why dont feel the earth rising up as we throw a stone upwards and falling down?

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

[tex]C=2\pi r[/tex]    (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

[tex]r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m[/tex]

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

[tex]L=Lo[1+\alpha \Delta T][/tex]         (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

[tex]L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m[/tex]

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

[tex]r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m[/tex]

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

Explanation:

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las personas junto a él respiran para calentarlo con su aliento y aumentar su temperatura 1 grado Celsius. El tubo se hace más largo. También ya no queda ajustado. ¿A qué distancia sube sobre sobre el nivel del suelo? (solo tomar en cuenta la expansión radial al centro de la tierra, y aplicar la fórmula geométrica que relaciona la circunferencia C con el radio r: C= 2πr).

Answer:

the water is going to boil and the mercury ill melt and shoot the cork out the bottom of the tube

Explanation:

Answer:

a. Convert  120 × 10⁸ g to i mg = 1.2 × 10¹³ mg ii. to g = 1.2 × 10⁷ kg

b. Convert 200000 × 10³ m to i. micrometers = 0.2 × 10³ μm  ii. to cm = 2 × 10⁶ cm iii. to km = 2 × 10⁵ km

Explanation:

a. i. To convert the mass = 120 × 10⁸ g to mg, We know that 1000 mg = 10³ mg = 1 g, Since we are converting to mg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10³ mg = 120 × 10¹¹ mg = 1.2 × 10² × 10¹¹ mg = 1.2 × 10¹³ mg

ii. To convert the mass = 120 × 10⁸ g to kg, We know that 1000 g = 10³ g = 1 kg, 1 g = 10⁻³ kg. Since we are converting to kg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10⁻³ kg = 120 × 10⁵ kg = 1.2 × 10² × 10⁵ kg = 1.2 × 10⁷ kg

b. i.To convert the length = 200000 × 10³ m to micrometers, We know that 1/1000000 μm = 10⁻⁶ mg = 1 m, Since we are converting to micrometers, μm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ ×  1/1000000 μm = 200000/1000000 × 10³ μm = 0.2 × 10³ μm

ii. To convert the length = 200000 × 10³ m to cm, We know that 100 cm = 10² cm = 1 m, 1 m = 10⁻² cm = 1/100 cm. Since we are converting to cm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ × 1/100 cm = 200000/100 × 10³ cm = 2000 × 10³ cm  = 2 × 10³ × 10³ cm = 2 × 10⁶ cm

iii. To convert the length = 200000 × 10³ m to km, We know that 1000 m = 10³ m = 1 km, 1 m = 10⁻³ km = 1/1000 km Since we are converting to km, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ ×  1/1000 km = 200000/1000 × 10³ km = 200 × 10³ km = 2 × 10² × 10³ km = 2 × 10⁵ km

Answer:

force = 45.12N

Explanation:

f = ma

f = 0.06 x 752 = 45.12N

Answer:

d. Both kinetic and sliding friction

Explanation:

Kinetic friction, commonly known as sliding friction, happens when a body with its surfaces in contact is in relative motion with another. It's the frictional force slowing it down, and finally stopping a moving body. One can describe sliding friction as the resistance any two objects create while sliding against each other. It is often documented as the force required to hold a surface moving along another surface. It is determined by two variables- one is material of the object and another is its weight.

Answer:

Depending on the relative position of the Earth the Sun and Neptune in the Earths orbit the distances are;

The closest (minimum) distance of Neptune from the Earth is 29 AU

The farthest (maximum) distance of Neptune fro the Earth is 31 AU

Explanation:

The following parameters are given;

The distance from the Earth to the Sun = 1 AU

The distance of Neptune from the Earth = 30 AU

We have;

When the Sun is between the Earth and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 + 1 = 31 AU

When the Earth is between the Sun  and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 - 1 = 29 AU

Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU

The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.

Answer:

29 AU

Explanation:

Answer:

Explanation:

We shall apply the concept of impulse which is given as follows .

