In This Example, The System Gains The Ability To Cause Change The Higher The Brick Is Raised Above The (2024)

The left hand side and the right hand side of the given match the following questions are arranged correctly in order.

A possible explanation for a problem using what you know is called as hypothesis. A representation of an idea, event, structure or object is called a model. Physics is the branch of science that studies matter and energy.

The factor that is changed or manipulated during an experiment is called as independent variable. A dependent variable changes according to the changed that happen in an independent variable. A relationship in which the dependent variable varies is called as linear relationship.

Therefore, the left hand side and the right hand side of the given match the following questions are arranged correctly in order.

To know more about independent variable

https://brainly.com/question/1479694

#SPJ1

Given data:

* The initial angular speed of the car is,

[tex]\omega_i=0.54\text{ rad/s}[/tex]

* The final angular speed of the car is,

[tex]\omega_f=0.96\text{ rad/s}[/tex]

* The angular displacement of the car is,

[tex]\theta=1.4\text{ radians}[/tex]

Solution:

By the kinematics equation, the angular acceleration of the car in terms of the angular displacement is,

[tex]\omega^2_{\text{f}}-\omega^2_i=2\alpha\theta[/tex]

where,

[tex]\alpha\text{ is the angular acceleration,}[/tex]

Substituting the known values,

[tex]\begin{gathered} 0.96^2-0.54^2=2\times\alpha\times1.4 \\ \alpha=\frac{0.96^2-0.54^2}{2\times1.4} \\ \alpha=\frac{0.9216-0.2916}{2.8} \\ \alpha=\frac{0.63}{2.8} \end{gathered}[/tex]

By simplifying,

[tex]\alpha=0.225rads^{-2}[/tex]

Thus, the angular acceleration of the car is 0.225 radians per second squared.

ANSWER

[tex]C.250,000N[/tex]

EXPLANATION

Parameters given:

Weight of rocket, W = 220,000 N

Thrust, F = 500,000N

Angle of thrust, θ = 20°

The vertical forces acting on the rocket are the thrust (acting at an angle) and its weight. This means that the sum of vertical forces is:

[tex]F_y=F\cos \theta-W[/tex]

Note: W is negative since it acts opposite the motion of the rocket.

Therefore, we have that the net vertical force (vertical component of forces) is:

[tex]\begin{gathered} F_y=500000\cos 20-220000 \\ F_y=469,846.31-220000 \\ F_y=249,846.31N \\ F_y\approx250,000N \end{gathered}[/tex]

The answer is option C.

[tex]P(t)=1*0.5^{\frac{t}{h}}[/tex]

Where:

P(t): percentage of initial remaining

t: time elapsed

h: half life of substance

Given the variables we know, we can solve for P(t).

[tex]P(t)=0.5^{\frac{100}{1599}}[/tex]

Plugging this into a calculator, we get that

P(t) = 0.95757733 = 95.757733% remaining

Given,

The mass of the block moving to the right, M=3.10 kg

The speed of the block, u=2.60 m/s

The mass of the block at rest, m=1.00 kg

The total momentum of the two-block system before the collision is given by

[tex]p_i=Mu_{}[/tex]

On substituting the known values,

[tex]\begin{gathered} p_i=3.10\times2.60 \\ =8.06\text{ kg}\cdot\frac{m}{s} \end{gathered}[/tex]

From the law of conservation of energy, the total momentum of a system always remains the same.

Thus the momentum after the collision is equal to the total momentum of the blocks before the collision.

Thus the total momentum of the two blocks after the collision is 8.06 kg·m/s

Using the concept of Harmonic-wave, we got the desired graph which is shown in the image.

It must be an open-closed pipe.

Open-closed pipe

Let the fundamental frequency of f be

The second harmonic is 3f.

