a. If 1gcm²S-¹ = x Js. Find the value of x.​

The car that starts from rest and travels 243 m in 7.86 s with constant acceleration has a velocity when it crosses the finish line of: 61.82 m/s

The formulas and procedures we will use to solve this exercise are:

Where:

Information about the problem:

Applying the acceleration formula we have:

a = 2 * [x - (vi * t)] / t²

a = 2 * [243 m - (0 m/s * 7.86 s)] / (7.86 s)²

a = 2 * [243 m - (0 m)] / 61.779 s²

a = 2 * [243 m] / 61.779 s²

a = 486 m/ 61.779 s²

a = 7.866 m/s²

Applying the final velocity formula we have:

vf = vi + (a * t)

vf = 0 m/s + (7.866 m/s² *7.86 s)

vf = 0 m/s + 61.82 m/s

vf = 61.82 m/s

It is a physical quantity that indicates the variation of velocity as a function of time, it is expressed in units of distance per time squared e.g.: m/sec2 ; km/h2

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It is frequently necessary to produce volume measurements in quantitative chemistry with an error of 0.1%, or one part per thousand.

Utilizing volumetric glassware with a known capacity down to a few hundredths of a milliliter, or approximately 0.01 mL, is required for this.

Five different types of glassware, including graduated cylinders, beakers, volumetric pipets, burets, and volumetric flasks, are frequently used to measure out precise volumes.

The most exact glassware are volumetric pipets, flasks, and burets; the glassware manufacturers calibrate these to a high degree of accuracy.

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A piano string having a mass per unit length equal to 5.20 10^-3 kg/m is under a tension of 1 250 n, then speed with which the wave travels is 490.29 m/sec

Wavelength x Frequency equals speed. When the wavelength and frequency values are known, this equation can be used to determine the wave speed.

If the wave speed and the other value are known, the equation for wave speed can be written to solve for wavelength or frequency. Unit = meters/second

The distance a wave shape travels divided by the time it takes to cover that distance is its velocity. Also known as wave velocity V. Symbol meters/second are the units (ms-1).

The speed of the wave's doubles when the frequency of a wave source is doubled.

V= sqrt T/u = sqrt [1250/5.20 x 10^-3]

= 490.29 m/sec

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By using uniform motion, the distance is 66.64 meters.

We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule

vt = vo + a . t

vt² = vo² + 2a . s

s = vo . t + 1/2 . a . t²

where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.

From the question above, we know that

vo = 0 m/s

vt = 100 km/h = 100000/3600 m/s =  27.78 m/s

t = 4.8 s

Calculate the acceleration

vt = vo + a . t

27.78 = 0 + a . 4.8

4.8a = 27.78

a = 27.78 / 4.8

a = 5.79 m/s²

Find the distance

vt² = vo² + 2a . s

27.78² = 0² + 2 . 5.79 . s

11.58s = 771.73

s = 66.64 m

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Answer:

None of the choices is appropriate.

Explanation:

By using uniform motion, the car displacement is 225 meters.

We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule

vt = vo + a . t

vt² = vo² + 2a . s

s = vo . t + 1/2 . a . t²

where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.

From the question above, we know that

a = 2 m/s²

vo = 0 m/s

t = 15 s

Use the third equation to calculate the displacement

s = vo . t + 1/2 . a . t²

s = 0 . 15 + 1/2 . 2 . 15²

s = 0 + 225

s = 225 m

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The maximum speed at which a car can negotiate a flat curve with radius 205m at maximum acceleration of 5.0m/s2 is 32m/s.

Given data

Radius of curve = 205m

Speed (v) =?

Acceleration (a) = 5.0m/s2

The expression for the centripetal force acting on the car

F_c=(mv^2)/r

The expression for the force acting on the car.

F=ma

             Equating the force acting on the car to centripetal force,

F_c=F

(mv^2)/r= ma

            Making v the subject of the equation above

v=√ar  

          Putting the values of a and r into the expression above

v=√(5×205)  

v=√1025  

v=32.02 m/s

Approximately,

v=32 m/s

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The tension in the rope which a climber, that has a mass of 60 kg, is hang motionless is 588 N

The tension formula and the procedure we will use is:

T = m * g

Where:

Information of the problem:

Knowing that the body is in equilibrium the tension of the rope is equal to the gravity force. Applying the tension formula and substituting values we have:

T = Fg

Fg = m * g

Fg = 60 kg * 9.8 m/s²

Fg = 588 N

It is the physical magnitude that expresses the effort necessary to move the mass of one kilogram with an acceleration of one meter per second squared, this is expressed in units of the international system in Newton.

