Poor dog his treat! Move the items around to make a Rube Goldberg device that will give this dog his treat. Use all of the items.​

N₂O: Nitrous oxide

oxygen

metal and hair I think hope this helps

Answer:

cars and trains

Explanation:

i hope that good

Answer:

\(\boxed {\tt 14.7 \ Newtons}\)

Explanation:

Weight is force due to gravity.

The formula for weight is:

\(W=m*g\)

where m is the mass and g is the acceleration of gravity.

On Earth, the acceleration due to gravity is 9.8 meters per seconds squared.

The mass of the book is 1.5 kilograms. Therefore:

\(m= 1.5 \ kg \\g=9.8 \ m/s^2\)

Substitute the values into the formula.

\(W= 1.5 \ kg *9.8 \ m/s^2\)

Multiply.

\(W=14.7 \ kg*m/s^2\)

1 kg meter per second squared is equal to 1 Newton, so our answer of 14.7 kg*m/s² is equal to 14.7 Newtons.

\(W=14.7 \ N\)

The weight of a 1.5 kilogram book on Earth is 14.7 Newtons.

Answer:

Answer:

Explanation:

Weight is force due to gravity.

The formula for weight is:

where m is the mass and g is the acceleration of gravity.

On Earth, the acceleration due to gravity is 9.8 meters per seconds squared.

The mass of the book is 1.5 kilograms. Therefore:

Substitute the values into the formula.

Multiply.

1 kg meter per second squared is equal to 1 Newton, so our answer of 14.7 kg*m/s² is equal to 14.7 Newtons.

The weight of a 1.5 kilogram book on Earth is 14.7 Newtons.

Explanation:

Based on the given diagram, we can only reasonably conclude statement c, that the upper plate is positive and the lower plate is negative

What is Electric Field?

Electric field is a concept used in physics to describe the influence that an electric charge exerts on other charges in its vicinity. It is defined as the force experienced by a positive test charge placed in the vicinity of an electric charge, divided by the magnitude of the test charge

The upper plate is positive and the lower plate is negative: This statement is a common convention for charged parallel metal plates in a capacitor. The positive plate is usually considered as the plate with electric field lines originating from it (in this case, the upper plate), and the negative plate is usually considered as the plate with electric field lines terminating on it (in this case, the lower plate). So, this statement is likely to be true based on convention.

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Answer:

0.98 [sec].

Explanation:

1) the time when the player is moving up equals the time the player falls, it can be 't'. Then

2) the equation of moving down can be written as

\(\frac{g*t^2}{2} =h, \ where\)

g=10 [m/s²], t - time, h - the given height.

3) according to the formula above the time spent to moving down is:

\(t=\sqrt{\frac{2h}{g}};\)

\(t=\sqrt{\frac{2.4}{10}}=\sqrt{0.24} =0.49[s].\)

4) finally, the total time is t*2=0.98[sec].

Answer: 1968 cm3

Explanation: You first need to convert all of your dimensions to the same units.

240 mm = 24 cm

20.5 cm = 20.5 cm

0.04 m = 4 cm

Then volume is Length x width x height = 24 cm x 20.5 cm x 4 cm =

The appliance is run with a 12 V car battery and the value of the electric current will be 0.3A.

The pace at which electrons travel through a conductor is known as electric current. The ampere is the SI unit for electric current.

From ohm's law;

Case 1;

V=IR

120 V = 3 A × R

R = 40 ohm

For the same appliances, the value of the resistance is the same;

Case 2;

V=IR

12 V =  I × 40 ohm

I = 0.3 A

Hence the value of the current for case 2 will be 0.3 A.

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Frequency, f, is how many cycles of an oscillation occur per second and is measured in cycles per second or hertz (Hz). The period of a wave, T, is the amount of time it takes a wave to vibrate one full cycle. These two terms are inversely proportional to each other: f = 1/T and T = 1/f.

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Hope It Helps!

Answer:

Inverse

Explanation:

Frequency is the number of cycles in a second. Frequency is the inverse of a period

frequency = 1 / period

Answer:

Velocity

Explanation:

Speed does not have a direction, Acceleration means that it is speeding up and not at a constant 15m/h, so the answer is B.

