How are reflectivity and solubility related?

The object requires the MOST power to lift  a 2 kg box 2 m in 1 s. Hence, option (B) is correct.

The quantity of energy moved or converted per unit of time is known as power in physics. The watt, or one joule per second, is the unit of power in the International System of Units. Power is also referred to as activity in ancient writings. A scalar quantity is power.

Power required for lifting a 2 kg box 1 m in 1 s = (2×9.8×1)/1 watt = 19.6 watt.

Power required for lifting a 2 kg box 2 m in 1 s = (2×9.8×2)/1 watt = 39.2 watt.

Power required for lifting a 2 kg box 1 m in 2 s = (2×9.8×1)/2watt =9.8 watt.

Power required for lifting a 2 kg box 2 m in 2 s = (2×9.8×2)/2watt =19.6 watt.

So, The most power required in  lifting a 2 kg box 2 m in 1 s.

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The moment of inertia (I) will changes and net torque (Tnet) will also change, while the angular acceleration (a) remains constant.

The formula for net torque acting on an object is given as;

T(net) = Ia

where;

The  moment of inertia of an object is given as;

I ∝ MR²

where;

So mass, M changes, the moment of inertia (I) changes and net torque will also change, while the angular acceleration remains constant.

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a. the resistance of the parallel pair of resistors is  4.71Ω.

b.  the total resistance of the circuit is 7.71Ω.

c. the total current flow of the circuit is  1.55A.

d. The voltage drop across each resistor is 12V.

e. he current flowing through the 2Ω resistor is 1.55A.

f. the current flowing through the 6Ω resistor is 1.55A.

Equivalent resistance :

1/Req = 1/R1 + 1/R2

1/Req = 1/22Ω + 1/6Ω

1/Req = (6 + 22)/(22 * 6)

1/Req = 28/132

Equivalent resistance = 132/28

Equivalent resistance = 4.71Ω

b) The total resistance of the circuit:

total  resistance   = equivalent resistance  + R3

total  resistance = 4.71Ω + 3Ω

total resistance  = 7.71Ω

c) The total current flow of the circuit:

We use Ohm's law

I = V / R

I = 12V / 7.71Ω

I =  1.55A

d) The voltage drop across each resistor is the same as the total voltage

e) The current flowing through the 2Ω resistor is same as all resistors.

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if you pump air into a 100ml balloon, the distance is that the Gases will be ale to spread out very quickly to fill all the space that has been available to them in the balloon.

The capacity of a container increases as the gas content rises. A container's capacity reduces as the gas content is reduced. Of course, this assumes that the container's walls are expandable.

Therefore, Gases swiftly disperse to occupy all available area. Imagine blowing up a balloon. When you blow air into a balloon, the entire balloon is filled. A gas will take up all of the available space. This is due to the fact that a gas's particle structure is random.

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BTU = Flow Rate In GPM (of water) x (Temperature Leaving Process - Temperature Entering Process) x 500.4*Formula changes with fluids others than straight water.

Sensible heat is the type of heat that causes a change in temperature when you heat or cool an object. For instance, when you heat an ice cube, its temperature will start to rise until it reaches 0 °C, at which time it will begin to melt.

Similar to this, when we heat water from room temperature, it gets hotter and hotter until it begins to boil. Sensible heat is the heat that causes a change in temperature. In other words, the sensation you have is heat.

Latent heat, in contrast, is a type of heat that solely affects phase transitions. Use our latent heat calculator to learn more. The equation for sensible heat is:

=Q=mc p (T f −Ti )

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Answer:

Explanation:

1. First draw a free body diagram of the scenerio (a block sliding down a a slant surface).
2. Then we analyze the forces and write equations that satisfy Fnet = ma. This will give us the acceleration as the block slides down the surface.

3. Last, we can use the kinematic equation (vf^2 = vi^2 + 2as) and to solve the final speed of the block.

Answer:

A. length, diameter, material, temperature

Answer:

increasing the temperature of the object

The height to which the toy gun can shoot a 0.0125 Kg toy bullet is 1.29 m

We'll begin by determine the energy of the spring. This can be obtained as follow:

E = ½Ke²

E = ½ × 350 × 0.03²

E = 175 × 0.0009

E = 0.1575 J

Finally, we shall determine the height. This is illustrated below:

E = mgh

0.1575 = 0.0125 × 9.8 × h

0.1575 = 0.1225 × h

Divide both sides by 0.1225

h = 0.1575 / 0.1225

h = 1.29 m

Thus, the height is 1.29 m

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A particle with the potential energy diagram shown is located at point A and is moving to the right with a kinetic energy of 10.0 Joules. When the particle reaches point F, the speed of the particle has decrease

As the sum of potential and kinetic energy remains constant

Kinetic energy decreases when potential energy increases and vice versa.