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Answer:

Collateral

Explanation:

The final velocity of a snowball dropped from a cliff in the Rocky Mts. that falls for 17s is 166.6 m/s

When a  body is dropped from a height, its initial velocity is 0. A body executing free fall moves with constant acceleration i.e acceleration due to gravity

Given data -

By using 1st equation of motion:

   v = u + at

Equation of motion under gravity :

   v = u + gt ( g = 9.8 m/s² )

      = 0 + 9.8 × 17

      = 166.6 m/s

Thus we can conclude that the final velocity of a snowball dropped from the cliff is 166.6 m/s

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The approximately time that takes to stop the stunt driver once she hit the cushion is: 0.4 s

To solve this exercise, the formulas and procedures to be applied are:

Where:

Information about the problem:

Applying the force formula and isolating the acceleration, we get:

F = m * a

a = F/m

a = -2500 N /50kg

a = -2500 kg * m/s² / 50 kg

a = -50 m/s²

Applying the time formula we get:

t = (vf – vi) / a

t = (0 m/s – 20 m/s) / 50 m/s²

t = (– 20 m/s) / -50 m/s²

t = 0.4 s

It is the physical magnitude that expresses the effort necessary to move the mass of one kilogram with an acceleration of one meter per second squared, this is expressed in units of the international system in Newton.

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The car will travel 69.3m while accelerating, using the equation of motion.

Using the equation of motion:

[tex]v^{2}- u^{2} = 2as[/tex]

[tex]s = \frac{v^{2}-u^{2} }{2a}[/tex]

where, s = displacement

u = initial velocity

v = final velocity

a = acceleration

Plugging in the values we get,

[tex]s = \frac{21^{2} - 5^{2} }{2*3.0}[/tex]

s = 69.3meters

Therefore, the car will travel 69.3m while accelerating, using the equation of motion.

The complete question is: A car accelerates from 5.0 m/s to 21 m/s at a constant rate of 3.0 m/s2, How far does it travel while accelerating?

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The velocity the bullet used in leaving the gun is 586 m/s (Option 1)

We'll begin by obtaining the time taken for the bullet to hit the ground. This is illustrated below::::

h = ½gt²

47 = ½ × 10 × t²

47 = 5 × t²

Divide both side by 5

t² = 47 / 5

Take the square root of both side

t = √(47 / 5)

t = 3.07 s

Finally, we shall determin the the velocity. This is illustrated below:

s = ut

1800 = u × × 3.07

Divide both sides by 3.07

u = 1800 / 3.07

u = 586 m/s

Thus, the velocity the bullet used in leaving the gun is 586 m/s (Option 1)

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The  given coefficient of linear expansion of glass is [tex]9 \times 10^{-6}[/tex]/°C.

change in area =[tex]a_1[/tex][tex]\times[/tex]β[tex]\times[/tex]ΔT

where β=2α

so,

[tex]60\times 150\times2\times9\times10^{-6}\times36 = 5.832cm^{2}[/tex]

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a) The velocity when it return to the ground is 9 m/s

b) The total time taken is 1.84 s

c) The highest distance covered is 4.2 m

We have to know that when an object has been projected upwards or downwards that the movement of the objects could be described by the use of the equations of motion under gravity. In that case we have to look at that tennis ball is thrown vertically upward with an initial velocity of +9.0

m/s as an object that is undergoing motion under gravity.

We now have;

The time taken for the ball to reach the starting point;

v = u - gt

At the maximum height the ball is stationary thus v = 0 m/s

t =u/g

t = 9.0 m/s / 9.8 m/s^2

t = 0.92 s

Given that the ball went up and came down the time taken to return to starting point = 2(0.92) = 1.84 s

The highest distance covered by the ball is obtained from;

h = ut - 1/2gt^2

h = 9 * 0.92 - (0.5 * 9.8 * (0.92)^2)

h = 8.3  - 4.1

h = 4.2 m

The highest distance covered is 4.2 m

The balls speed when it returns;

Recall that it now falls from a height thus u = 0 m/s

v = u + gt

v = gt

v = 9.8 * 0.92

v = 9 m/s

The velocity when it return to the ground is 9 m/s

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A boost chopper is powered by a 12-V DC power supply.

The output voltage range of this chopper if the duty cycle vary between 20 and 95% will be 15 - 240V

How to calculate the output voltage?