As the board is raised by an angle θ, the block will be held in place by 3 forces:

• its weight, pointing downward (magnitude w )

• the normal force of the board pushing up on the block, pointing in the direction perpendicular to the board (mag. n)

• static friction, pointing in the direction parallel to the board, opposite the direction the block would slip (mag. f )

Decompose the vectors into components that are parallel and perpendicular to the board (taking the direction of n to be the positive direction perpendicular to the board, and the direction of f to be the negative directoin parallel to it), so that by Newton's second law, we have

• net parallel force:

∑ F = w (//) + f

∑ F = m g sin(θ) - µ n = 0

where µ = 0.45 is the coefficient of static friction, g = 9.80 m/s² is the mag. of the acceleration due to gravity, and m = 15.0 kg is the mass of the block.

• net perpendicular force:

∑ F = n + w (⟂)

∑ F = n - m g cos(θ) = 0

Solve for n in the equation for net perpendicular force:

n = m g cos(θ)

Substitute this into the equation for net parallel force:

m g sin(θ) - 0.45 m g cos(θ) = 0

Solve for θ :

sin(θ) - 0.45 cos(θ) = 0

sin(θ) = 0.45 cos(θ)

tan(θ) = 0.45

θ = tan⁻¹(0.45)

θ ≈ 24.2°

So the maximum angle the board can be lifted before the block starts to slide is about 24.2°, since the coefficient of static friction µ is such that

f = µ n

where f is the maximum magnitude of the static friction force.

Explanation:

To determine the effort needed to lift a load of 200N, given a velocity ratio of 5 and an efficiency of 80%, we can use the formula:

Efficiency = (Output Work / Input Work) * 100

Efficiency can also be calculated as the ratio of the output force to the input force. In this case, the output force is the load being lifted (200N), and the input force is the effort required.

Given that the velocity ratio is 5, it means that for every 5 units of distance the effort moves, the load moves 1 unit of distance. This implies that the effort is exerted over a greater distance than the load.

Let's denote the effort force as "E" and the distance moved by the effort as "dE." Similarly, the load force is "L," and the distance moved by the load is "dL."

Using the velocity ratio, we have the following relationship:

dE / dL = 5

Now, we can calculate the input work (Wi) and the output work (Wo):

Input Work (Wi) = Effort (E) * Distance moved by the effort (dE)

Output Work (Wo) = Load (L) * Distance moved by the load (dL)

Given that the efficiency is 80%, we can rewrite the formula for efficiency as:

0.80 = (Wo / Wi) * 100

Now, let's solve for the effort (E) using the given values:

Load (L) = 200N

Efficiency = 0.80

Velocity Ratio = 5

First, calculate the output work (Wo):

Wo = Load (L) * Distance moved by the load (dL)

Since the velocity ratio is 5, the distance moved by the load (dL) will be 1/5 of the distance moved by the effort (dE):

dL = (1/5) * dE

Wo = L * (1/5) * dE

Wo = 200N * (1/5) * dE

Wo = 40N * dE

Next, calculate the input work (Wi):

Wi = Effort (E) * Distance moved by the effort (dE)

Wi = E * dE

Now, substitute the values into the efficiency formula:

0.80 = (Wo / Wi) * 100

0.80 = (40N * dE) / (E * dE) * 100

0.80 = 40 / E * 100

0.80 * E = 40

E = 40 / 0.80

E = 50N

Therefore, the effort needed to lift a load of 200N with a velocity ratio of 5 and an efficiency of 80% is 50N.

Answer:

Explanation:

If the mass of the probe is 500kg, its weight W = mass  acceleration due to gravity.

Weight of the probe = 500*9.81

Weight of the probe = 4,905N

If its average density =  1400kg/m³

Volume = Mass/Density

Volume = 500/1400

Volume = 0.3571m³

According to the floatation principle, the volume of the probe is equal to the volume of liquid displaced. Hence the volume of water displaced is 0.357m³.

Since density of water is 1000kg/m³, we can find the mass of the water using the formula;

Mass of water = Density of water * Volume of water

Mass of water = 1000*0.3571

Mass of water = 357.1kg

Weight of water displaced = 3571 * 9.81 = 3503.15N

The tension in the cable will be the difference between the weight of the probe and weight of the displaced fluid.

