A 80 kg parent and a 20 kg child meet at the center of an ice rink. They place their hands together and push. The parent pushes the child with a force of 25 N, what is the force acting on the parent?

Answer:small differences in models can lead to large differences in complex systems

Explanation:   this is the  most accurate phrase  

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:

options B,C,D are true

Explanation:

Answer:

Explanation:

Step one:

given data

speed = 123m/s

radius of circle= 4330m

Step two:

We need to find the circumference of the circle, it represents the distance traveled

C=2πr

C= 2*3.142*4330

C= 27209.72m

Step three:

We know that velocity= distance/time

time= distance/velocity

time= 27209.72/123

time=221.2 seconds

in minute = 221.2/60

time= 3.7 min

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) R n = ( 1 − α ) R si − L ↑ + L ↓ where Rn is the net radiation (W m−2), Rsi is the solar radiation (W m−2), α is the soil surface albedo (α = 0–1

Explanation:

Answer:

a) [tex]H_{in}=39.34 kJ[/tex]

b) Efficiency=76.77%

Explanation:

a)

In order to solve this problem, we can use the following formula:

[tex]H_{in}=H_{out}+W[/tex]

the problem provides us with all the necessary information so we can directly use the formula:

[tex]H_{in}=9.14kJ+30.2kJ[/tex]

[tex]H_{in}=39.34 kJ[/tex]

b) In order to find the efficiency, we can use the following formula:

[tex]Efficiency=\frac{W}{H_{in}}*100\%[/tex]

so we get:

[tex]Efficiency=\frac{30.2kJ}{39.34kJ}*100\%[/tex]

Efficiency=76.77%

Answer:

The fundamental resonance frequency is 172 Hz.

Explanation:

Given;

velocity of sound, v = 344 m/s

total length of tube, Lt = 1 m = 100 cm

height of water, hw = 50 cm

length of air column, L = Lt - hw = 100 cm - 50 cm = 50 cm

For a tube open at the top (closed pipe), the fundamental wavelength is given as;

Node to anti-node (N ---- A) : L = λ / 4

λ = 4L

λ = 4 (50 cm)

λ = 200 cm = 2 m

The fundamental resonance frequency is given by;

[tex]f_0 = \frac{v}{\lambda}\\\\f_0 = \frac{344}{2}\\\\f_0 = 172 \ Hz\\\\[/tex]

Therefore, the fundamental resonance frequency is 172 Hz.

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:

44 grams- 55 grams through the whole day. Probably about 14.6 grams per meal.

Explanation:

Answer:

2 thumbs (ISSA Guide)

Explanation:

Answer:

transition metals im sorry if this was too late

Answer:

Fx = 467.65N

Fy = 270N

Explanation:

Given

Force = 540N

angle of inclination = 40 degree

Horizontal component Fx = Fcos 30

Fx = 540cos30

Fx = 540(0.8660)

Fx = 467.65N

Hence the horizontal component is 467.65N

Vertical component Fy = Fsin 30

Fy = 540sin30

Fy = 540(0.5)

Fy = 270N

Hence the vertical component is 270N

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:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]

[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required:

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]

[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]

Answer:

The distance between the compact car and pickup truck is 0.96048 m

Explanation:

The gravitational force is directly proportional to the product of the masses of the interacting object, it is also inversely proportional to the square of the distance between them.  This is shown in equation 1;

[tex]F =G \frac{m_{1} X m_{2} }{d^{2} }[/tex]............ 1

Where F is the gravitational force = 0.0001 N

G is the gravitational constant = 6.673 x [tex]10^{-11} Nm^{2} kg^{-2}[/tex]

[tex]m_{1}[/tex]  is the mass of the compact car = 900kg

[tex]m_{2}[/tex] is the mass of the pickup truck = 1600kg

d is the distance and its unknown ?

Let us make d the subject formula in equation 1

[tex]d = \sqrt{G\frac{m_{1} m_{2} }{F } }[/tex] .... 2

Substituting into equation 2 we have

[tex]d = \sqrt{\frac{6.673x10^{-11} x 900 x 1600}{0.0001N} }[/tex]

d = 0.96048m

Therefore the distance between the compact car and pickup truck is 0.96048 m

Increase of momentum of the collision will have the car in a unsettling position, creating an unsettling spot for it. Moreover increasing this would most likely take worse effect, it is better than increasing the time, which will only create the car faster. Let me show you what I mean...

