A stone with a mass of 0.600 kg is attached to one end of a string 0.900 m long. The string will break if its tension exceeds 50.0 N. The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed.
A. Draw a free-body diagram of the stone.
B. Find the maximum speed the stone can attain without breaking the string.

Answer:

The what?

Explanation:

Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c)   Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

          Em₀ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₀ = Em_f

         Em₀ = m g h₀

b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

          Em₁ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₁ = Em_f

         Em₁ = m g h

d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

Answer:

a. i) The pressure drop per unit length is 52,151.89 Pa

ii) The atmospheric pressure ≈ 19.175 × The pressure drop along 10 cm length of aorta

b i) The power required for the heart to push blood along a 10 cm length of aorta, is 2.6075945 Watts

ii) The basal metabolic rate ≈ 38.35 × The power to push the blood along a 10 cm length of aorta

c. i) Please find attached the drawing for the velocity profile created with Microsoft Excel

ii) The velocity at the center is approximately 2.04 m/s

Explanation:

The given diameter of the aorta, D = 2.5 cm = 0.025 m

The maximum flow rate, Q = 500 cm³/s = 0.0005 m³/s

Assumptions;

The blood flow is laminar

The blood is a Newtonian fluid

The viscosity of water ≈ 0.01 poise = 1 cp

a. i) The pressure drop per unit length of pipe ΔP/L is given by the Hagen Poiseuille equation as follows;

[tex]Q = \dfrac{\pi \cdot R^4}{8 \cdot \mu} \cdot \left(\dfrac{\Delta p}{L} \right)[/tex]

Where;

Q = The flow rate = A·v

A = The cross sectional area

R = The radius = D/2

Δp/L = The pressure drop per unit length of the pipe

Therefore, we have;

[tex]\dfrac{\Delta p}{L} = \dfrac{Q\cdot 8 \cdot \mu }{\pi \cdot R^4} = \dfrac{0.0005 \times 8 \times 1}{\pi \times 0.0125^4 } = 52151.89[/tex]

The pressure drop per unit length ΔP/L = 52,151.89 Pa

ii) The pressure, ΔP, drop along 10 cm (0.1 m) length of aorta = ΔP/L × x;

∴ ΔP = 52,151.89 Pa × 0.1 m = 5,215.189 Pa

Given that the atmospheric pressure, [tex]P_{atm}[/tex] = 10⁵ Pa, we have;

[tex]P_{atm}[/tex]/ΔP = 10⁵/5,215.189 ≈ 19.175

Therefore, the atmospheric pressure is approximately 19.175 times the pressure drop along 10 cm length of aorta

b. i) The power, P = Q × ΔP

Therefore, the power required for the heart to push blood along a 10 cm length of aorta, is P₁₀ = 0.0005 m³/s × 5,215.189 Pa = 2.6075945 Watts

ii) Therefore compared to the basal metabolic rate of, 'P', 100 W, we have;

P/P₁₀ = 100 W/2.6075945 Watts = 38.349521 ≈ 38.35

The basal metabolic rate is approximately 38.35 times more powerful than the power to push the blood along a 10 cm length

c. i) The velocity profile across the aorta is given as follows;

[tex]v_m = \dfrac{1}{4 \cdot \mu} \cdot \dfrac{\Delta P}{L} \cdot R^2[/tex]

Where;

[tex]v_m[/tex] = The velocity at the center

We get;

[tex]v_m = \dfrac{1}{4 \times 1} \times 52,151.89 \times 0.0125^2 \approx 2 .04[/tex]

The velocity at the center, [tex]v_m[/tex] ≈ 2.04 m/s

ii) The velocity profile, v(r), is given by the following formula;

[tex]v(r) = v_m \cdot \left[1 - \dfrac{r^2}{R^2} \right][/tex]

Therefore, we have;

[tex]v(r) = 2.04 - \dfrac{2.04 \cdot r^2}{0.0125^2} \right] = 2.04 - 163\cdot r^2[/tex]

The velocity profile of the pipe is created with Microsoft Excel

Answer:

Vf = 0.0108 m³

Explanation:

Assuming there is no change in internal energy, we can calculate the changed volume by the following formula:

[tex]Energy\ Supplied = Work = W = P \Delta V \\W = P(V_{f}-V_{i})\\[/tex]

where,

W = Work = Energy Supplied = 1000 J

P = Pressure = 101325 Pa

Vf = Final Volume = ?

Vi = Initial Volume = 0.001 m³

Therefore,

[tex]1000\ J = (101325\ Pa)(V_{f}-0.001\ m^3)\\\\V_{f}-0.001\ m^3 = \frac{1000\ J}{101325\ Pa}\\\\V_{f} = 0.001\ m^3 + 0.0099\ m^3[/tex]

Vf = 0.0108 m³

Answer:

The Cell Membrane is found in both plant and animal cells.

Explanation:

Chloroplast, Large Vacuole and Cell Wall are all found in plant cells.

Answer:

it would be 83% in a fatal crash.

Answer:

1.11 m.

Explanation:

Why?