Tension in the cable = 4,905N -  3503.15N

Tension in the cable = 1,401.85N

Hence the tension in the cable is 1,401.85N

Answer:

1)  Approximately  \(4\,\,\,10^{13}\)  excess protons

2)  negative \(4\,\,\,10^{13}\)     (\(-4\,\,\,10^{13}\))

Explanation:

1)

Recall that the charge of an electron or proton is approximately: \(1.6\,\,\,10^{-19} \,C\)

Therefore, to find the number of protons transferred in 7 micro-Coulombs of charge, we do:

\(\frac{7\,\,10^{-6}\,}{1.6\,\,10^{-19}} \approx 4\,\,10^{13}\)

Approximately  \(4\,\,\,10^{13}\)  excess protons

2)

The sign and number of uncanceled elemental charges on plate A is therefore negative \(4\,\,\,10^{13}\)  , because the same number of positive charges were removed from it, changing its neutrality

Answer:

Unequal Wavelengths

Explanation:

Got it right on the exam

Unequal wavelengths limit interference between waves because the waves will have different frequencies and will not be able to form a stable interference pattern. When waves of different wavelengths interact, they will interfere constructively and destructively at different points, creating an unpredictable pattern.

Answer:

Unequal wavelengths

Explanation:

Got it correct on the quiz.

Answer:

kinetic energy

Explanation:

Hope it helps you

Answer:

D.

Explanation

Air pressure.

Answer:air pressure

Explanation:

Answer:

The temperature, the density, and the element of stars.

Explanation:

The spectral line also can tell us about any magnetic field of the star. The width of the line can tell us how fast the material is moving. We can learn about winds in stars from this.

Explanation:

Stationary waves are still, they do not travel in a medium. Progressive waves travel in the medium in different phases.

hope this helps you

have a great day

Answer:

Steps and solution in the attached pictureSteps and solution in the attached picture.

Explanation:

Steps and solution in the attached pictureSteps and solution in the attached picture.

The velocity of the second ball just after it is struck will be 9.091 m/s.

To quantify the effect of force acting over time to change the momentum of an object is referred to as impulse.

To calculate the velocity in the given scenario,

Given that,

mV=1.5

\(V=\frac{impulse}{mass}\)

\(V=\frac{1.5}{0.165}\)

V=9.091 m/s

Thus, the velocity will be 9.091 m/s.

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This question is incomplete, the missing image is uploaded along this answer below.

Answer:

the speed of the 50-kg cylinder after it has descended is 3.67 m/s

Explanation:

 Given the data in the question and the image below;

relation between velocity of cylinder and velocity of the drum is;

V\(_D\) = ω\(_c\) × r\(_c\)  ----- let this be equ 1

where V\(_D\) is velocity of cylinder,  ω\(_c\) is the angular velocity of drum C and r\(_c\) is the radius of drum C

Now, Angular velocity of gear B is;

ω\(_B\) = ω\(_C\)

ω\(_B\) = V\(_D\) / r\(_c\)  -------- let this equ 2

so;

V\(_D\) / 0.1 m = 10V\(_D\)

Next, we determine the angular velocity of gear A;

from the diagram;

ω\(_A\)( 0.15 m ) = ω\(_B\)( 0.2 m )

from equation 2; ω\(_B\) = V\(_D\) / r\(_c\)

so

ω\(_A\)( 0.15 m ) = (V\(_D\) / r\(_c\) ) 0.2 m

substitutive in value of radius r\(_c\) (0.1 m)

ω\(_A\)( 0.15 m ) = (V\(_D\) / 0.1 m ) 0.2 m

ω\(_A\)( 0.15 ) = 0.2V\(_D\) / 0.1

ω\(_A\) =  2V\(_D\)  / 0.15

ω\(_A\) = 13.333V\(_D\)   ----- let this be equation 3

To get the speed of the cylinder, we use energy conversation;

assuming that the final position is;

T₁ + ∑\(U_{1-2\) = T₂

0 + m\(_D\)gh = \(\frac{1}{2}\)m\(_D\)V²\(_D\) + \(\frac{1}{2}I_A\)ω²\(_A\) + \(\frac{1}{2}I_B\)ω²\(_B\)

so

m\(_D\)gh = \(\frac{1}{2}\)m\(_D\)V²\(_D\) + \(\frac{1}{2}\)(m\(_A\)k\(_A\)²)(13.333V\(_D\))² + \(\frac{1}{2}\)(m\(_B\)k\(_B\)²)(10V\(_D\))²

we given that; m\(_D\) = 50 kg, h = 2 m, m\(_A\) = 10 kg, k\(_A\) 125 mm = 0.125 m, m\(_B\) = 30 kg, k\(_B\) = 150 mm = 0.15 m.