Impulse = force x time

Impulse = change in momentum

If u be the initial velocity of golf ball and v be the final velocity , m be the mass

change in momentum

= mu - ( - mv )

= mu+ mv

If F be the force applied and t be the duration of touch with the ball

Impulse = F x t

F x t = mu + mv

mv = Ft - mu

For given mu , greater the value of t , greater will be the value of v

so v is increased when t is increased .

Increased value of v will help in achieving greater distance attained by

golf ball

Answer:

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

Explanation:

To solve the problem/ we first write the differential equation governing the motion. So,

[tex]m\frac{d^{2} x}{dt^{2} } = -kx \\ m\frac{d^{2} x}{dt^{2} } + kx = 0\\\frac{d^{2} x}{dt^{2} } + \frac{k}{m} x = 0[/tex]

with m = 1 slug and k = 9 lb/ft, the equation becomes

[tex]\frac{d^{2} x}{dt^{2} } + \frac{9}{1} x = 0\\\frac{d^{2} x}{dt^{2} } + 9 x = 0[/tex]

The characteristic equation is

D² + 9 = 0

D = ±√-9 = ±3i

The general solution of the above equation is thus

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity  v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ    where k is an integer

3t  = ±π/6 + 2kπ - 4π/3

t  = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3  or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6. So,

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

The time interval at which the mass will head downward at the velocity of 3 ft/s is t = 5π/18 + 2nπ/3.

Given data:

The mass suspended from spring is, m = 1 slug.

The spring constant is, k = 9 lb/ft.

The magnitude of upward velocity is, v = 3 ft/s.

The magnitude of downward velocity is, v' = 3 ft/s.

The given problem can be resolved by framing a differential equation that governs the motion of spring. The differential equation governing the motion of spring is,

[tex]m \dfrac{d^{2}x}{dt^{2}}=-kx\\\\\\\dfrac{d^{2}x}{dt^{2}}+\dfrac{k}{m}x=0[/tex]

With m = 1 slug and k = 9 lb/ft, the equation becomes

[tex]\dfrac{d^{2}x}{dt^{2}}+\dfrac{9}{1}x=0\\\\\\\dfrac{d^{2}x}{dt^{2}}+9x=0[/tex]  

Now, the characteristic equation is,

D² + 9 = 0

D = ±√-9 = ±3i

And the general solution of the above equation is,

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂²

              = √[(-1)² + (-√3/3)²]

              = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3

and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ    

where k is an integer

3t  = ±π/6 + 2kπ - 4π/3

t  = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3  or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6.

t = 5π/18 + 2nπ/3

here, n is a natural number.

Thus, we can conclude that the time interval at which the mass will head downward at the velocity of 3 ft/s is t = 5π/18 + 2nπ/3.

Learn more about the differential equation here:

https://brainly.com/question/14620493  

Answer:

Force x time = Impulse which is a change in momentum. You can then use this to find the velocity (as you know the ball's initial momentum is 0)

1500 x 1.8 x 10^-3 = 0.045v

v = (1500 x 1.8 x 10^-3)/0.045 = 60 ms^-1

Explanation:

The distance the ball travels is directly related to the force that is applied from the head of the golf club, the angle at which the force is applied, gravity, as well as air and wind resistance. The force of the club head on the ball causes the ball to compress and apply an equal amount of force on the club.

use ke=1/2mv^2 rearrange for V

The golf club is in contact with the golf ball for 1.8 ms and exerts a force of 1500 N on the golf ball. The mass of the golf ball is 0.045kg then the velocity of the golf ball as it leaves the golf club. Velocity 60 m/s

Velocity can be defined as the rate at which the change in the position of the object with respect to time, it is mainly involve in speeding of the object in a specific direction.

Velocity is a vector quantity which shows both magnitude  and direction  and The SI unit of velocity is meter per second (ms-1), the change in magnitude or the direction of velocity of a body is said to be accelerating.

Finding the final velocity is simple but few calculations and basic conceptual knowledge are needed.

