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jasenka [17]
2 years ago
8

predict the reading of the spring balance when the wooden block is pulled on the sandpaper. Explain your answer. ​

Physics
1 answer:
igomit [66]2 years ago
4 0

Answer:

* if the spring force is greater than the maximum of the static friction force,  

               Fe = m (a + μ_k g)

* If the elastic force is less than or equal to the static friction force, the result is with static friction coefficient

              Fe =μ_s m g

Explanation:

For this exercise we must apply Newton's second law to the system

X axis

         Fe -fr = m a

         Fe = m a + fr

Y axis  

        N-W = 0

        N = W = mg

the roe force is given by

        fr = μ N

        fr = μ mg

we substitute

      Fe = m a + μ m g

      Ee = m (a + μ g)

Let's analyze the solution. We have several possibilities

* if the spring force is greater than the maximum of the static friction force, the system acquires an acceleration and the result is with the kinetic friction coefficient

               Fe = m (a + μ_k g)

* If the elastic force is less than or equal to the static friction force, the result is with static friction coefficient

              Fe =μ_s m g

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What is the largest possible magnitude of the acceleration of the electron due to the magnetic field?
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6 0
3 years ago
An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at
andrew-mc [135]

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

6 0
3 years ago
During the middle of a family picnic, Barry Allen received a message that his friends Bruce and Hal
weeeeeb [17]

The kinematics of the uniform motion and the addition of vectors allow finding the results are:

  • The  Barry's initial trajectory is 94.30 10³ m with n angles of θ = 138.8º
  • The return trajectory and speed are v = 785.9 m / s, with an angle of 41.2º to the South of the East

Vectors are quantities that have modulus and direction, so they must be added using vector algebra.

A simple method to perform this addition in the algebraic method which has several parts:

  • Vectors are decomposed into a coordinate system
  • The components are added
  • The resulting vector is constructed

 Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case

              v = \frac{\Delta d}{t}

              Δd = v t

Where  v is the average velocity, Δd the displacement and t the time

We look for the first distance traveled at speed v₁ = 600 m / s for a time

          t₁ = 2 min = 120 s

          Δd₁ = v₁ t₁

          Δd₁ = 600 120

          Δd₁ = 72 10³ m

Now we look for the second distance traveled for the velocity v₂ = 400 m/s    

  time t₂ = 1 min = 60 s

          Δd₂ = v₂ t₂

          Δd₂ = 400 60

          Δd₂ = 24 103 m

   

In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,

We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)

Let's use trigonometry for each distance

Route 1

          cos (180 -35) = \frac{x_1}{\Delta d_1}

          sin 145 = \frac{y_1}{\Delta d1}

          x₁ = Δd₁ cos 125

          y₁ = Δd₁ sin 125

          x₁ = 72 103 are 145 = -58.98 103 m

          y₁ = 72 103 sin 155 = 41.30 10³ m

Route 2

          cos (90+ 30) = \frac{x_2}{\Delta d_2}

          sin (120) = \frac{y_2}{\Delta d_2}

          x₂ = Δd₂ cos 120

          y₂ = Δd₂ sin 120

          x₂ = 24 103 cos 120 = -12 10³ m

           y₂ = 24 103 sin 120 = 20,78 10³ m

             

The component of the resultant vector are

              Rₓ = x₁ + x₂

              R_y = y₁ + y₂

              Rx = - (58.98 + 12) 10³ = -70.98 10³ m

              Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m

We construct the resulting vector

Let's use the Pythagoras' Theorem for the module

             R = \sqrt{R_x^2 +R_y^2}

             R = \sqrt{70.98^2 + 62.08^2}   10³

             R = 94.30 10³ m

We use trigonometry for the angle

             tan θ ’= \frac{R_y}{R_x}

             θ '= tan⁻¹ \frac{R_y}{R_x}

             θ '= tan⁻¹ \frac{62.08}{70.98}

             θ ’= 41.2º

Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction

            θ = 180 - θ'

            θ = 180 -41.2

            θ = 138.8º

Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.

The average speed of the trip should be

             v = \frac{\Delta R}{t_3}  

             v = \frac{94.30}{120}  \ 10^3

              v = 785.9 m / s

in the opposite direction, that is, the angle must be

               41.2º to the South of the East

In conclusion, using the kinematics of the uniform motion and the addition of vectors, results are:

  • To find the initial Barry trajectory is 94.30 10³ m with n angles of  138.8º
  • The return trajectory and speed is v = 785.9 m / s, with an angle of 41.2º to the South of the East

Learn more here:  brainly.com/question/15074838

4 0
3 years ago
A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 n that is parallel to the slope. the coefficient of kinetic
Drupady [299]
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.

∑Fy = Fn - mgcos30° = 0
           Fn = (50)(9.81)(cos 16) = 471.5 N

When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:

Fnet = F - μk*Fn - mgsin30° = ma
          250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
          a = 2.84 m/s²

8 0
3 years ago
A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cable
kompoz [17]

Answer:

4.44s

Explanation:

A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables

since the mass of the child and that of the swing is negligible, the masses wont be involved in the calculation

T=2π√L/g

g=acceleration due to gravity which is 9.81m/s2

the length of the supporting cable is 4.9m

T the period

period is the time required to make a complete oscillation

T=2*π√4.9/9.81

T=2*π*0.706

T=4.44s

4.44s

5 0
3 years ago
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