we know that; g = 9.81 m/s²

so we substitute

50 × 9.81 × 2 = ( \(\frac{1}{2}\) × 50 × V\(_D\)²) + \(\frac{1}{2}\)( 10 × (0.125)² )(13.333V\(_D\))² + \(\frac{1}{2}\)( 30 × (0.15)²)(10V\(_D\))²

981 = 25V\(_D\)² + 13.888V\(_D\)² + 33.75V\(_D\)²

981 = 72.638V\(_D\)²

V\(_D\)² = 981 / 72.638

V\(_D\)² = 13.5053

V\(_D\) = √13.5053

V\(_D\) = 3.674955 ≈ 3.67 m/s

Therefore,  the speed of the 50-kg cylinder after it has descended is 3.67 m/s

Answer:

Fx = 2.5 N

Fy = 5 N

|F| = 5.59 N

Explanation:

Given:-

- The mass of puck, m = 4.0 kg

- The initial velocity of puck, u = 3.00 i m/s

- The final velocity of puck, v = ( 8.00 i + 10.00 j ) m/s

- The time interval for the duration of force, Δt = 8 seconds

Find:-

the components of the force and (b) its magnitude.

Solution:-

- We will set up a coordinate system ( x - y ) plane. With unit vectors i and j along x and y axes respectively.

- To model the situation we will seek help from Newton's second law of motion. Defined by the rate of change of linear momentum of the system.

                                \(F_net = \frac{m*( v - u ) }{dt}\)

Where,

                   Fnet: The net force that acts on the puck-rocket system

- Here we will assume that the mass of rocket is negligible compared to the mass of the puck. The only force ( F ) acting on the puck is due to the thrust produced of the rocket. The dry and air frictions are both neglected for the analysis.

- We will apply the newton's second law of motion in component forms. And determine the components of force F, as  ( Fx ) and ( Fy ) as follows:

                         \(F_x = \frac{m* ( v_x - u_x)}{dt} \\\\F_x = \frac{4* ( 8 - 3)}{8} \\\\F_x = 2.5 N\\\\F_y = \frac{m* ( v_y - u_y)}{dt} \\\\F_y = \frac{4* ( 10 - 0)}{8} \\\\F_y = 5 N\\\\\)

- We will apply the Pythagorean theorem and determine the magnitude of the thrust force produced by the rocket with which the puck accelerated:

                          \(| F | = \sqrt{( F_x)^2 + ( F_y)^2} \\\\| F | = \sqrt{( 2.5)^2 + ( 5)^2} \\\\| F | = \sqrt{31.25} \\\\| F | = 5.590\)            

Answer: the magnitude of the thrust force is F = 5.59 N

The nature of the image formed by the convex lens is virtual, the position of the image is 30 cm away from the lens on the same side as the object, and the size of the image is twice the size of the object. The magnification is 2, meaning the image is magnified.

Given:

Object height (h) = 5 cm

Focal length of the convex lens (f) = 10 cm

Object distance (u) = 15 cm (positive since it's on the same side as the incident light)

To determine the nature, position, and size of the image, we can use the lens formula:

1/f = 1/v - 1/u

Substituting the given values:

1/10 = 1/v - 1/15

To simplify the equation, we find the common denominator:

3v - 2v = 2v/3

Simplifying further:

v = 30 cm

The image distance (v) is 30 cm. Since the image distance is positive, the image is formed on the opposite side of the lens from the object.

To find the magnification (M), we use the formula:

M = -v/u

Substituting the values:

M = -30 / 15 = -2

The magnification is -2, indicating that the image is inverted and twice the size of the object.

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A is the oscillation's amplitude, or the maximum displacement of the object from equilibrium, in either the positive or negative x-direction. Simple harmonic motion is cyclical.

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Given what we know, as with any binary star system, the life cycle of these stars will end with one of them forming a white dwarf star, and the other becoming a red giant that will engulf the white dwarf.

Taking information from the diagram provided, we can infer that the star Sabik is in its white dwarf state at the current time. This means that before long, the remaining star in this binary system will become a red giant. When this happens, it will engulf star Sabik, and that will be the end of this star's life cycle.

Therefore, we can confirm that as with any binary star system, the life cycle of these stars will end with one of them forming a white dwarf star, and the other becoming a red giant that will engulf the white dwarf.

