1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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>

You might be interested in
Suppose you are at the center of a large freely-rotating horizontal turntable in a carnival funhouse. As you crawl toward the ed
I am Lyosha [343]

Answer:

c. remains the same, but the RPMs decrease.

Explanation:

Because there aren't external torques on the system composed by the person and the turntable it follows that total angular momentum (I) is conserved, that means the total angular momentum is a constant:

\overrightarrow{L}=constant

The total angular momentum is the sum of the individual angular momenta, in our case we should sum the angular momentum of the turntable and the angular momentum of a point mass respect the center of the turntable (the person)

\overrightarrow{L_{turnatble}}+\overrightarrow{L_{person}}=constant (1)

The angular momentum of the turntable is:

\overrightarrow{L_{turnatble}}=I\overrightarrow{\omega} (2)

with I the moment of inertia and ω the angular velocity.

The angular momentum of the person respects the center of the turntable is:

\overrightarrow{L_{person}}=\overrightarrow{r}\times m\overrightarrow{v} (3)

with r the position of the person respects the center of the turntable, m the mass of the person and v the linear velocity

Using the fact v=\omega r:

\overrightarrow{L_{person}}=\overrightarrow{r}\times rm\overrightarrow{\omega}(3)

By (3) and (2) on (1) and working only the magnitudes (it's all that we need for this problem):

I\omega+r^{2}m\omega=constant

\omega(I+r^{2}m)=constant

Because the equality should be maintained, if we increase the distance between the person and the center of the turntable (r), the angular velocity should decrease to maintain the same constant value because I and m are constants, so the RPM's (unit of angular velocity) are going to decrease.

4 0
3 years ago
You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen
Paladinen [302]

Answer:

(A)

\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s  \right )D+t_t^2v_s^2=0

<em>(B)  D=54.71 m</em>

Explanation:

<u>Free Fall</u>

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

\displaystyle D=\frac{gt^2}{2}

Solving for t

\displaystyle t=\sqrt{\frac{2D}{g}}

If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

\displaystyle t=\frac{D}{v_s}

(A)

The total measured time is the sum of both times and it's given as t_t=3.5\ seconds

\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}

From this equation we'll manage to compute D

First, we isolate the square root

\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}

Let's square both sides

\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}

Multiplying by v_s^2

\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2

Rearranging and factoring

\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}

Now, let's put in numbers:

g=9.8\ m/s^2,\ v_s=345\ m/s,t_t=3.5\ sec

\displaystyle D^2-\left (\frac{2(345)^2}{9.8}+2(3.5)(345)\right )D+(12.25)345^2=0

Computing all the coefficients:

\displaystyle D^2-26,705.82D+1,458,056.25=0

Solving for D, we have two possible solutions:

D=54.71,\ D=26,651.11

The second solution is called "extraneous", since it comes from squaring an equation, which can introduce non-valid (or external) solutions. It's impossible, given the conditions of the problem, that the well could be 26.5 km deep. So we'll keep the only solution as.

<em>D=54.71 m</em>

Let's prove our calculations by computing both times:

\displaystyle t_1=\sqrt{\frac{2(54.71)}{9.8}}=3.34\ sec

\displaystyle t_2=\frac{54.71}{345}=0.16\ sec

We can see their sum is 3.5 seconds, 3.34 of which were taken to reach the bottom of the well, and 0.16 sec took the sound to reach the top.

3 0
3 years ago
I Need help please with this question..
pantera1 [17]
D............................
4 0
3 years ago
According to the law of conservation of mass, which of these is true.
dimaraw [331]
C because mass is how much space an object takes up
8 0
3 years ago
When a cyclist moves downhill without pedalling, what type of energy does he gain?
Vesna [10]

Answer:

He is gaining kinetic energy and losing potential energy

Explanation:

3 0
2 years ago
Read 2 more answers
Other questions:
  • (HELP!!! 30 pts if answered right. )What formula gives the strength of an electric field, E, at a distance from a known source c
    11·1 answer
  • A laser beam of wavelength 439.4 nm is inci- dent on two slits 0.306 mm apart.
    6·1 answer
  • A person runs at a speed of 4 meters per second. How far will the person travel in 40 seconds?
    10·1 answer
  • A lamp has a current of 2.17 A. In hours, how long does it take for 1 mole of electrons to pass through the lamp?
    5·1 answer
  • A janitor comes across a spilled substance on the floor of a highschool chemistry lab. he wants to make sure that the liquid is
    12·1 answer
  • Measurements of two electric currents are shown in the chart.
    15·2 answers
  • Why did it take Kepler so long to discover the truth about the motions of the solar system?
    5·1 answer
  • Explain how the velocity of the aeroplane will change in the vertical direction
    15·1 answer
  • Three collisions are elastic and three are inelastic. Determine which collision type took place for each collision. Support your
    6·1 answer
  • What does properties mean
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!