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Answer

I Think Its 150

Answer:

O A. Freudian concepts

Freud (1905) stated that events in our childhood have a great influence on our adult lives, shaping our personality. He thought that parenting is of primary importance to a child's development, and the family as the most important feature of nurture was a common theme throughout twentieth-century psychology (which was dominated by environmentalists theories).

Explanation:

Hope this helped!

O A. Freudian concepts. Freud (1905) stated that events in our childhood have a great influence on our adult lives, shaping our personality.

He thought that parenting is of primary importance to a child's development, and the family as the most important feature of nurture was a common theme throughout twentieth-century psychology (which was dominated by environmentalists theories).

One of the case studies most closely linked with the Austrian psychotherapist Sigmund Freud is the hysterics and treatment of Anna O.

Despite the fact that Freud is intimately identified with Anna O, it is thought that he never actually treated her; instead, Breuer saw the patient.

Therefore, O A. Freudian concepts. Freud (1905) stated that events in our childhood have a great influence on our adult lives, shaping our personality.

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Answer:

Transparent or Translucent

Explanation:

Not my answer but nevertheless

Answer:

Electric potential develops in primary/dry cells through a chemical reaction between the cell plates of the cell. free electrons move from the zinc plate to the copper plate through a conducting material.  

Electric potential develops in Generators via magnetic induction i.e. the movement of a conducting rod through the magnetic field between the poles of the horseshoe magnets produces Electric potential in Generators.

Electric potential develops in thermocouples via heat transfer ; A heat source is applied to the connecting end of the thermocouple strips and this will cause the production electric charges ( potential ) at the free ends

3) Identifying effect of each scenerio

a) The resistance of an electric current will increase when the length of the conductor increases

b) The resistance of an electric current through a body will decrease when the conductor has a larger cross-section

c) The resistance of an electric current through a body will decrease when the temperature of the conductor is cooled

4) The amount of heat lost as heat in Btu = 2.73 Btu

amount of heat lost = 0.8 Wh

convert to Btu = 0.8 Wh / 0.293 = 2.73 Btu   ( note : 1 Btu = 0.293 Wh )

5) The amount of of energy used by a 40 W bulb for 365 days = 146 kWh

Power of bulb = 40 W

Run time = 10 hours * 365 days

∴ amount of energy used = 3650 * 40 = 146 * 10^3 Wh = 146 kWh

Complete Question

a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop? Assume that the car does not skid and that each tire has a radius of 0.307 m. Answer in units of rev.

Answer:

The value is  [tex]N = 109 \ rev[/tex]      

Explanation:

From the question we are told that

    The speed of the car is  [tex]u = 28.4 \ m/s[/tex]

     The constant deceleration experienced is  [tex]a = 1.92 \ m/s^2[/tex]

      The radius of the tire is  [tex]r = 0.307 \ m[/tex]

Generally from kinematic equation we have that

      [tex]v^2 = u^2 + 2as[/tex]

Here  v is the final velocity which is  0 m/s

   So

         [tex]0^2 = 28.4^2 + 2 * 1.92 * s[/tex]

=>      [tex]s = 210.04 \ m[/tex]

Generally the circumference of the tire is mathematically represented as

         [tex]C = 2 \pi r[/tex]

=>      [tex]C = 2 * 3.142 * 0.307[/tex]    

=>      [tex]C = 1.929 \ m[/tex]

Generally the number of revolution is mathematically represented as

         [tex]N = \frac{ s}{C}[/tex]    

=>     [tex]N = \frac{210.04}{1.929}[/tex]

=>     [tex]N = 109 \ rev[/tex]      

Answer: Plates shifting

Explanation: After years and years of plates colliding into solid rock, they slowly become closer together. As recent studies have shown, Africa is currently moving closer to Europe one centimeter every year (one inch every 2.5 years).

Answer:

nvudbwasivnjlscv bwbfvsz

Explanation:

Answer:

6.94446 min

Explanation:

t=s/v

Answer:

light refraction

Explanation:

Answer: Plant A

Explanation: The control group is participants who do not receive the experimental treatment.

Answer:

The distance is [tex]r = 55430496 \ m[/tex]  

Explanation:

From the question we are told that

   The period of the moon [tex]T = 1.2 days = 1.2 * 24 * 3600 = 103680 \ s[/tex]

    The mass of the planet is  [tex]m_p = 9.38*10^{24} kg[/tex]

Generally the period of the moon is mathematically represented as

       [tex]T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }[/tex]

Here G is the gravitational constant with value

        [tex]G = 6.67 *10^{-11} \ N \cdot m^2/kg^2[/tex]

=>   [tex]T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }[/tex]

=>   [tex]103680 = 2 * 3.142 * \sqrt{ \frac{r^3 }{ 6.67*10^{-11} * 9.38*10^{24} } }[/tex]

=>    [tex]272218492.31 = \frac{r^3}{ 6.67 *10^{-11} * 9.38*10^{24}}[/tex]

=>    [tex]r = \sqrt[3]{ 1.7031241*10^{23}}[/tex]j

=>   [tex]r = 55430496 \ m[/tex]        

Answer:

That is false. Take a look at this way. You can push a ball with your own breath, you just need to blow it. And you can pull something from afar with a magnet. It is possible to do both.

