PHYSICS COLLEGE

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. If g-9.8 m/s, what is the coefficient of static friction? a. 3.0 b. 0.15 c. 0.28 d. 0.31

Answers

Answer 1
Answer:

Answer:

0.31

Explanation:

horizontal force, F = 750 N

mass of crate, m = 250 kg

g = 9.8 m/s^2

The friction force becomes applied force = 750 N

According to the laws of friction,

Friction force = μ x Normal reaction of the surface

here, μ be the coefficient of friction

750 = μ x m g

750 = μ x 250 x 9.8

μ = 0.31

Thus, the coefficient of static friction is 0.31.

Answer 2
Answer:

Based on the calculatins, the coefficient of static friction is equal to: D. 0.31.

Given the following data:

Force = 750 Newton.

Mass = 250 kg.

Acceleration due to gravity = 9.8 m/s^2

How to calculate the coefficient of static friction.

Mathematically, the staticfrictional force acting on an object is giving by this formula:

F_s=\mu N=\mu mg\n\n\mu =(F_s)/(mg)

Where:

  • \mu is the coefficient of static friction.
  • N is the normal force.
  • m is the mass.
  • g is the acceleration due to gravity.

Substituting the given parameters into the formula, we have;

\mu =(750)/(250 * 9.8)\n\n\mu =0.31

Read more on force here: brainly.com/question/1121817

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A block (0.50 kg) is attached to an ideal spring with a spring constant of 80 N/m, oscillating horizontally on a frictionless surface. The total mechanical energy is 0.12 J. (a) What is the greatest extension of the spring from its equilibrium length? (b) Now the block is traveling 2.00 m/s, and brought to rest by compressing a very long spring of spring constant 800.0 N/m. How much does the spring compress?

Answers

Answer:

a) x =  5.48 10⁻² m and b)  0.05 m

Explanation:

a) For a system in oscillatory motion the mechanical energy conserves and is described by the equation

     Em = ½ k A²

Where k is the spring constant and at the amplitude of the movement

When the spring has the greatest extent, the kinetic energy is zero

     Em = U = ½ k x²

Therefore, the amplitude of the movement is the same amplitude of the spring

Let's calculate

    A = √ (2Em / k)

    A = √ (2 0.12 / 80)

   A = 0.0548 m = 5.48 10⁻² m

b) In this case the spring has kinetic energy that becomes elastic potential energy, let's calculate the mechanical energy before and after compressing the spring

Initial

      Em = K = ½ m v²

Final

     Em = Ke = ½ k x²

     ½ m v² = ½ k x²

     x = √(m/k) v

     x = 2 √(0.50 /800.0)

     x = 0.05 m

Answer:

a) The greatest extension of the spring is 0.055 m

b) The spring compress 0.05 m

Explanation:

Please look at the solution in the attached Word file

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Answers

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Answers

The measurement will be significantly affected.

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A flat circular coil having a diameter of 25 cm is to produce a magnetic field at its center of magnitude, 1.0 mT. If the coil has 100 turns how much current must pass through the coil?

Answers

Answer:

The current pass through the coil is 6.25 A

Explanation:

Given that,

Diameter = 25 cm

Magnetic field = 1.0 mT

Number of turns = 100

We need to calculate the current

Using the formula of magnetic field

B =(\mu_(0)NI)/(2\pi r)

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Where, N = number of turns

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Put the value into the formula

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The height h (in feet) of an object shot into the air from a tall building is given by the function h(t) = 650 + 80t − 16t2, where t is the time elapsed in seconds. (a) Write a formula for the velocity of the object as a function of time t.

Answers

Answer:

80 - 32t

Explanation:

The height, h, in terms of time, t, is given as:

h(t) = 650 + 80t − 16t²

Velocity is the derivative of distance with respect to time:

v(t) = dh(t)/dt = 80 - 32t

Final answer:

The velocity of the object as a function of time is given by the derivative of the height function, which is v(t) = 80 - 32t.

Explanation:

The height h(t) of an object is given by the equation h(t) = 650 + 80t − 16t2. To find the velocity v(t), we need to take the derivative of h(t) with respect to time t. Using the power rule, we get:

v(t) = dh/dt = 0 + 80 - 32t.

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Learn more about Velocity Function here:

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#SPJ11

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?

Answers

Answer:

v=3.564\ m.s^(-1)

\Delta v =2.16\ m.s^(-1)

Explanation:

Given:

  • mass of John, m_J=30\ kg
  • mass of William, m_W=30\ kg
  • length of slide, l=3\ m

(A)

height between John and William, h=1.8\ m

Using the equation of motion:

v_J^2=u_J^2+2 (g.sin\theta).l

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

v_J^2=0^2+2* (9.8* (1.8)/(3))* 3

v_J=5.94\ m.s^(-1)

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

m_J.v_J+m_w.v_w=(m_J+m_w).v

where: v = velocity with which they move together after collision

30* 5.94+0=(30+20)v

v=3.564\ m.s^(-1) is the velocity with which they leave the slide.

(B)

  • frictional force due to mud, f=105\ N

Now we find the force along the slide due to the body weight:

F=m_J.g.sin\theta

F=30* 9.8* (1.8)/(3)

F=176.4\ N

Hence the net force along the slide:

F_R=71.4\ N

Now the acceleration of John:

a_j=(F_R)/(m_J)

a_j=(71.4)/(30)

a_j=2.38\ m.s^(-2)

Now the new velocity:

v_J_n^2=u_J^2+2.(a_j).l

v_J_n^2=0^2+2* 2.38* 3

v_J_n=3.78\ m.s^(-1)

Hence the new velocity is slower by

\Delta v =(v_J-v_J_n)

\Delta v =5.94-3.78= 2.16\ m.s^(-1)
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