When kinetic energy decreases speed will also decrease.

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Between Kinetic energy and speed: The relationship is quadratic or parabolic. According to the kinetic energy formula, KE = 1/2 mv^2, the kinetic energy is proportional to the square of the speed.

Between potential energy and mass: The relationship is linear. The potential energy is directly proportional to the mass. In simple cases, the potential energy is given by PE = mgh, where mass (m) and height (h) are directly proportional.

Between pressure and volume: The relationship is inverse. According to Boyle's law, the pressure and volume of a gas are inversely proportional when temperature is constant. Mathematically, P1V1 = P2V2.

Between pressure and temperature: The relationship is linear. According to Charles's law, the pressure and temperature of a gas are directly proportional when volume is constant. Mathematically, P1/T1 = P2/T2.

Between displacement and velocity: The relationship is linear. Velocity is the rate of change of displacement with respect to time, so the two are directly proportional.

Non-Malleable and Ductile: Non-metals are very brittle, and cannot be rolled into wires or pounded into sheets. Conduction: They are poor conductors of heat and electricity. Luster: These have no metallic luster and do not reflect light.

Group 15, the nitrogen family, contains two nonmetals: nitrogen and phosphorus. These non-metals usually gain or share three electrons when reacting with atoms of other elements. Group 16, the oxygen family, contains three nonmetals: oxygen, sulfur, and selenium.

Elements: Nitrogen; Oxygen; Phosphorus; Selenium...

Answer:

3.94 s

Explanation:

Take down to be positive.  Given:

Δy = 76.0 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

76.0 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 3.94 s

Answer:

C

Explanation:

Usually when you are at the bottom you are at peak speed. It also shows that Kinetic Energy is the green bar and in picture C the green bar is highest.

In terms of work produced, the gravitational force acting on block 2 is equal to that acting on block 1.

The quantity of labor is equal to zero when an object moves a fixed distance. The movement of the item as well as the applied load define the quantity of work. It rises in tandem with displacement and vice versa.

Blocks 2 and 1 are the same and are frictionlessly raised to the identical height, h = Z.

As a result, both blocks will have the same mass and gravitational pull.

m1 = m2

FG = m1g = m2g

Gravitational acceleration (g) is indicated here.

the labor put in while block 2 is subject to the gravitational pull.

W2 = m2gh

When gravity pulls on block 1, what labor is done on block 1?

W1 = m1gh

W1 = m1gZ

W1 = m2gZ , [m1=m2]

W1 = W2

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The question is -

Two identical blocks are pushed up frictionless inclines as shown. Consider the portion of the motion of block 1 during which it moves a distance of L 1 and the portion of the motion of block 2 during which it moves a distance of L 2. Each block is displaced the same vertical distance Z. The hand pushes parallel to the incline with the same magnitude of force F B H in both cases. Block 2 moves at a decreasing speed. Is the absolute value of the work done by the gravitational force on block 2 greater than, less than, or equal to the work done by the gravitational force on block 1? Explain.

Answer:

5.0 s

Explanation:

The equation for time is:

\(t = \frac{d}{s} \)

We plug in the given values into the equation

\(t=\frac{16\ m}{3.2\ m/s^}\)

\(t=5.0 \ s\)

Which of the following substances is formed during photosynthesis?

⠀ ⠀A. Carbon dioxide

⠀ ⠀B. Light

⠀ ⠀C. Sugar

⠀ ⠀D. Water

Sugar is formed during photosynthesis.

So, correct option is C. Sugar.

During photosynthesis, Carbon dioxide and water are used to produce sugar with help of light.

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:

B

Explanation:

The height of the jump having a mass of 67kg with the vertical velocity of 6.4m/s is 2.089m.

Potential energy is defined as the energy stored in an object or system by virtue of its position or arrangement of parts while kinetic energy is defined as the energy of the moving particles of an object or system.