For more details regarding velocity, visit

brainly.com/question/12109673

#SPJ2

Answer:

0.03 Joules have been converted into other forms of energy as the direct result of the collision.

Explanation:

Let's start studying the conservation of momentum for the system:

[tex]P_i=P_f\\(0.25\,kg)\,{0.6\,m/s)+(0.5\,kg)\,(0\,m/s)=(0.25\,kg+0.5\,kg)\, v_f \\\\\\ 0.15\,kg\,m/s=0.75\,kg\,\,v_f\\v_f=0.15/0.75\,\,m/s\\v_f=0.2\,\,m/s[/tex]

Now that we know the speed of the newly created object, we can calculate how the final kinetic energy differs from the initial one:

[tex]K_i=\frac{1}{2} (0.25)\,(0.6)^2+\frac{1}{2} (0.5)\,(0)^2=0.045\,\,J\\ \\K_f=\frac{1}{2} (0.75)\,(0.2)^2=0.015\,\,J\\[/tex]

Then, when we subtract one from the other, we can estimate how much kinetic energy has been converted into other forms of energy in the collision:

0.045 J - 0.015 J = 0.03 J

Answer:

The correct option is;

A. The component of the woman's weight along the hillside is larger than the kinetic friction between the woman and the ground

Explanation:

Given that the coefficient of static friction is always larger than the coefficient of kinetic friction, we have that before the wind blew, the component of the woman's weight along the hillside was lesser than the static friction between  the woman and the ground, when the wind blew the total force of the wind and the component of the woman's weight put her into motion such that the acting frictional force was then the kinetic frictional force which was lesser than the kinetic frictional force so the woman continues to slide down the hillside without the wind.

Answer:

The black lines represent the energy absorbed by the electrons.

Explanation:

Atoms emit lights when they are excited. These lights are of particular wavelengths that match with different colors. A series of colored lines appear along with spaces in the middle of the two colors. The middle of the colors is filled with dark spaces. Each spectral line of an element represents a specific characteristic of the element. These colored lines appearing in the form of series are termed to be the atomic spectrum of the element. Identification of the elements is done through the line of the spectrum they possess.

Answer:

(B) The black lines represent the energy absorbed by the electrons.

Explanation:

Answer: Because of the fine bore of the tube.

Explanation:

Temperature is the degree of hotness and coldness. And thermometer is the instrument use to measure temperature.

The two most common types of themometric fluids for thermometer are alcohol and mercury.

What makes a clinical thermometer suitable for measuring small changes in body temperature is because of the fine bore of the tube which makes it possible for small temperature changes to cause large changes in the length of mercury columns, making the thermometer very sensitive to temperature changes.

The most prominent feature of the thermometer is the kink or constriction of bore near the bulb.

Answer:

xxx

Explanation:

Answer:

The magnitude of the average force on the wall during the collision is 6 N.

Explanation:

Given;

mass of snowball, m = 120 g = 0.12 kg

velocity of the snowball, v = 7.5 m/s

duration of the collision between the snowball and the wall, t = 0.15 s

Magnitude of the average force can be calculated by applying Newton's second law of motion;

F = ma

where;

a is acceleration = v / t

a = 7.5 / 0.15

a = 50 m/s²

F = ma

F = 0.12 x 50

F = 6 N

Therefore, the magnitude of the average force on the wall during the collision is 6 N.

Answer:

1. it helps to change the direction.

2. it helps us to walk on ground.

3. it helps the vechils to break while moving.

4. helps in changing one form of enegry to another form. eg when we rub our hands we feel heat energy.

5. it opposites the force.

6. it helps us to change shape of objects.eg we roll the dough to make it roti.

7. it changes the state of body from rest motion.eg when we push any obj from inclined plane it moves.

i all know is just 7..

Answer:

option d is answer because pd is measured in volt.

Answer:

Time, t = 3.2 ms

Explanation:

It is given that,

Mass of basketball, m = 0.622 kg

Initial velocity, u = 4.23 m/s

Final velocity, v = 3.85 m/s

Average force acting on the ball, F = 72.9 N

We need to find the time of contact of the ball with the floor. Let t is the time of contact. So,

[tex]F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-4.23)}{72.9}\\\\t=0.0032\ s\\\\\text{or}\\\\t=3.2\ ms[/tex]

So, the ball is in contact with the floor for 3.2 ms.

