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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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A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed <img src="https:
s344n2d4d5 [400]

Answer:

a) v_com= Rω

b) -2.254 m/s^{2}

c) 51.2 rad/s^{2}

d) t=1.08 seconds

e) x=7.865m

f) v_roll=6.07m

Explanation:

Initially, the ball is travelling with v_com=v_0

Wen not rotating, at the initial stage the ball must be sliding along the surface.

This motion therefore generates a frictional force F_r at the point of contact.

Let the velocity at the point of contact be v_bottom

v_bottom=v_com-Rω

Therefore when ω=0, v_bottom=v_com

So when the ball begins rolling

v_com= Rω

F_r=μ_rmg

〖-F〗_r=ma_com

a_com=(〖-μ〗_r mg)/m

a_com=-μ_rg

a_com=-(0.23)(9.8)

a_com=-2.254m/s^2

Te negative sow decrearse  

\alpha=(μ_r mgR)/I  =  (〖5μ〗_r mgR)/2mRR

=(〖5μ〗_r g)/2R

=(5*(0.23)*(9.8))/(2*0.11)

=51.2 rad/s^2

t=v_0/(〖-a〗_com+Rα)

=8.5/(2.255+0.11*(51.2))

=8.5/7.886

=1.08 seconds

X=v_0 t+1/2 a_com t^2

X=8.5*(2.254) -  1/2 (2.254)*〖1.08〗^2

=7.865m

v_roll=v_0+a_com t_r

=8.5-(2.254)(1.08)

        =6.07m/sec

Initially, the ball is travelling with v_com=v_0

Wen not rotating, at the initial stage the ball must be sliding along the surface.

This motion therefore generates a frictional force F_r at the point of contact.

a) Let the velocity at the point of contact be v_bottom

v_bottom=v_com-Rω

Therefore when ω=0, v_bottom=v_com

So when the ball begins rolling

v_com= Rω

b)    F_r=μ_rmg

〖-F〗_r=ma_com

a_com=(〖-μ〗_r mg)/m

a_com=-μ_rg

a_com=-(0.23)(9.8)

a_com=-2.254m/s^2

Te negative sow decrearse  

c) α=(μ_r mgR)/I  =  (〖5μ〗_r mgR)/2mRR

=(〖5μ〗_r g)/2R

=(5*(0.23)*(9.8))/(2*0.11)

=51.2 rad/s^2

d) t=v_0/(〖-a〗_com+Rα)

=8.5/(2.255+0.11*(51.2))

=8.5/7.886

=1.08 seconds

e) X=v_0 t+1/2 a_com t^2

X=8.5*(2.254) -  1/2 (2.254)*〖1.08〗^2

=7.865m

f) v_roll=v_0+a_com t_r

=8.5-(2.254)(1.08)

        =6.07m/sec

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<em>Magnetite is a Natural magnet</em>

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At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c
fiasKO [112]

Answer:

497.00977 N

3742514.97005

Explanation:

\rho = Density of water = 1000 kg/m³

C = Drag coefficient = 0.09

v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N

The drag force on the dolphin's nose is 497.00977 N

at 20°C

\mu = Dynamic viscosity = 1.002\times 10^{-3}\ Pas

Reynold's Number

Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005

The Reynolds number is 3742514.97005

8 0
3 years ago
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