Potential energy is represented as :

P.E.= mgh

Kinetic energy  is represented as :

\(K.E.= 1/2mv^2\)

where, m = mass of the object

h= height

v= velocity

g= acceleration due to gravity which is \(9.8m/s^2\)

For above given example,

m= 67kg, v= 6.4m/s

we first find out Kinetic energy,

K.E.=\(\frac{1}{2} 67* (6.4)^2\)= 2744.32/2= 1372.16 J

As we know, K.E.= P.E.

So, P.E.= 1372.16J = mgh

1372.16= 67* 9.8* h

height= 2.089 m

Thus, the height of the jump having a mass of 67kg with the vertical velocity of 6.4m/s is 2.089m.

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Answer:

b) 0 = y₀ + 0 - ½ g t² ,  v₀ₓ = x / t

c)  v₀ₓ = 30.31 m / s

Explanation:

This is a projectile launching exercise, in your statement you have units of several systems, we are going to reduce everything to the SI system

   v = 35 mph (0.447 m / s / 1 mph) = 15.645 m / s

   y₀ = 75 m

A) In the attachment we can see a diagram of the movement of the vehicle in its fall

B) let's find the time for the fall to the base of the cliff (y = 0)

        y = y₀ + \(v_{oy}\) t - ½ g t²

when the vehicle leaves the cliff it goes horizontally, so its vertical speed is zero (v_{oy} = 0)

         0 = y₀ + 0 - ½ g t²

         t = √2y₀ / g

with this time we can use the equation of motion on the x axis

         x = v₀ₓ t

         v₀ₓ = x / t

C) we perform the calculations

            t = √ (2 30 / 9.8)

            t = 2.474 s

           v₀ₓ = 75 / 2.474

           v₀ₓ = 30.31 m / s

D) as we can see, the vehicle speed when leaving the cliff is almost twice the allowed speed (15.6 m / s)

therefore the city is not responsible for the accident

It is measured in the airless tube because  there is no air to offer resistance to the fall of the object.

The acceleration due to gravity refers to the acceleration that is imparted to a body because of its movement in the earth's gravitational field. Irrespective of mass, all objects move with the same acceleration in the earth gravitational field.

If air is present, air resistance makes it appear as if different masses are accelerated to different amounts. Thus, the acceleration due to gravity is measured in an airless tube the where there is no air to offer resistance to the fall of the object.

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Answer: The answer is A, energy

Explanation: Did the test

a. It takes 13.94 seconds to change your displacement by 18.0m east.

b. It takes 28.9 seconds to change your displacement by 32.0m north,

In geometry and mechanics, a displacement is described as a vector whose length is the shortest distance from the initial to the final position of a point P undergoing motion.

To change your displacement by 18.0m east, we use the

X direction = \(\frac{18}{1.70 cos 22.7}\)

x direction = 13.94 seconds

To change your displacement by 32.0m north, we use the

Y direction = \(\frac{32}{1.70 sin 22.7}\)

y direction= 28.9 seconds

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Answer:

Therefore, the 10 kg mass should be placed at x = 5.7 m along the x-axis to achieve a center of mass located at y = 4.5 m.

Explanation:

To find the position along the x-axis where a 10 kg mass should be placed such that the center of mass is located at y = 4.5, we can use the formula for the center of mass:

x_cm = (m1 * x1 + m2 * x2) / (m1 + m2)

Here, m1 and x1 represent the mass and position of the 8 kg mass, respectively. m2 is the mass of the 10 kg mass, and we need to find x2, its position.

Given:

m1 = 8 kg

x1 = 3 m

x_cm = unknown (to be found)

m2 = 10 kg

y_cm = 4.5 m

Since the center of mass is at y = 4.5, we only need to consider the y-coordinate when calculating the center of mass position along the x-axis.

To solve for x2, we can rearrange the formula as follows:

x2 = (x_cm * (m1 + m2) - m1 * x1) / m2

Substituting the given values:

x2 = (x_cm * (8 kg + 10 kg) - 8 kg * 3 m) / 10 kg

Simplifying:

x2 = (x_cm * 18 kg - 24 kg*m) / 10 kg

Now, we can set the y-coordinate of the center of mass equal to 4.5 m and solve for x_cm:

4.5 m = (8 kg * 3 m + 10 kg * x2) / (8 kg + 10 kg)

Simplifying:

4.5 m = (24 kg + 10 kg * x2) / 18 kg

Multiplying both sides by 18 kg:

81 kg*m = 24 kg + 10 kg * x2

Subtracting 24 kg from both sides:

10 kg * x2 = 81 kg*m - 24 kg

Dividing both sides by 10 kg:

x2 = (81 kg*m - 24 kg) / 10 kg

Simplifying:

x2 = 8.1 m - 2.4 m

x2 = 5.7 m

(brainlest?) ples:(

Answer:

the 10 kg mass should be placed at x = -2.4 m to achieve a center of mass at y = 4.5 m.