The speed for a wave is done by the equation:   v = f * w. Because the frecuency tells us about how many cycles the wave makes each time, but for each cycle the wave runs certain distance, given for the wavelenght. If you isolate the letter w you get the value just doing a ratio.

v = speed

f = frecuency

w = wavelenght

w = v / f

Answer:

ask internet?

Explanation:

easy .....

..

Answer: 13.9 m

Explanation:

Answer:

13.9m

Explanation:

Answer on Edge

Answer:

[tex]0.005847\ \text{s}[/tex]

Explanation:

Radio waves travel at the speed of light

[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]

[tex]d_r[/tex] = Distance between two radios = 46 km

[tex]v[/tex] = Speed of sound in air = 348 m/s

[tex]d_s[/tex] = Distance sound travels across the newsroom = 2.1 m

Time taken by the radio signal to reach the required location is

[tex]t_r=\dfrac{d_r}{c}\\\Rightarrow t_r=\dfrac{46\times 10^3}{3\times 10^8}\\\Rightarrow t_r=0.000153\ \text{s}[/tex]

Time taken by sound to reach the required location is

[tex]t_s=\dfrac{d_s}{v}\\\Rightarrow t_s=\dfrac{2.1}{348}\\\Rightarrow t_s=0.006\ \text{s}[/tex]

The time difference is

[tex]t_s-t_r=0.006-0.000153=0.005847\ \text{s}[/tex]

The difference in time that the message is received is [tex]0.005847\ \text{s}[/tex].

Answer

A. Half elementary charge

Explanation:

Answer:

Walking –We can walk only if we apply frictional force. Friction is what holds your shoe to the ground. The friction present on the ice is very little, this is the reason why it is hard to walk on the slippery surface of the ice.

Writing – A frictional force is created when the tip of the pen comes in contact with the surface of the paper. Rolling friction is what comes into play while writing with a ballpoint pen while sliding friction arises when one writes with a pencil.

Skating – A thin film of water under the blade is necessary to make the skate slide. The heat generated by the skate blade rubbing against the surface of ice causes some of the ice to melt right below the blade where the skater glides over the ice. This water acts as a lubricant reducing friction.

Lighting a matchstick – When the head of the matchstick is rubbed against a rough surface, heat is generated and this heat converts red phosphorous to white phosphorous. White phosphorous is highly inflammable and the match stick ignites. Sometimes, matchsticks fail to ignite due to the presence of water. Water lowers friction.

Driving of the vehicle on a surface- While driving a vehicle, the engine generates a force on the driving wheels. This force initiates the vehicle to move forwards. Friction is the force that opposes the tyre rubber from sliding on the road surface. This friction avoids skidding of vehicles.

Applications of breaks in the vehicle to stop it- Friction braking is the most widely used braking method in vehicles. This process involves the conversion of kinetic energy to thermal energy by applying friction to the moving parts of a vehicle. The friction force resists the motion and in turn, generates heat. This conversion of energy eventually bringing the velocity to zero.

Flight of aeroplanes- Drag is the force that opposes the forward motion of the aeroplane. The friction which resists the motion of an object moving through a fluid or immobile in a moving fluid, as occurs when we fly a kite. The friction of the air is created as it meets and passes over an aeroplane and its components. Drag is generated by air impact force, skin friction and displacement of the air.

Drilling a nail into the wall- Friction is responsible for fixing of nails in a wall. As the nail is driven into the wall, the nearby material to the nail of the wall gets compressed. This exerts a force on the nail. This force is the friction that converts the normal force exerted by the compressed layers of the wall into the resisting shear force. In this manner the friction cause nails and screws to hold on to walls.

The dusting of the carpet by beating it with a stick- When the carpet is beaten with the stick, the dust comes out. The dust is carried off by the wind or falls on the floor. The carpet exhibits a little static friction that holds the dust to the carpet.  When the carpet is beaten, it will overcome the friction and the carpet will move away from the dust making the carpet free from dust.

Sliding on a garden slide- We know that friction is a force that is present whenever two objects rub against each other. In case of a slide in the garden such as a slide and a person’s backside rub each other’s surface. Without friction, a slide would accelerate the rider too quickly, resulting in possible injury due to the fall. The friction reduces the velocity of the sliding person and makes him stop.

Hope that helps! :D Sorry if it's too lengthy...

Explanation:

Answer:

D. 24.0m

Explanation:

formula

S=d/t

s=speed

d=distance

t=time

s=3.0m/s

d=x

t=8.0s

s=d/t

d=s*t

d=3.0*8.0

d=24.0m

Answer: D. When both objects reach the same temperature.

Explanation: When will heat flow between the objects stop? Heat will always flow from the warmer object to the colder object. The heat transfer will stop when the two objects are at the same temperature and reach thermal equilibrium.

Answer: Heat will always flow from the warmer object to the colder object. The heat transfer will stop when the two objects are at the same temperature and reach thermal equilibrium.

so the answer to your question is c

Explanation:

Answer:

Did you ask the question and answer it for yourself?