Answer:

Non-uniform velocity as the laser light beam has got reflected by the mirror

And as the light got reflected there is a change in velocity making it non-uniform velocity

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:

i don't know the correct answer

but this is what i found on web

Explanation:

Changes of state are physical changes in matter. They are reversible changes that do not change matter's chemical makeup or chemical properties. Processes involved in changes of state include melting, freezing, sublimation, deposition, condensation, and evaporation. Energy is always involved in changes of state.

Answer:

a). a = F/m

Explanation:

Formula is F=ma

Potential energy+Kinetic energy=Total energy

When you release a spring the velocity increases, therefore the kinetic energy increases ke=1/2*mv^2 and the displacement decreases therefore the potential energy decreases pe=1/2*kx^2.

Answer:

light refraction

Explanation:

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.

Answer:

A) 2.650 km

Explanation:

The relationship between acceleration of gravity and gravitational constant is:

[tex]g = \frac{Gm}{R^2}[/tex] ---- (1)

Where

[tex]R = 6,400 km[/tex] -- Radius of the earth.

From the question, we understand that the gravitational field of the rocket is 50% of its original value.

This means that:

[tex]g_{rocket} = 50\% * g[/tex]

[tex]g_{rocket} = 0.50 * g[/tex]

[tex]g_{rocket} = 0.5g[/tex]

For the rocket, we have:

[tex]g_{rocket} = \frac{Gm}{r^2}[/tex]

Where r represent the distance between the rocket and the center of the earth.

Substitute 0.5g for g rocket

[tex]0.5g = \frac{Gm}{r^2}[/tex] --- (2)

Divide (1) by (2)

[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}/\frac{Gm}{r^2}[/tex]

[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}*\frac{r^2}{Gm}[/tex]

[tex]\frac{1}{0.5} = \frac{1}{R^2}*\frac{r^2}{1}[/tex]

[tex]2 = \frac{r^2}{R^2}[/tex]

Take square root of both sides

[tex]\sqrt 2 = \frac{r}{R}[/tex]

Make r the subject

[tex]r = R * \sqrt 2[/tex]

Substitute [tex]R = 6,400 km[/tex]

[tex]r = 6400km * \sqrt 2[/tex]

[tex]r = 6400km * 1.414[/tex]

[tex]r = 9 049.6\ km[/tex]

The distance (d) from the earth surface is calculated as thus;

[tex]d = r - R[/tex]

[tex]d = 9049.6\ km - 6400\ km[/tex]

[tex]d = 2649.6\ km[/tex]

[tex]d = 2650\ km[/tex] --- approximated

Answer:

2008

Explanation:

2000+3+5======2008

Answer:

[tex]8\hat i-2\hat j-2\hat k[/tex]

Explanation:

Vectors in 3D

Given a vector

[tex]\vec P = P_x\hat i+P_y\hat j+P_z\hat k[/tex]

A vector [tex]\vec Q[/tex] parallel to [tex]\vec P[/tex] is:

[tex]\vec Q = k.\vec P[/tex]

Where k is any constant different from zero.

We are given the vectors:

[tex]\vec A = \hat i+4\hat j-2\hat k[/tex]

[tex]\vec B = 3\hat i-5\hat j+\hat k[/tex]

It's not specified what the 'resultant' is about, we'll assume it's the result of the sum of both vectors, thus:

[tex]\vec A +\vec B = \hat i+4\hat j-2\hat k + 3\hat i-5\hat j+\hat k[/tex]

Adding each component separately:

[tex]\vec A +\vec B = 4\hat i-\hat j-\hat k[/tex]

To find a vector parallel to the sum, we select k=2:

[tex]2(\vec A +\vec B )= 8\hat i-2\hat j-2\hat k[/tex]

Thus one vector parallel to the resultant of both vectors is:

[tex]\mathbf{8\hat i-2\hat j-2\hat k}[/tex]