3rd harmonic = 5f

Therefore, the difference between harmonics is 3f-f = 2f

So, 2f = (1120-800) = 1440-1120 = 320

Also, f = 160Hz

800Hz = 5th Harmonic

1120 Hz = 7th

1440 = 9th

For an open- or closed-pipe,

f = v/4L

where v is the speed of sound in the air = 343 M/s

So, 160 = 343/4L

Also, L = 0.5336 m = 53.6cm

Hence for the frequency of 1120Hz, we got the desired graph of a wave which is shown in the below graph.

To know more about the mechanical wave, visit here:

https://brainly.com/question/8885243

#SPJ9

Given:

the height of the object is

[tex]h_0=5\text{ cm}[/tex]

The distance of the object is

[tex]d_0=-20\text{ cm}[/tex]

The focal length of the lens is

[tex]f=15\text{ cm}[/tex]

Required: the distance of the image and height of the image.

Explanation:

the lens formula is given by

[tex]\frac{1}{f}=\frac{1}{d_i}-\frac{1}{d_0}[/tex]

Plugging all the values in the above relation, we get:

[tex]\begin{gathered} \frac{1}{15\text{ cm}}=\frac{1}{d_i}-\frac{1}{-20\text{ cm}} \\ \frac{1}{d_i}=\frac{1}{15\text{ cm}}-\frac{1}{20\text{ cm}} \\ \frac{1}{d_i}=\frac{4-3}{60\text{ cm}} \\ d_i=60\text{ cm} \end{gathered}[/tex]

Thus, the distance of the image is 60 cm.

now calculate the height of the image

we know that

[tex]\frac{h_i}{h_0}=\frac{d_i}{d_0}[/tex]

substitute all the values in the above relation, we get:

[tex]\begin{gathered} h_i=5\text{ cm}\times\frac{60}{-20} \\ h_i=-15\text{ cm} \end{gathered}[/tex]

Thus, the height of the image is 15 cm.

The average density of a planet is given by:

[tex]\rho=\frac{m}{V}[/tex]

where m is the mass of the planet and V is its volume. We know the mass of the planet but we don't know its volume, to find it we will need find its radius.

To find the radius of the planet we can use Newton's Law of gravitation:

[tex]F=G\frac{mM}{d^2}[/tex]

where G is the gravitational constant, m is the mass of the object, M is the mass of the planet and d is the distance between the planet and the object. Let r be the radius of the planet, and x be the distance from the surface of the planet to the object (x=13660 in this case); then we have:

[tex]\begin{gathered} F=G\frac{mM}{(r+x)^2} \\ (r+x)^2=\frac{GmM}{F} \\ r+x=\pm\sqrt[]{\frac{GmM}{F}} \\ r=-x\pm\sqrt[]{\frac{GmM}{F}} \end{gathered}[/tex]

Plugging the values given we have:

[tex]\begin{gathered} r=-13660\times10^3\pm\sqrt[]{\frac{(6.67\times10^{-11})(407)(7.9\times10^{24})}{585}} \\ \text{ Using the positive root we have:} \\ r=5.49\times10^6 \\ \text{ Using the negative root we have:} \\ r=-3.28\times10^7 \end{gathered}[/tex]

Since the radius of the planet has to be positive we choose the positive solution.

Now, that we know the radius of the planet we can calculate its volume; assuming the planet is spherical we have that:

[tex]V=\frac{4}{3}\pi r^3[/tex]

then we have:

[tex]\begin{gathered} V=\frac{4}{3}\pi(5.49\times10^6)^3 \\ V=6.92\times10^{20} \end{gathered}[/tex]

Finally we can calculate the density:

[tex]\begin{gathered} \rho=\frac{7.9\times10^{24}}{6.92\times10^{20}} \\ \rho=11398 \end{gathered}[/tex]

Therefore, the average density is 11398 kg/m^3

The following equation represents the value (y) of a car based on its age (x).

[tex]y=-800x+12000[/tex]

We are told that the value of a car of the given model, based on its age, is less than $6000.

So, we can write

[tex]6000<-800x+12000[/tex]

Let us solve the above inequality for x.

