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
A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.
Diano4ka-milaya [45]

solution:

As Given plane is flying in east direction.

It throws back some supplies to designated target.

Time taken by the supply to reach the target =10 seconds

g = Acceleration due to gravity = - 9.8 m/s²[Taken negative as object is falling Downwards]

As we have to find distance from the ground to plane which is given by d.

d = \frac{1}{2}\times g\times t^2

 = \frac{1}{2}\times (9.8) \times(100) =50\times 9.8=490 meters

Distance from the ground where supplies has to be land  to plane  =  Option B =490 meters

6 0
3 years ago
Read 2 more answers
Incident rays parallel to the axis of a concave mirror reflect parallel to the axis.
coldgirl [10]
No they don't.  Incident rays parallel to the axis of a concave mirror
reflect from the mirror's surface and converge at its focal point.
3 0
3 years ago
Suppose Group A runs their experiment 3 times and calculates that the best estimate of the distance slid by red blocks is 15 + 3
Alex73 [517]
Sorry I just need points
7 0
3 years ago
How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33××10âˆ
AlekseyPX

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, F=3.33\times 10^{-21}\ N

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

F=k\dfrac{q^2}{r^2}

q=\sqrt{\dfrac{Fr^2}{k}}

q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}

q=1.21\times 10^{-16}\ C

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

n=\dfrac{q}{e}

n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

8 0
3 years ago
A mass of 1 kg is moving in a circle of radius 3.3 meters. What is the liner velocity v m/s that would give a Centripetal Force
Ghella [55]

Answer:

The linear velocity of the object is 8.71 m/s.

Explanation:

Given;

mass of the object, m = 1 kg

radius of the circle, r = 3.3 meters

centripetal force, F = 23 N

Centripetal force is given by;

F_c = \frac{mv^2}{r}\\\\

where;

v is the linear velocity of the object

F_c = \frac{mv^2}{r}\\\\mv^2 = F_cr\\\\v^2 = \frac{F_cr}{m}\\\\v= \sqrt{\frac{F_cr}{m}} \\\\v= \sqrt{\frac{23*3.3}{1}}\\\\v = 8.71 \ m/s

Therefore, the linear velocity of the object is 8.71 m/s.

4 0
3 years ago
Other questions:
  • What are two ways air resistance can be increased?
    9·1 answer
  • A 15 cm3 block of gold weighs 2.8 N. It is carefully submerged in a tank of mercury. One cm3 of mercury weighs 0.13 N.
    7·1 answer
  • The Sojourner Mars rover has a weight of 42.7 N on Mars where the acceleration due to gravity is 3.72 m/s2. What is Sojourner's
    12·1 answer
  • Which of these statements is most likely correct about the Nebular theory of formation of planets?
    7·1 answer
  • How do different lense types connect to helping locate sharks through mechanical radio waves?​
    6·1 answer
  • How does natural selection produce change in a population of mice?
    7·1 answer
  • Imagine that person B is more massive than person A in the picture above.
    12·1 answer
  • What happens as lakes get older?
    12·1 answer
  • HELP ME PLEASE BRAINIEST IF CORRECT!!!!!
    8·2 answers
  • 005 (part 1 of 2) 10.0 points
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!