Answer:

Your amazing thank you, A was right.

Answer:

it is for sure B. Quantitative data !

Explanation:

As I learned from quizlet!. there your welcome

Answer:

B. Quantity

Explanation:

This is easy

Answer:

Thermal Expansion

Explanation:

These liquid thermometers are based on the principal of thermal expansion. When a substance gets hotter, it expands to a greater volume. Nearly all substances exhibit this behavior of thermal expansion. It is the basis of the design and operation of thermometers.

Answer:

I think it's theoretical.

molecular weights are written in the picture.

CH4<NH3<H2O<Cl2

Answer:

B cus is in the south but that pfp tho

sorry bro but i don't know his i am also bad

Explanation:

but maybe you can you the formula of volatage as well as

Answer:

a) Please see below as the answer is self-explanatory.

b) 1. vertically downward

   2. vertically upward

   3  N/A

   4  vertically downward.

Explanation:

a)

Answer:

Once we place a positive test at a point close to the sphere, we find that an electrostatic force is applied to the outside of the sphere. Therefore, at any point around the sphere, the electric field vector is radially outward.

Explanation:

Finger bones have joints.

Answer: true , finger bones have joints

Explanation:

The energy present in pond water and kerosene are potential energy and chemical energy respectively.

Energy is a quantifiable quantity in physics that can be transferred from an item to do work. Thus, we might characterise energy as the capacity to engage in any kind of physical action. So, to put it simply, we can define energy as: Energy is the capacity to carry out work.

Energy "can neither be created nor destroyed but can only be changed from one form to another," according to the rules of conservation of energy. The SI unit for energy is called a joule.

In pond, water is stored during rain and stored potential energy whereas in kerosene, carbon molecules have chemical energy and when it burns, chemical energy revels as thermal energy.

Learn more about energy here:

https://brainly.com/question/1932868

#SPJ2

Answer:

a)     v₀ = 2.5 m / s,   heading south.

b)  W = 1,219 10⁵ J

Explanation:

a) For this exercise we can use the conservation of the moment, we create a system formed by all the 4 cars, in this case when the last one separates the forces are intense and the moment is conserved

initial instant. Before separation

        p₀ = M v₀

final instant. When uncoupling the last car

        p_f = 3m v₁ + m v₂

where they indicate that the speed of the wagons is v₁ = 2.00 m / s and the speed of the last wagon is v₂ = 4.00 m / s

the total mass is M = 4m

how the moment is preserved

           p₀ = p_f

         4m v₀ = 3m v₁ + mv₂

         v₀ = ¾ v₁ + v₂ / 4

let's calculate

           v₀ = ¾ 2 + ¼ 4

           v₀ = 2.5 m / s

heading south.

b) work is equal to the change in kinetic energy

              W = ΔK = K_f -K_o

              W = ½ m v_f² - ½ m v₀²

               W = ½ m (v_f² -v₀²)

               W = ½ 2.50 10⁴ (4² - 2.5²2)

               W = 1,219 10⁵ J

Answer:

a)  v = 2.9 m / s, b) A = 0.350 m, c)    v = 2.9 m / s, d)   A = 1.00 m

Explanation:

The oscillatory motion is described by the expression

          x = A cos (wt + Ф)

the wavelength which is the distance for the wave to repeat and the frequency which is the number of times a wave oscillates per unit of time

a) In this part they ask us for the speed of the wave.

Let's use the relationship between speed, wavelength and frequency

          v = λ f

For the wavelength they indicate that the distance between two crest is 6.1 m

        λ / 2 = 6.10

        λ = 12.20 m

They give us the period of the wave is the time it takes to return to the same point, in this case they give half a period

       A / 2 = 2.10

       A = 4.20 me

        f = 1 / t

        f = ¼, 2

        f = 0.238 Hz

let's calculate

         v = 12.20 0.238

         v = 2.9 m / s

b) the amplitude of the wave, is the distance from zero to some maximum

                 2A = 0.700

                   A = 0.350 m

c) the speed of the wave is not function of the amplitude, so the speed is the same

           v = 2.9 m / s

d) the amplitude is

           2A = 0.50

             A = 1.00 m

Answer:

Work= Fcos∆×S

W=60N×cos 30⁰×100

W=60×0.866×100=5196.1J

PLEASE GIVE BRAINLIEST

Answer:

e)

Explanation:

i)

        [tex]V = I*r_{i} + I*R_{1} (1)[/tex]

      [tex]I_{i} = \frac{V}{r_{int} + R_{i}} = \frac{1.5V}{0.1 \Omega + 1.4 \Omega} = 1.0 A (2)[/tex]

ii)

       [tex]I_{ii} =\frac{V_{term} }{R_{ii} } =\frac{3.6V}{1.8 \Omega} = 2.0 A (3)[/tex]

iii)

      [tex]I_{iii} =\frac{V-V_{term}}{r_{int} } =\frac{12.0 V- 11.0 V}{0.2 \Omega} =\frac{1.0V}{0.2\Omega} = 5.0 A (4)[/tex]