[tex]\begin{gathered} 6000<-800x+12000 \\ 6000-12000<-800x \\ -6000<-800x \\ \frac{-6000}{-800}<\frac{-800x}{-800} \\ 7.5>x \\ x<7.5 \end{gathered}[/tex]

This means that the age of the car is less than 7.5 years.

Therefore, the possible range of values that could represent the age of the car is 0 to 7.5 years.

[tex]0

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

[tex]\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)[/tex]

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

[tex]r_{M\perp}(F_M)-r_{W\perp}(W)=0[/tex]

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

[tex]r_{M\perp}(F_M)=r_{W\perp}(W)[/tex]

Now, we divide by the distance of the muscle:

[tex](F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}[/tex]

Now, we substitute the values:

[tex]F_M=\frac{(2.4cm)(50N)}{5.1cm}[/tex]

Now, we solve the operations:

[tex]F_M=23.53N[/tex]

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

[tex]\Sigma F_v=F_j-F_M-W[/tex]

Since there is no vertical movement the sum of vertical forces is zero:

[tex]F_j-F_M-W=0[/tex]

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

[tex]F_j=F_M+W[/tex]

Now, we plug in the values:

[tex]F_j=23.53N+50N[/tex]

Solving the operations:

[tex]F_j=73.53N[/tex]

Therefore, the force is 73.53 Newtons.

Given:

• Mass of bag, m =25.0 kg

,

• Height, h = 15.0 m

,

• c = 0.20 cal/g•°C

Let's find by how much the temperature will increase.

Apply the Law of Conservation of Energy:

[tex]mc\Delta T=mgh[/tex]

Where:

• m is the mass

,

• c is the specific heat capacity

,

• g is acceleration due to gravity

,

• h is the height.

,

• ΔT is the temperature change.

Thus, we have:

[tex]\begin{gathered} mc\Delta T=mgh \\ \\ Eliminate\text{ m on both sides:} \\ c\Delta T=gh \end{gathered}[/tex]

Now, plug in the values and solve for ΔT:

[tex]\begin{gathered} \Delta T=\frac{gh}{c} \\ \\ \Delta T=\frac{9.8*15.0}{0.20\times10^3\times4.2} \\ \\ \Delta T=\frac{147}{840} \\ \\ \Delta T=0.175^o\text{ C} \end{gathered}[/tex]

Therefore, the temperature change is 0.175° C.

• ANSWER:

0.175° C

[tex]\begin{gathered} \text{For car} \\ v_1=11.6\text{ m/s} \\ \Delta x=0.287m \\ v_2=\text{ 0 m/s} \\ t=\text{?} \\ To\text{ find t} \\ v^2_2=v^2_1+2a\Delta x \\ \text{Solving a} \\ v^2_2-v^2_1=2a\Delta x \\ a=\frac{v^2_2-v^2_1}{2\Delta x} \\ a=\frac{(0m/s)^2-(11.6m/s)^2}{2(0.287m)} \\ a=\frac{0m^2/s^2-134.56m^2/s^2}{0.574m} \\ a=\frac{-134.56m^2/s^2}{0.574m} \\ a=-234.42m/s^2 \\ \text{Then} \\ t=\frac{v_2-v_1}{a} \\ t=\frac{0\text{ m/s-11.6m/s}}{-234.42m/s^2} \\ t=\frac{\text{-11.6m/s}}{-234.42m/s^2} \\ t=0.0495\text{ s} \\ \text{For child} \\ m=21.2\text{ kg} \\ v_1=11.6\text{ m/s} \\ v_2=\text{ 0 m/s} \\ t=0.0495\text{ s} \\ F=\text{?} \\ F=\frac{P_2-P_1}{t} \\ P_2-P_1=mv_2-mv_1=m(v_2-v_1),\text{ then} \\ F=\frac{m(v_2-v_1)}{t} \\ F=\frac{(21.2kg)(0\text{ m/s-11.6m/s})}{0.0495\text{ s}} \\ F=\frac{(21.2kg)(-11.6\text{ m/s})}{0.0495\text{ s}} \\ F=\frac{-245.92\operatorname{kg}\text{ m/s}}{0.0495\text{ s}} \\ F=-4968\text{ N} \\ \text{The force to stop the child is 4968 N. The negative means } \\ \text{the force is opposite to the movement} \end{gathered}[/tex]