Answer:

Liquids

Explanation:

Liquids take up the shape of the container it is poured into but will never change its volume.

Answer:

MARK me brainliest please and follow my page

Explanation:

All you have to do to get the average speed is to calculate the total distance covered and divide it by the total time taken

= 16/18 = 0.88m/s

Average speed = (distance covered) / (time to cover the distance)

For the full 18 seconds described by the graph . . .

Average speed = (16 meters) / (18 seconds)

Average speed = (16 / 18) m/s

Average speed =  0.89 m/s

Answer:

Over tens of millions of years, the giant supercontinent Pangaea began to slowly drift apart, forming the continents as they are known today.

Explanation:

i took the active

Answer:

the continents were once connected as one landmass.

Explanation:

right on edge 2022

Answer:

Explanation:

Force equal to resistive force will be applied for movement . So force applied

F = 220 N .

displacement = 2800 m

work done against resistive force

= force x displacement

= 220 x 2800 J

= 6.16 x 10⁵ J .

Answer:

The answer is Revising

Answer:

D

Explanation:

Revising

Answer:

World Wrestling Entertainment, Inc., d/b/a WWE, is an American integrated media and entertainment company that is primarily known for professional wrestling. WWE has also branched out into other fields, including movies, football, and various other business ventures.

Answer:

This means WORLD WRESTLING ENTERTAINMENT.

Explanation:

This is an american integrated media and entertainment company that is primarily known for PROFESSIONAL WRESTLING.

Answer:

Yes it is normal

Explanation:

When you exercise, your heart beat goes up, resulting in people saying that their heart feels like it is "beating out of their chests".

Answer:

E≅1.2×10^7 N/C

Explanation:

First off I'd like to say that I'm taking "net electric field" to mean that they don't want this answer to be put into vector component form and instead want magnitudes. Sometimes the wording of these questions throws me off, so sorry ahead of time if that's what they want from you!

Edit: I ended up adding it anyways ;P

Since we are observing the net electric field acting at q1, we need to use the formula:  [tex]E=k\frac{q}{r^{2} }[/tex]

And since we are observing the effects of multiple charges at once...

E=ΣE, which just means wee need to add all the observed electric fields together:

ΣE= [tex]k\frac{q2}{r^{2} } +k\frac{q3}{r^{2} }[/tex]

Since we are observing [static] electric fields here, we don't actually need q1's charge. (Though if you wanted to find the net force you would.) Now, before we start plugging values in, let's acknowledge what we know. We know that:

These are actually really nice to have, because now we can simplify our expression to:

[tex]E=k\frac{2q}{r^{2} }[/tex]

Now let's plug in our values and get an answer out.

E= 2(8.99×10^9)(4×10^-5)/(0.24)

Plugging all that in, I get:

E≅1.2×10^7 N/C

If you end up needing the net force, F=(q1)(E). That is, you just multiply the electric field by the value of q1. And again, if your teacher wants the answer in vector component form, then the answer will look different.

Let me know what doesn't make sense, or if I got something wrong. Good luck with AP Phy.!

Edit: I put the component form for my answer in the attachment. I also noticed a small calculator related error in my original answer. I updated that to match the new one.

Explanation:

Given the conditions A,B and C when the pendulum is released, at point A the initial velocity of the pendulum is zero(0), the potential energy stored is maximum(P.E= max),

the conditions can be summarized bellow

point A

initial velocity= 0

final velocity=0

P.E= Max

K.E= 0

point B

initial velocity= maximum

final velocity=maximum

P.E=K.E

point C

initial velocity= min

final velocity=min

P.E= 0

K.E= max

Answer:

a) El caudal de salida del chorro es [tex]1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex].

Explanation:

a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:

[tex]\Delta z = \frac{v_{out}^{2}}{2\cdot g}[/tex]

Donde:

[tex]\Delta z[/tex] - Diferencia de altura, medida en metros.