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1) The starting temperature of the gas is 288 K

b) When the pressure is doubled, the temperature is 576 K

We know that we have to use the ideal gas equation so as to be able to obtain the temperature of the starting of the gas and we know that;

Pressure = 200 kPa or 1.97 atm

Volume =  3000 cm3  or 3 L

Temperature = ?

Number of moles = mass/molar mass = 7 g/28 g/mol = 0.25 moles

Then we have;

PV = nRT

T = PV/nR

T =  1.97 * 3/0.25 * 0.082

T = 5.91/0.0205

T = 288 K

If the pressure doubles and we have;

P1/T1 = P2/T2

P1T2 = P2T1

T2 = 2(1.97) * 288/1.97

T2 = 576 K

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or the 3500-lb car

W= 3500lb

A). \(W= mg\)

\(m=\frac{3500}{32.2} = 108.69lbm\)

B). \(W=3500lb*4.448\frac{N}{lb}=W= 15568N\)

C).

\(W= mg\\15568= m*9.81\\m= \frac{15568}{9.81} \\m= 1586.95kg\)

Mass is an intrinsic property of a body. It was traditionally believed to be related to the quantity of being counted in a bodily body, till the invention of the atom and particle physics. It changed into determined that distinctive atoms and one-of-a-kind elementary debris, theoretically with an identical amount of depend, have nevertheless specific masses. Mass in contemporary physics has a couple of definitions that are conceptually awesome, but physically equivalent. Mass can be experimentally defined as a measure of the frame's inertia, which means the resistance to acceleration (trade-off speed) when an internet pressure is implemented. The object's mass also determines the energy of its gravitational attraction to different bodies.

In physics, mass is not the same as weight, despite the fact that mass is frequently determined by measuring the object's weight through the use of a spring scale, in place of a balance scale evaluating it without delay with regarded loads. An item on the Moon might weigh less than it does on earth because of the decreased gravity, however, it would nonetheless have the same mass.

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Answer:

161.938 Hz

Explanation:

the computation of the fundamental resonant frequency is shown below

p = 1100 kg/m^3

A = 130 mm^2

= 130 ×10^-6 m^2

T = 600 N

L = 20 cm

= 0.2 m

Now the linear density of tendon is

=  1100 kg/m^3 ×  130 ×10^-6

= 0.143 kg/m

Now the wave of the string is

= √600  ÷ √0.143

= 64.775 m/s

Now finally the fundamental resonant frequency is

= 64.775 ÷ (2 × 0.2)

=161.938 Hz

The magnetic field in the region is into the paper and has a magnitude of 60 mT and the tension in either wire is 0.096 N.

To find the tension in either wire, we can apply the equation for the force experienced by a current-carrying wire in a magnetic field.

The force experienced by a current-carrying wire in a magnetic field is given by the equation F = B * I * L * sin(θ), where B is the magnetic field strength, I is the current, L is the length of the wire, and θ is the angle between the wire and the magnetic field.

In this case, the wire is suspended horizontally by two vertical wires, and the magnetic field is into the paper. Since the wire is horizontal, the angle between the wire and the magnetic field is 90 degrees, so sin(θ) = 1.

The force experienced by the wire due to the magnetic field is F = B * I * L.

Given:

Current (I) = 8.0 A

Magnetic field (B) = 60 mT = 60 * 10^(-3) T

Length of the wire (L) = 40 cm = 40 * 10^(-2) m

Substituting the given values into the equation, we get:

F = (60 * 10^(-3) T) * (8.0 A) * (40 * 10^(-2) m)

Simplifying the expression, we find:

F = 0.192 N

Since the wire is suspended by two vertical wires, the tension in each wire will be half of the total force. Therefore, the tension in either wire is 0.192 N / 2 = 0.096 N.

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Unreliable personality tools on the market, inaccurate test results, anxious candidates as a result of linguistic and cultural barriers, pricey personality tools, and lengthy assessments are some of the issues with personality tests.

Some aspects of a person's personality are overemphasized by personality tests, while others are underemphasized, misplaced, or completely ignored. Personality tests don't accurately represent a particular aspect of a person's personality; instead, they present a skewed, false impression of the whole thing.

Projective tests and self-report inventories are the two main categories of personality tests. Test-takers must read questions and then rate how well the question or statement applies to them in self-report inventories.

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