • Height from which it has dropped; h = 15m
  • Gravitational field strength; g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2

Final speed of brick as it hits the ground; v =  \ ?

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

v^2 = u^2 + 2gh

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

8 0
2 years ago
Four objects are situated along the y axis as follows: a 1.99-kg object is at 2.99 m, a 2.96-kg object is at 2.57 m, a 2.43-kg o
Dominik [7]

Answer:

The center of mass for the object is  y_c = 1.063 \  m from the origin

Explanation:

From the question we are told that

   The mass of the first object is  m_1 =  1.99 \  kg

   The position of first object with respect to origin y_1 =  2.99 \ m

   The mass of the second object is  m_2 =  2.96 \  kg

   The position of second object with respect to origin y_2 =  2.57 \ m

   The mass of the third object is  m_3 =  2.43  \  kg

   The position of third object with respect to origin y_3 =  0 \ m

   The mass of the fourth object is  m_3 =  3.96  \  kg

   The position of fourth object with respect to origin y_3 =  -0.502  \ m

Generally the center of mass of the object along the x-axis is  zero  because all the mass lie on the y axis

Generally the location of the center mass of the object is mathematically represented as

    y_c = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3 + m_4 * y_4}{m_1 + m_2 + m_3 + m_4}

=>y_c = \frac{1.99 * 2.99 + 2.96 * 2.57 + 2.43 * 0 + 3.96 * (-0.502)}{1.99+ 2.96  + 2.43 + 3.96}

=>y_c = 1.063 \  m

3 0
3 years ago
According to the Law of the Conservation of Matter, if you dissolve 25 grams of sugar into 150 grams of water, the mixture shoul
levacccp [35]
19grams becauce if u
7 0
2 years ago
2.0 mol of monatomic gas A initially has 5000 J of thermal energy. It interacts with 3.0 mol of monatomic gas B, which initially
Naddika [18.5K]

Answer:

 E_particle = 1,129 10⁻²⁰ J / particle

  T= 817.5 K

Explanation:

Energy is a scalar quantity so it is additive, let's look for the total energy of each gas

Gas a

         E_a = 2 5000 = 10000 J

Gas b

         E_b = 3 8000 = 24000 J

When the total system energy is mixed it is

          E_total = E_a + E_b

          E_total = 10000 + 24000 = 34000

The total mass is

           M = m_a + m_b

           M = 2 +3 = 5

The average energy among the entire mass is

           E_averge = E_total / M

            E_averago = 34000/5

            E_average = 6800 J

One mole of matter has Avogadro's number of atoms 6,022 10²³ particles

Therefore, each particle has an energy of

                E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³

                E_particle = 1,129 10⁻²⁰ J / particle

For  find the temperature let's use equation

               E = kT

               T = E / k

     

               T = 1,129 10⁻²⁰ / 1,381 10⁻²³

               T = 8.175 102 K

               T= 817.5 K

5 0
3 years ago
The pressure of liquid varies as per<br>its depth​
Slav-nsk [51]

Answer:

Yes.

Explanation:

The pressure varies as per the depth of the container

4 0
3 years ago
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