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Assoli18 [71]
3 years ago
10

Each of 100 identical blocks siting on a frictionless surface is connected to the next bloc by a massless string. The first bloc

k is pulled with a force of 100 ????. a) What is the tension in the string connecting block 100 to block 99? b) What is the tension in the string connecting block 50 to block 51?

Physics
1 answer:
Anit [1.1K]3 years ago
6 0

Answer:

A) 1 N

B) 50 N

Explanation:

Let us consider that the string does not deform.

To solve this problem lets consider the whole system as two parts. In the initial case, the first part will be de 100N being exterted to the whole system and in the second the 100 blocks system.

In this case we can imagine as the whole system being pulled by 100 N, and therefore its acceleration will be:

a = 100 N /(100 m)

where m stantds for the mass of one block

Now, the whole system and its individual parts must move with the same acceleration otherwise the string would stretch.

Now lets consider the first part of the system as the first block, and the second part as the other 99 blocks.

The Tension of the string pulling the 99 blocks must be so that it exterts the enough force to move that 99blocks-system at an acceleration a, since that sub-system has a mass of 99m

T1 = 99 m * a = (99 m) * (100 N/ 100 m) = 99 N

Now lets consider an intermidiate sub-system, where the first part is made of n blocks and the second susbsystem is made of (100 -n) blocks

Following the same logic, the tension of the corresponding string must be the acceleration of the whole systems times the mass of the second subsystem:

Tn = (100 -n)m * ( 100 N / 100 m ) = (100 -n) N

a)

Therefore the tension in the string connecting block 100 to block 99 must be

<u>T99 = 1 N</u>

<u />

b)

And

<u>T50 = 50 N</u>

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Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

  • The statement that indicates how the relationship between <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>

Reasons:

The energy given  to the block by the spring = \mathbf{0.5  \cdot k  \cdot x^2}

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = 0.5  \cdot k  \cdot x^2 = kinetic energy of

block = 0.5  \cdot m  \cdot v^2

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = \mathbf{0.5  \cdot k  \cdot x^2}

  • The force of the weight of the block on the string, F = m \cdot g  \cdot sin(\theta)

The energy given to the block = 0.5 \cdot k \cdot x^2 - m \cdot g  \cdot sin(\theta) = The kinetic energy of block as it leaves the spring = \mathbf{0.5  \cdot m  \cdot v^2}

Which gives;

0.5 \cdot k \cdot x^2 - m \cdot g  \cdot sin(\theta) = 0.5  \cdot m  \cdot v^2

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and <em>c</em> are constants

The graph of the equation a·x² + c·v² = b  is an ellipse

Therefore;

  • As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.

<em>Please find attached a drawing related to the question obtained from a similar question online</em>

<em>The possible question options are;</em>

  • <em>As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x</em>
  • <em>The relationship is no longer linear and v will be more for the same value of x</em>
  • <em>The relationship is still linear, with lesser value of v</em>
  • <em>The relationship is still linear, with higher value of v</em>
  • <em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>

<em />

Learn more here:

brainly.com/question/9134528

6 0
2 years ago
A person holding a lunch bag is moving upward in a hot air balloon at a constant speed of 7.3 m/s . When the balloon is 24 m abo
Kitty [74]

Explanation:

Given that,

Initial speed of the bag, u = 7.3 m/s

Height above ground, s = 24 m

We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :

v^2=u^2+2as

v^2=(7.3)^2+2\times 9.8\times 24  

v=\sqrt{523.69}

v = 22.88 m/s

So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.

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3 years ago
Alexandra is attempting to drag her 32.6 kg golden retriever across the wooden floor by applying a horizontal force. What force
vichka [17]

Answer:

230.26 N

Explanation:

Since the speed is constant, acceleration is zero hence the net force will be given by the product of mass, coefficient of friction and acceleration due to gravity

F=0.72*32.6*9.81=230.26 N

7 0
3 years ago
Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light
motikmotik

Answer:

n_cladding = 1.4764

Explanation:

We are told that θ_max = 5 °

Thus;

θ_max + θ_c = 90°

θ_c = 90° - θ_max

θ_c = 90° - 5°

θ_c = 85°

Now, critical angle is given by;

θ_c = sin^(-1) (n_cladding/n_core)

sin θ_c = (n_cladding/n_core)

n_cladding = (n_core) × sin θ_c

Plugging in the relevant values, we have;

n_cladding = 1.482 × sin 85

n_cladding = 1.4764

7 0
3 years ago
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
4 years ago
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