Output voltage can be calculated by the following formula:

[tex]V_0=\frac{V_i}{1-D}[/tex]

Here,

[tex]V_i[/tex] is the initial power supply.

D = Duty cycle

Calculating output voltage for 20%

[tex]V_0=\frac{12}{1-0.20} \\V_0=-\frac{12}{0.8} \\V_0=15[/tex]

Calculating output voltage for 95%

[tex]V_0=\frac{12}{1-0.95} \\V_0=-\frac{12}{0.05} \\V_0=240[/tex]

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You need to analyse what is possible and get the answer.

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The height of the cliff is [tex]$90.69 \mathrm{~m}$[/tex]

The initial speed of the stone vi = 10m/s

The stone is also moving downward vertically under the effect of the gravitational acceleration [tex]$g=9.81 \mathrm{m} / \mathrm{s}^2$[/tex]

Thus, the stone will have two velocity components while moving. One component is the horizontal speed which will be constant during motion and the other component is the vertical speed which will increase under the effect of gravitational acceleration.

Hence, the final vertical speed when the stone hits the ground is

[tex]$$\begin{aligned}&V_y=V_{y o}+g t \\&V_y=0+9.81(4.3)=42.183 \mathrm{~m} / \mathrm{s}\end{aligned}$$[/tex]

The height of the cliff where the stone is thrown is

[tex]$$\begin{aligned}&h=\frac{\left(V_y\right)^2-\left(V_{y o}\right)^2}{2 g} \\&h=\frac{42.183^2-(0)^2}{2(9.81)} \\&h=90.69 \mathrm{~m}\end{aligned}$$[/tex]

So, the height of the cliff is [tex]$90.69 \mathrm{~m}$[/tex]

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A low hydrostatic pressure combined with high oncotic pressure will result in fluid overload.

Hydrostatic Pressure is a specified pressure that refers to a particular force that is exerted by the fluid present inside the blood capillaries as it is running against the capillary wall. Oncotic Pressure is also a type of pressure that refers to the force which is exerted by albumin as well as other proteins existing in the same blood vessel.

Osmotic agents are involved in the process of increasing oncotic pressure existing in the blood. This results in the pulling of water from tissues which helps in increasing the volume of the blood more acutely. Then the increased blood volume will result in inhibition of renin release as it increases renal blood flow.

During a situation of prolonged water restriction or when it is after water loss due to sweat condition, the blood concentration will increase so that its oncotic pressure will also increase. Since this pressure is opposing the Glomerular filter rate, dehydration results in reducing the GFR due to the reduction taking place in net driving force. Thus fluid overload will be caused by a low hydrostatic pressure combined with high oncotic pressure.

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Answer:

Доброе утро, ответ 5+5="4

Explanation:

[tex]\frac{-2}{15} \\radians per secondExplanation:Denoting the distance in feet between the wall and the base of the ladder by x and the angle in radians between the ladder and the ground by y, it is notedcos(y)=x/10which impliesy=arccos(x/10)Denoting time in seconds by t, it is further noted that[/tex]

[tex]\frac{dy}{dt}= \frac{dy}{dx}\frac{dx}{dt} (chain rule)\\ Noting (using standard table of derivatives for convenience)\\\frac{dy}{dx} = -\frac{0.1}{\sqrt{1-0.01x^{2} } } \\\frac{dx}{dt} =0.8 f/sec\\so,\\\frac{dy}{dt}(8)=\frac{dy}{dx}\frac{dx}{dt} = -\frac{0.08}{\sqrt{1-0.01(64)} } \\ = -\frac{0..8}{\sqrt{0.36} } =\frac{8}{60} = \frac{-2}{15 } radian/sec[/tex]

The radian is a term of angular measurement established so that an arc with an arc length of 1 is produced by an angle over one radian area enclosed from the center of a unit circle.

When measuring angles, the radian is frequently employed in physics. For instance, angular velocity is frequently expressed in terms of radians per second (rad/s). The reciprocal of one revolution per second is two radians per second.

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By ideal gas approximation, the final pressure is 75.76 atm.

We need to know about the ideal gas theory to solve this problem. The ideal gas is assumed that there is no interaction between particles in a gas. It can be determined by the equation

P . V = n . R . T

where P is pressure, V is volume, n is the number of moles gas, R is the ideal gas constant (8.31 J/mol.K) and T is temperature.