[tex]g[/tex] - Constante gravitacional, medida en metros por segundo al cuadrado.

[tex]v_{out}[/tex] - Rapidez de salida del chorro, medida en metros por segundo.

Se despeja la rapidez de salida del chorro:

[tex]v_{out} = \sqrt{2\cdot g \cdot \Delta z}[/tex]

Si [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] y [tex]\Delta z = 0.3\,m[/tex], entonces la rapidez de salida del chorro es:

[tex]v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}[/tex]

[tex]v_{out} \approx 2.426\,\frac{m}{s}[/tex]

Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:

[tex]\dot V_{out} = v_{out}\cdot A_{t}[/tex]

Donde:

[tex]v_{out}[/tex] - Rapidez de salida del chorro, medida en metros por segundo.

[tex]A_{t}[/tex] - Área transversal del orificio, medido en metros cuadrados.

[tex]\dot V_{out}[/tex] - Caudal de salida del chorro, medido en metros cúbicos por segundo.

Dado que [tex]v_{out} = 2.426\,\frac{m}{s}[/tex] y [tex]A_{t} = 5\,cm^{2}[/tex], el caudal de salida del chorro es:

[tex]\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}} \right)[/tex]

[tex]\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex]

El caudal de salida del chorro es [tex]1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex].

Answer:

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

Explanation:

Puesto que la bola gira en un círculo vertical, existe claramente una diferencia entre las tensiones debido a la influencia de la gravedad y la tensión que resulta de la aceleración centrípeta experimentada por la masa. La máxima tensión ocurre cuando la bola se encuentra en el nadir (o la sima) del trayecto circular, la cual se describe por la Segunda Ley de Newton:

[tex]T_{max} - m\cdot g = m\cdot \frac{v^{2}}{L}[/tex]

En cambio, la mínima tensión aparece cuando la bola se encuentra en el cénit (o la cima) del trayecto circular, descrita por la misma ley de Newton:

[tex]T_{min} + m\cdot g = m\cdot \frac{v^{2}}{L}[/tex]

Donde:

[tex]T_{min}[/tex], [tex]T_{max}[/tex] - Tensiones mínima y máxima, medidas en newtons.

[tex]m[/tex] - Masa de la bola, medida en kilogramos.

[tex]g[/tex] - Constante gravitacional, medida en metros por segundo al cuadrado.

[tex]L[/tex] - Distancia con respecto al eje de rotación, medida en metros.

[tex]v[/tex] - Rapidez tangencial, medido en metros por segundo.

Se elimina la aceleración centrípeta de ambas expresiones por igualación:

[tex]T_{min} + m\cdot g = T_{max} - m\cdot g[/tex]

Ahora, la diferencia entre las tensiones máxima y mínima es:

[tex]T_{max} - T_{min} = 2\cdot m \cdot g[/tex]

Si [tex]m = 1\,kg[/tex] y [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], entonces:

[tex]T_{max} - T_{min} = 2\cdot (1\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]T_{max}-T_{min} = 19.614\,N[/tex]

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

Answer:

[tex]C_{pb}=0.501\ kJ/kg.K[/tex]

Explanation:

Given that

[tex]m_1=0.35 kg[/tex]

[tex]T_1=-27.5^oC[/tex]

[tex]m_2=0.214 kg[/tex]

[tex]T_2=25^oC[/tex]

[tex]T=16.4^oC[/tex]

We know that

[tex]C_{pw}=4.187 kJ/kg.K[/tex]

By using energy conservation

Heat lost by water = Heat gain by block

[tex]m_2\times C_{pw}\times (T_2-T)=m_1\times C_{pb}\times (T-T_1)[/tex]

[tex]0.214\times 4.187\times (25-16.4)=0.35\times C_{pb}\times (16.4+27.5)[/tex]

[tex]C_{pb}=0.501\ kJ/kg.K[/tex]

Therefore the specific heat of the block will be 0.501 kJ/kg.K