Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.311 rad/s2 for 4.13 s. What is the drill's angular displacement during that time interval?

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:

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:

The value is [tex]KE = 259.6 \ J[/tex]

Explanation:

From the question we are told that

     The weight of the horizontal solid disk is  [tex]W = 805 \ N[/tex]

      The radius of the horizontal solid disk is  [tex]r = 1.58 \ m[/tex]

      The force applied by the child is  [tex]F = 49.5 \ N[/tex]

       The time considered is  [tex]t = 2.95 \ s[/tex]

Generally the mass of the  horizontal solid disk is mathematically represented as

          [tex]m_h = \frac{W}{ g}[/tex]

=>       [tex]m_h = \frac{805}{ 9.8 }[/tex]

=>       [tex]m_h = 82.14 \ N[/tex]

Generally the moment of inertia  of the horizontal solid disk is mathematically represented as  

         [tex]I = \frac{1}{2} * m * r^ 2[/tex]

=>      [tex]I = \frac{1}{2} * 82.14 * 1.58^ 2[/tex]  

=>      [tex]I = 102.5 \ kg \cdot m^2[/tex]

Generally the net torque experienced by the horizontal solid disk is mathematically represented as

           [tex]T = I * \alpha = F * r[/tex]

=>         [tex]\alpha = \frac{ F * r }{ I }[/tex]

=>         [tex]\alpha = \frac{ 49.5 * 1.58 }{ 102.53 }[/tex]

=>         [tex]\alpha = 0.7628[/tex]

Gnerally from kinematic equation we have that

         [tex]w = w_o + \alpha t[/tex]

Here  [tex]w_o[/tex] is the initial angular velocity velocity of the horizontal solid disk  which is  [tex]w_o = 0\ rad/s[/tex]

So

           [tex]w = 0 + 0.7628 * 2.95[/tex]

=>        [tex]w = 2.2503 \ rad/s[/tex]

Generally the kinetic energy is mathematically represented as

        [tex]KE = \frac{1}{2} * I * w^2[/tex]

=>      [tex]KE = \frac{1}{2} * 102.53 * 2.2503 ^2[/tex]

=>      [tex]KE = 259.6 \ J[/tex]

Answer:

organ

Explanation:

Answer:

an organ

Explanation:

cell -> tissue -> organ -> organ system -> organism

We know, [tex]1\ rpm = \dfrac{2\pi}{60} \ rad/s[/tex] .

[tex]64\ rpm\ is = \dfrac{2\pi}{60}\times 64\ rad/s\\\\= \dfrac{32\pi}{15}\ rad/s[/tex]

We know, kinetic energy is given by :

[tex]K.E = \dfrac{I\omega^2}{2}\\\\I = \dfrac{2(K.E)}{\omega^2}\\\\I = \dfrac{2\times 10^6}{\dfrac{32}{15}\times \pi}\\\\I = 298415.52 \ kg \ m^2[/tex]

Hence, this is the required solution.

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:

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:

Pt 1: [tex]W=39J[/tex]

Pt 2: θ = [tex]19.7[/tex]

Explanation:

Part 1:

[tex]W=FS[/tex]

[tex]W=F_{_xs_x}+F{_ys_y}[/tex]

imagine the "i"and the "j" with the hat

[tex]W=(10i-1j)(4i+3j)[/tex]

[tex]W=(10*4)-(1*1)[/tex]

[tex]W=40-1[/tex]

[tex]W=39J[/tex]

Part 2:

[tex]|F|=\sqrt{10^2+(-1)^2}[/tex]

[tex]|F|=\sqrt{101}[/tex]

[tex]|S|=\sqrt{4^2+1^2}[/tex]

[tex]|S|=\sqrt{17}[/tex]

[tex]W=|F||S|cos[/tex]θ

[tex]39=(\sqrt{101})(\sqrt{17} )cos[/tex]θ

θ = [tex]19.7[/tex]

Hope it helps

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:

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:

(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:

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

Explanation:

Answer:

2 thumbs (ISSA Guide)

Explanation:

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:

light refraction

Explanation:

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

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:

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.

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:

Explanation:

The speed of the horse can be found by using the formula

[tex]v = \frac{p}{m} \\ [/tex]

p is the momentum

m is the mass

From the question we have

[tex]v = \frac{1200}{313} \\ = 3.83386..[/tex]

We have the final answer as

Hope this helps you

Answer:

transition metals im sorry if this was too late

Answer:

A. 28.42 m/s

B. 41.21 m.

Explanation:

A. Determination of the initial velocity of the ball:

Time (t) to reach the maximum height = 2.9 s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)

Initial velocity (u) =?

Thus, we can obtain the initial velocity of the ball as follow:

v = u + gt

0 = u + (–9.8 × 2.9)

0 = u – 28.42

Collect like terms

u = 0 + 28.42

u = 28.42 m/s

Therefore, the initial velocity of the ball is 28.42 m/s.

B. Determination of the maximum height reached.

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)

Initial velocity (u) = 28.42 m/s.

Maximum height (h) =?

Thus, we can obtain the maximum height reached by the ball as follow:

v² = u² + 2gh

0² = 28.42² + (2 × –9.8 × h)

0 = 807.6964 + (–19.6h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = –807.6964 / –19.6

h = 41.21 m

Therefore, the maximum height reached by the ball is 41.21 m

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

Explanation:   this is the  most accurate phrase  

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:

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]