From the question above, we know that

P1 = 50 atm

T1 = 330 K

T2 = 500 K

When the initial and final volume is the same, we can use the ratio of pressure and temperature as

P1 / P2 = T1 / T2

50 / P2 = 330 / 500

P2 = 50 x 500 / 330

P2 = 75.76 atm

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d(u, v) = [tex]\sqrt{34}[/tex]

An inner product of a real vector space V is an assignment that for any two vectors u, v ∈ V, there is a real number hu, vi,

d(u, v) = ║u - v║ = [tex]\sqrt{(u_{1} - v_{1} )^{2} + (u_{2} - v_{2} )^{2}[/tex]

<u = v> = 3u₁v₁ + u₂v₂

u = (-3,9)

v = (0,4)

So, d(u, v) = [tex]\sqrt{(-3 0)^{2} + (9 - 4)^{2}[/tex]

                = [tex]\sqrt{9 + 25}[/tex] = [tex]\sqrt{34}[/tex]

d(u, v) = [tex]\sqrt{34}[/tex]

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This suggests the percentage of those who come lower back on your blog each and every week after their first visit.

As an example, if this week you had one hundred human beings go to your blog and inside the week one column, it suggests seventeen percent. which means of the initial a hundred human beings, seventeen got here returned. underneath week two in case you see eight percent, that means of the preliminary a hundred human beings, eight humans came lower back in two weeks.

This quantity will keep getting smaller, but the goal is to get humans back as often as viable. That increases belief, social shares, and the capability of people to link to you, and it even will increase the odds that the vacationer will convert right into a customer. The Pageviews column tells you ways generally each individual blog published was regarded in the decided on time-frame. The specific Pageviews column simply to the right of it tells you ways a lot of those pageviews were from precise visitors

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Answer:

a phase change occurs when the pressure on the system changes. during pressure change on the system changes the object is going through the phase change of melted then solid, melting, liquid, and then vaporization. sense it is going from a solid state to a melting state. the particles are Freezing: liquid to solid. Melting: solid to liquid. Condensation: gas to liquid. Vaporization: liquid to gas. Sublimation: solid to gas. Deposition: gas to solid. during the a phase change the energy of the particles are The energy that is changing during a phase change is potential energy. During a phase change, the heat added increases or released decreases will allow the molecules to move apart or come together. a phrase change occurs because a phase change is when matter changes to from one state to another example: solid to liquid. another reason why a phase change occurs is because water is a substance that has many interesting properties that influence its phase changes. Most people learn from an early age that water melts from ice to liquid at 0°C, and boils from a liquid to a gas at 100°C  but this isn't true in all circumstances. The pressure affects these transition points, so for water, the boiling point actually decreases as the pressure decreases.

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The notion that every stimulus pattern is seen in such a way that the resulting structure is as simple as possible is called the law of pragnanz

The law of prägnanz is also known as the law of simplicity or the law of good figure . This law states that when presented with a group of ambiguous or complicated objects, your brain will simplify them as much as possible. When presented with the Olympic logo, for example, you see overlapping circles rather than a jumble of curved, connected lines.

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The maximum height to which the potato will rise is 13.06 m

From the question given above, the folloing data were obtained:

The maximum height can be obtaibed as illustarted below

v² = u² – 2gh (since the ball is going against gravity)

0² = 16² – (2 × 9.8 × h)

0 = 256 – 19.6h

Collect like terms

0 – 256 = –19.6h

–256 = –19.6h

Divide both side by –19.6

h = –256 / –19.6

h = 13.06 m

Thus, the potato will rise to a height of 13.06 m

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Answer:

SYRINGE B PLUNGER FORCE=3*2=6

Force produced by plunger B is 6 Newtons

The correct option is A.

Sour receptors help in the detection of the presence of acids in substances, whereas salt receptors help in the detection of sodium in substances.

A sour type of taste is considered as a quality of primary task that makes an innate rejection type of response in humans as well as in many other animals. Generally, Acidic stimuli are certain unique sources of a sour taste as whenever a  rejection response might serve a discouragement of ingestion of foods which is spoiled by microorganisms that produce acids. The mechanisms by which an acid excites receptor cells of taste (TRCs) are actually complicated by the variability of wide species and are within species that are apparently different kinds of mechanisms for the case of strong and weak acids in substances.

The presence of epithelial sodium channels mediates the mechanism of salt taste. Usually, these receptors are open, and whenever foods are ingested with a high concentration of salt, sodium starts to flow into the cell which causes depolarization. This change occuring in the membrane potential then opens to voltage-gated sodium as well as calcium channels. The calcium influx is in an increased state that causes releasing of serotonin which is filled by vesicles. The serotonin aids in the acting of afferent taste as by axon which causes depolarization and action potentials.

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