Explanation:

Not all physical things can be done only physically. Like I just said, it is possible to use other forces (no, not the dark side one),  such as a magnetic force, displayed by a magnet or anything with a force like so.

Answer:

true

Explanation:

A gene mutation is a permanent change in the DNA sequence of a gene. Mutations can occur in a single base pair or in a large segment of a chromosome and even span multiple genes. Mutations can result from endogenous (occurring during DNA replication) or exogenous (environmental) factors. Germline mutations occur in gametes. Somatic mutations occur in other body cells. Chromosomal alterations are mutations that change chromosome structure. Point mutations change a single nucleotide.

Two major categories of mutations are germline mutations and somatic mutations.

Germline mutations occur in gametes. These mutations are especially significant because they can be transmitted to offspring and every cell in the offspring will have the mutation.

Somatic mutations occur in other cells of the body.

Answer:25 N

Explanation:

in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s is 25N

Speed is the ratio of distance with respect to the time in which the distance was covered. Speed is a scalar quantity as it does not have magnitude only have direction

The formula of speed can be represented as s=d/t, Where, s is the speed in m.s-1, d is the distance traveled in m, t is the time taken in s

Uniform speed is defined when the object covers equal distance at equal time intervals, variable speed is defined as when the object covers a different distance at equal intervals of times.

Average speed is defined as the total distance travelled by an object to the total time taken by the object.

Instantaneous speed is defined as when the object is move with variable speed, then the speed at any instant of time is known as instantaneous speed.

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Answer:

1.2 kg

__________________________________________________________

We are given:

Mass of the block = 2 kg

Coefficient of Static Friction = 0.6

__________________________________________________________

Friction Force on the Block:

Finding the Normal Force:

We know that the normal force will be equal and opposite to the weight of the 2 kg block

So, Normal Force = mg

replacing the variables with the given values

Normal Force = (2)(9.8)                    [Taking g = 9.8]

Normal Force = 19.6 N

Friction force on the Block:

We know that:

Coefficient of Static Friction =  Static Friction Force/Normal Force

replacing the variables

0.6 = Static Friction force / 19.6

Static Friction force = 0.6*19.6 N                 [Multiplying both sides by 19.6]

Static Friction force = 11.76 N

__________________________________________________________

Largest Mass that can Hang:

We know that the Static Friction force is 11.76 N, this means that a force of 11.76 N will be applied to keep the object at rest

So, if the weight of the second block is less than the static friction force, it will hang

Weight of the second block ≤ 11.76

We know that weight = mg

mg ≤ 11.76

m(9.8) ≤ 11.76                                                   [since g = 9.8]

m ≤ 1.2 kg                                                        [dividing both sides by 9.8]

From this, we can say that the maximum mass of the second block is 1.2 Kg

Answer:

The displacement is 6.4m

Explanation:

Step one:

given

we are told that Lucy runs 4 meters to the east,

then 5 meters south.

let the distance east be the displacement in the x-direction, and south be the y-direction

Step two:

The resultant of the x and y displacement is the magnitude of the total displacement z

applying Pythagoras theorem we have

z=√x^2+y^2

z=√4^2+5^2

z=√16+25

z=√41

z=6.4m

Answer:

Sensitivity of biosensor

The biosensor showed good linear correlation in the wide detection range of 0.001–2000 ng/mL with good sensitivity. In addition, it retained its biosensing property for seven days with high reproducibility

Explanation:

Split up the forces into components acting parallel to and perpendicular to the slope. See the attached picture for the reference axes.

The box stays on the surface of the plane, so that the net force acting perpendicular to it is 0, and the only acceleration is applied in the parallel direction.

Let m be the mass of the box, θ the angle the plane makes with the ground, and a the acceleration of the box. By Newton's second law, we have

• net parallel force

∑ Force (//) = W (//) - F = m a

(that is, the net force in the parallel direction is the sum of the parallel component of the weight W and the friction F which acts in the negative direction)

• net perpendicular force

∑ Force (⟂) = W (⟂) + N = 0

Notice that

W (//) = W sin(θ) … … … which is positive since it points down the plane

W (⟂) = -W cos(θ) … … … which is negative since it points opposite the normal force N

So the equations become

W sin(θ) - F = m a

-W cos(θ) + N = 0

Solving for a gives

a = (W sin(θ) - F ) / m

which is good enough if you know the magnitude of the friction force.

If you don't, you can write F in terms of the coefficient of kinetic friction between the box and plane, µ, as

F = µ N

so that

a = (W sin(θ) - µ N ) / m

and the normal force itself has a magnitude of

N = W cos(θ)

so that

a = (W sin(θ) - µ W cos(θ) ) / m

The weight W has magnitude m g, where g is the magnitude of the acceleration due to gravity, so

a = (m g sin(θ) - µ m g cos(θ) ) / m

a = g (sin(θ) - µ cos(θ))

Answer:

The location of the center of gravity of the fourth mass is [tex]\vec r_{4} = (-0.961\,m,-0.779\,m)[/tex].

