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dexar [7]
3 years ago
10

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

ock is pulled with a force of 100 N. What is the tension in the string connecting block 100 to block 99? What is the tension in the string connecting block 50 to block 51?
Physics
1 answer:
o-na [289]3 years ago
8 0

Answer:

The tension in the string connecting block 50 to block 51 is 50 N.

Explanation:

Given that,

Number of block = 100

Force = 100 N

let m be the mass of each block.

We need to calculate the net force acting on the 100th block

Using second law of newton

F=ma

100=100m\times a

ma=1\ N

We need to calculate the tension in the string between blocks 99 and 100

Using formula of force

F_{100-99}=ma

F_{100-99}=1

We need to calculate the total number of masses attached to the string

Using formula for mass

m'=(100-50)m

m'=50m

We need to calculate the tension in the string connecting block 50 to block 51

Using formula of tension

F_{50}=m'a

Put the value into the formula

F_{50}=50m\times a

F_{50}=50\times1

F_{50}=50\ N

Hence, The tension in the string connecting block 50 to block 51 is 50 N.

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Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

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Mass of the person  standing on platform = 65.7 kg

Distance from the center of platform = 1.07 m

Mass of the dog on the platform = 27.3 kg

Distance from center of platform = 1.31 m

We have to calculate the moment of inertia.

Formula:

MOI of disk = \frac{MR^2}{2}

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia (I_y_y )  of the system with respect to the axis yy.

⇒ I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g

⇒ I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2

⇒ I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2

⇒ I_y_y=289.088\ kg.m^2

The moment of inertia of the system is 289.088 kg.m^2

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When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amp
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Answer:

796.18 Hz

Explanation:

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make f the subject of the equation

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From the question,

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Substitute these values into equation 2

f = 3.6×10⁻³/( 7.2×10⁻⁷×2×3.14)

f = 796.18 Hz

6 0
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