Given data

*Kevin exerted a force is F = 165 N

*The given acceleration is

[tex]a=28.5m/s^2[/tex]

The formula for the mass of the shot put is given as

[tex]m=\frac{F}{a}[/tex]

Substitute the values in the above expression as

[tex]\begin{gathered} m=\frac{165}{28.5} \\ =5.78\text{ kg} \end{gathered}[/tex]

Thus, the mass of the shot put is m = 5.78 kg

Answer:

72 cm³

Explanation:

The volume of a textbook can be calculated as

Volume = Length x Width x Height

Replacing the length by 6 cm, the width by 4 cm, and the height by 3 cm, we get:

Volume = 6 cm x 4 cm x 3 cm

Volume = 72 cm³

Therefore, the volume of the textbook is 72 cm³

Answer:

m V^2 / R = G M m / R^2 balancing forces

R = G M / V^2 (I)

V = 2 π R / (24 * 3600) to complete 1 rev in 24 hrs

V = π R / (12 * 3600) = R * 7.27E-5

V^2 = R^2 * 5.29E-9

R = G M / (R^2 * 5.29E-9) using (I)

R = (G M / 5.29E-9)^1/3

R = (6.67E-11 * 6.0E24 / 5.29E-9)^1/3

R = (66.7 * 6.0 / 5.29)^1/3 E7

R = 4.22E7 meters = 42200 km about 7 earth radii

ANSWER:

304.3 Pa

STEP-BY-STEP EXPLANATION:

We have the poiseuille law, which would be the following equation:

[tex]v=\frac{\pi\cdot\Delta P\cdot r^4\cdot t}{8\cdot\eta\cdot L}[/tex]

Where,

v = volume of the liquid

r = radius

t: time

n = coefficiente of viscosity

Δp = change of pressure

L : lenght

We solve for Δp, and we would have:

[tex]\begin{gathered} \Delta P=\frac{8\cdot\eta\cdot L\cdot v}{\pi\cdot\cdot r^4\cdot t} \\ \frac{v}{t}=Q \\ \text{ therefore:} \\ \Delta P=\frac{8\cdot\eta\cdot L\cdot Q}{\pi\cdot r^4} \\ \text{ replacing:} \\ L=52.5\text{ cm = 0.525 m} \\ r=2\text{ mm = 0.002 m} \\ \Delta P=\frac{8\cdot1.3\cdot10^3\cdot0.525\cdot2.8\cdot10^{-6}}{3.14\cdot(0.02)^4} \\ \Delta P=304.3\text{ Pa} \end{gathered}[/tex]

The pressure difference is 304.3 Pa

Answer:

14.6 m/s²

Explanation:

First, we will make the free body diagram

Since the net vertical force is equal to 0 because the block is not moving up, we can write the following equation

[tex]\begin{gathered} F_{nety}=F_n+F_{Ty}-mg=0 \\ F_n+F_T\sin15-mg=0 \end{gathered}[/tex]

Then, we can solve for the normal force and replace Ft = 1197 N, m = 71 kg and g = 10 m/s²

[tex]\begin{gathered} F_n=mg-F_T\sin15 \\ F_n=(71\text{ kg\rparen\lparen10 m/s}^2)-(1197N)(0.26) \\ F_n=400.2\text{ N} \end{gathered}[/tex]

Now, we can write the following equation for the net horizontal force

[tex]\begin{gathered} F_{net}=F_{Tx}-F_f=ma \\ F_T\cos15-\mu F_n=ma \end{gathered}[/tex]

Where μ is the coefficient of friction and a is the acceleration of the block. Solving for a and replacing Ft = 1197 N, m = 71 kg, Fn = 400.2 N and μ = 0.3, we get