Explanation:

Vectorially speaking, the center of gravity with respect to origin ([tex]\vec r_{cg}[/tex]), measured in meters, is defined by the following formula:

[tex]\vec r_{cg} = \frac{m_{1}\cdot \vec r_{1}+m_{2}\cdot \vec r_{2}+m_{3}\cdot \vec r_{3}+m_{4}\cdot \vec r_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}[/tex] (1)

Where:

[tex]m_{1}[/tex], [tex]m_{2}[/tex], [tex]m_{3}[/tex], [tex]m_{4}[/tex] - Masses of the objects, measured in kilograms.

[tex]\vec r_{1}[/tex], [tex]\vec r_{2}[/tex], [tex]\vec r_{3}[/tex], [tex]\vec r_{4}[/tex] - Location of the center of mass of each object with respect to origin, measured in meters.

If we know that [tex]\vec r_{cg} = (0,0)\,[m][/tex], [tex]\vec r_{1} = (0,0)\,[m][/tex], [tex]\vec r_{2} = (0, 4.1)\,[m][/tex], [tex]\vec r_{3} = (1.9,0.0)\,[m][/tex], [tex]m_{1} = 6\,kg[/tex], [tex]m_{2} = 1.5\,kg[/tex], [tex]m_{3} = 4\,kg[/tex] and [tex]m_{4} = 7.9\,kg[/tex], then the equation is reduced into this:

[tex](0,0) = \frac{(6\,kg)\cdot (0,0)\,[m]+(1.5\,kg)\cdot (0,4.1)\,[m]+(4.0\,kg)\cdot (1.9,0)\,[m]+(7.9\,kg)\cdot \vec r_{4}}{6\,kg+1.5\,kg+4\,kg+7.9\,kg}[/tex]

[tex](6\,kg)\cdot (0,0)\,[m]+(1.5\,kg)\cdot (0,4.1)\,[m]+(4\,kg)\cdot (1.9,0)\,[m]+(7.9\,kg)\cdot \vec r_{4} = (0,0)\,[kg\cdot m][/tex]

[tex](7.9\,kg)\cdot \vec r_{4} = -(6\,kg)\cdot (0,0)\,[m]-(1.5\,kg)\cdot (0,4.1)\,[m]-(4\,kg)\cdot (1.9,0)\,[m][/tex]

[tex]\vec r_{4} = -0.759\cdot (0,0)\,[m]-0.190\cdot (0,4.1)\,[m]-0.506\cdot (1.9,0)\,[m][/tex]

[tex]\vec r_{4} = (0, 0)\,[m] -(0, 0.779)\,[m]-(0.961,0)\,[m][/tex]

[tex]\vec r_{4} = (-0.961\,m,-0.779\,m)[/tex]

The location of the center of gravity of the fourth mass is [tex]\vec r_{4} = (-0.961\,m,-0.779\,m)[/tex].

Answer:

effects of gravity on the Moon and Earth

Explanation:       Newton found the Moon's inward acceleration in its orbit to be 0.0027 metre per second per second, the same as (1/60)2 of the acceleration of a falling object at the surface of Earth.

Absolute zero, technically known as zero kelvins, equals −273.15 degrees Celsius, or -459.67 Fahrenheit, and marks the spot on the thermometer where a system reaches its lowest possible energy, or thermal motion.

Answer:

diverging lens

Explanation:

Since the nature of the problem of nearsightedness is that the light is focused in front of the retina, a diverging lens will serve to diverge light before it reaches the eye. This light will then be converged by the cornea and lens to produce an image on the retina.

Answer:

The answer is 493.05kJ

Explanation:

given

mass m= 3kg

t1=  10°C

t2= 60°C

Specific heat capacity of ham =3287 cal/kgºC

Required,

The quantity of heat, Q

Step two:

the formula is

Q= mcΔT

substitute

Q= 3*3287(60-10)

Q=9861*50

Q=493050 J

Q= 493.05kJ

Answer:

Answer:

724.3J/Kg.K

Explanation:

CHECK THE COMPLETE QUESTION BELOW

Compute the specific heat capacity at constant volume of nitrogen (N2) gas.and compare with specific heat of liquid water. The molar mass of N2 is 28.0 g/mol.

The specific heat capacity can be computed by using expression below

c= CV/M

Where c= specific heat capacity

M= molar mass

CV= molar hear capacity

Nitrogen is a diatomic element, the Cv can be related to gas constant with 5/2R

Where R= 8.314J/mol.k

Molar mass= 28 ×10^-3Kg/mol

If we substitute to the expression, we have

c= (5R/2)/(M)

=5R/2 × 1/M

=(5×8.314) /(2×28 ×10^-3)

=724.3J/Kg.K

Hence, the specific heat capacity at constant volume of nitrogen (N2) gas is

724.3J/Kg.K

The specific heat of liquid water is about 4182 J/(K kg) which is among substance with high specific heat, therefore specific heat of Nitrogen gas is 724.3J/Kg.K which is low compare to that of liquid water.