[tex]\begin{gathered} a=\frac{F_T\cos15-\mu F_n}{m} \\ \\ a=\frac{(1197\text{ N\rparen}\cos15-0.3(400.2N)}{71\text{ kg}} \\ \\ a=14.6\text{ m/s}^2 \end{gathered}[/tex]

Therefore, the acceleration of the block is 14.6 m/s²

The diagram representing this scenario is shown below

From the information given,

Number of molecules in hydrogen, NH = 1

Number of molecules in oxygen, NO = 15.99

y = 0.1cos53

y = 0.06 nm

x = 0.1Sin53

x = 0.08 nm

r1 = xi + yj = 0.08i + 0.06j

r2 = = 0.08i - 0.06j

r3 = 0

rcm = 2miri/2mi = (m1r1 + m2r2 + m3r3)/(m1 + m2 + m3)

r = 1(0.06i + 0.08j + 0.06i - 0.08j)/2(1 + 1 + 15.99)

r = (6.7 x 10^-3i) nm

the center of mass of the molecule = (6.7 x 10^-3i) nm

We are given that Amanda made 141 pounds of coffee. If "x" is the pounds of type A coffee and "y" is the amount of type b coffee then we can write this mathematically as:

[tex]x+y=141,(1)[/tex]

We are also given that the cost of type A is $5.6 per pound and that the cost of type B is $4.55 per pound and that the total cost is $728.70, this can be written mathematically as:

[tex]5.6x+4.55y=728.7,(2)[/tex]

Now, we solve for "x", first by subtracting "y" from both sides:

[tex]y=141-x[/tex]

Now, we substitute the value of "y" in equation (2):

[tex]5.6x+4.55(141-x)=728.7[/tex]

Now, we apply the distributive property in the parenthesis:

[tex]5.6x+641.55-4.55x=728.7[/tex]

Now, we add like terms:

[tex]1.05x+641.55=728.7[/tex]

Now, we subtract 641.55 from both sides:

[tex]\begin{gathered} 1.05x=728.7-641.55 \\ 1.05x=87.15 \end{gathered}[/tex]

Now, we divide both sides by 1.05:

[tex]x=\frac{87.15}{1.05}[/tex]

solving the operations:

[tex]x=83[/tex]

Now, we substitute the value of "x" in equation (1):

[tex]\begin{gathered} y=141-83 \\ y=58 \end{gathered}[/tex]

Therefore, the amount of coffee type A is 83 pounds and type B is 58 pounds.

If a man tried to move a truck by pushing it, but he is not able to move it, then the work done by the man will be equal to zero.

What is Work?

In physics, the word "work" involves measurement of energy transfer that takes place when an item is moved over a range by an externally applied, at least a portion of which is applied with in the direction of the displacement. The length of the path is multiplied by the element of a force acting all along the path to calculate work if the force is constant. The work W is theoretically equivalent towards the force f times the length d, or W = fd, to portray this concept.

As per the given question,

The man tries to move the truck, but he is not able to move. It means that the total displacement is zero, which means according to the formula of work done the work is also zero.

To know more about Work:

https://brainly.com/question/18094932

#SPJ1

Given data:

Comcentration of H+ ion in water,

[tex]\lbrack H^+\rbrack=10^{-7}\text{ }\frac{mol}{l}[/tex]

The pH value is given as,

[tex]pH=-\log _{10}\lbrack H^+\rbrack_{}[/tex]

Substituting all known values,

[tex]pH=-\log _{10}\lbrack10^{-7}\rbrack[/tex]

Since,

[tex]\log _{10}(a^b)=b\times\log _{10}(a)[/tex]

Therefore,

[tex]\begin{gathered} pH=-7\times(-\log _{10}\lbrack10\rbrack) \\ =7\log _{10}\lbrack10\rbrack \end{gathered}[/tex]

Since,

[tex]\log _{10}\lbrack10\rbrack=1[/tex]

Therefore,

[tex]\begin{gathered} pH=7\times1 \\ =7 \end{gathered}[/tex]

Therefore, pH of a pure water solution is 7.

To isolate v₁ from the given equation, subtract aΔt from both members of the equation and simplify:

[tex]\begin{gathered} v_2=v_1+a\cdot\Delta t \\ \Rightarrow v_2-a\cdot\Delta t=v_1+a\cdot\Delta t-a\cdot\Delta t \\ \Rightarrow v_2-a\cdot\Delta t=v_1 \\ \therefore v_1=v_2-a\cdot\Delta t \end{gathered}[/tex]

Therefore, the formula for v₁ is:

[tex]v_1=v_2-a\cdot\Delta t[/tex]

The answer is 10.00

in series circuit , the fo

Given,

[tex]{ \blue{ \tt{m = 560g}}}[/tex]

[tex]{ \blue{ \tt{g = 9.8 m/s²}}}[/tex]

[tex]{ \blue{ \tt{W = ?}}}[/tex]

[tex]{ \purple{ \tt{W = mg}}}[/tex]

[tex]{ \purple{ \sf{W= 560 × 9.8}}}[/tex]

[tex]{ \boxed{ \red{ \sf{W = 5488N}}}}[/tex]

The relationship between the force and area of the material is,

[tex]F_f=\frac{P}{A}[/tex]

where P is the load and A is the area.

As the tensile strength of the material is same for all the material tested.

As the area of the material is directly proportional to the diameter.

Thus, the force of failure is inversely proportional to the diameter.

ANSWER:

6.9 x 10^-9 W

STEP-BY-STEP EXPLANATION:

Given:

Area (A) = 2.61× 10^-3 m²

Intensity (I) = 2.65 x 10^-6 W/m²

We can determine the power by the following formula:

[tex]\begin{gathered} P=A\cdot I \\ \\ \text{ We replacing:} \\ \\ P=2.61\cdot10^{-3}\cdot2.65\cdot10^{-6} \\ \\ P=6.9\cdot10^{-9}\text{ W} \end{gathered}[/tex]

The power is 6.9 x 10^-9 W

1)

R is the required distance from the starting point. The right triangle representing this scenario is shown below

We would apply pythagorean theorem which is expressed as

hypotenuse^2 = one leg^2 + other leg^2

hypotenuse = R

one leg = 25

other leg = 18

Thus,

R^2 = 25^2 + 18^2 = 949

R = √949

R = 30.81

You're 30.81m from your starting point

2) We would find θ by applying the tangent trigonometric ratio which is expressed as

tanθ = opposite side/adjacent side

opposite side = 25

adjacent side = 18

tanθ = 25/18

θ = tan^-1(25/18)

θ = 54.25 degrees

Position = 54.25 degrees west of north

Given the area is:

A=4.0 cmx4.0 cm

The seperation distance is d=5.0 mm

The capacitance is :

[tex]\begin{gathered} C=\frac{\epsilon_oA}{d} \\ \Rightarrow C=8.85\times10^{-12}\frac{4.0\times10^{-2}\times4.0\times10^{-2}}{5\times10^{-3}} \\ \Rightarrow C=2.83\times10^{-12}F \end{gathered}[/tex]

Thus the answer is

[tex]2.8\times10^{-12}F[/tex]

Given,

The mass of the moving block, M=2.10 kg

The velocity of the moving block, u=2.00 m/s

The mass of the block that was at rest, m=1.00 kg

The velocity of the second block before the collision, v=0 m/s

From the law of conservation of momentum, the total momentum of a system always remains constant. That is, the total momentum of the two blocks before the collision is equal to the total momentum of the blocks after the collision.

Therefore, the total momentum of the two blocks after the collision is given by,

[tex]p=Mu+mv[/tex]

On substituting the known values,

[tex]\begin{gathered} p=2.10\times2.00+1.00\times0 \\ =4.2\text{ kg. m/s} \end{gathered}[/tex]

Therefore, the total momentum of the blocks after the collision is 4.2 kg· m/s

ANSWER:

False

STEP-BY-STEP EXPLANATION:

Even without friction, the efficiency cannot be greater than 100%, because the output cannot be greater than the input, therefore, the statement is false.