Explanation:

To find the position along the x-axis where the 10 kg mass should be placed so that the center of mass is located at y = 4.5, we can use the principle of the center of mass.

The center of mass of a system is given by the equation:

x_cm = (m1x1 + m2x2) / (m1 + m2),

where x_cm is the x-coordinate of the center of mass, m1 and m2 are the masses, and x1 and x2 are the positions along the x-axis.

Given:

m1 = 8 kg,

x1 = 3 m,

m2 = 10 kg,

y_cm = 4.5 m.

To solve for x2, we need to find the x-coordinate of the center of mass (x_cm) by using the y-coordinate:

y_cm = (m1y1 + m2y2) / (m1 + m2),

where y1 and y2 are the positions along the y-axis.

Rearranging the equation and substituting the given values:

4.5 = (83 + 10y2) / (8 + 10).

Simplifying the equation:

4.5 = (24 + 10*y2) / 18.

Multiplying both sides by 18:

81 = 24 + 10*y2.

Rearranging the equation:

10*y2 = 81 - 24,

10*y2 = 57.

Dividing both sides by 10:

y2 = 5.7.

Therefore, the y-coordinate of the 10 kg mass should be 5.7 m.

To find the x-coordinate of the 10 kg mass, we can use the equation for the center of mass:

x_cm = (m1x1 + m2x2) / (m1 + m2).

Substituting the given values:

x_cm = (83 + 10x2) / (8 + 10).

Since the center of mass is at x_cm = 0 (the origin), we can solve for x2:

0 = (83 + 10x2) / (8 + 10).

Rearranging the equation:

83 + 10x2 = 0.

24 + 10*x2 = 0.

10*x2 = -24.

Dividing both sides by 10:

x2 = -2.4.

I think I have already answer this question. Anyways please do let me know whether it is useful or not

To solve this problem, we can use Coulomb's law, which states that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.

We can then use the electric force to find the electric field at the location of q3 and the initial acceleration of q3.

a) To find the electric field at the location of q3, we can first find the electric force on q3 due to q1 and q2 and then use the definition of the electric field, which is the electric force per unit charge. The electric force on q³ due to q¹ and q² is:

F1 = k x q¹ x q³/ r1²

F2 = k x q² x q³ / r2²

where r¹ and r² are the distances from q¹ and q² to q³, respectively, and k is Coulomb's constant.

Since q³ is equidistant from q¹ and q², we have r¹ = r² = 0.20 m. Substituting the given values, we get:

F1 = (9.0 x 10⁹ N-m²/C²) x (4.0 x 10⁻⁶ C) x (2.0 x 10⁻⁶C) / (0.20 m)² = 1.8 N

F2 = (9.0 x 10⁹ N-m⁻²/C²) x (-6.0 x 10⁻⁶ C) x (2.0 x 10⁻⁶C) / (0.20 m)² = -5.4 N

The negative sign of F2 indicates that the force on q³ due to q² is in the opposite direction to the force due to q¹.

The net electric force on q3 is the vector sum of the forces due to q1 and q2:

Fnet = F1 + F2 = 1.8 N - 5.4 N = -3.6 N

The electric field at the location of q³ is then:

E = Fnet / q³ = (-3.6 N) / (2.0 x 10⁻⁶ C) = -1.8 x 10⁻⁶N/C

The negative sign of the electric field indicates that the field is directed towards q².

b) To find the initial acceleration of q³, we can use Newton's second law, which states that the net force on an object is equal to its mass times its acceleration:

Fnet = ma

where m is the mass of q³ and a is its initial acceleration.

Substituting the given values, we get:

-3.6 N = (2.0 x 10⁻⁶ kg) x a

Solving for a, we get:

a = -1.8 x 10³ m/s²

The negative sign of the acceleration indicates that it is directed towards q².

c) The direction of the initial acceleration of q³ is towards q².

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Answer:

B, the internet serves to provide people with more insightful explanations on things that they have not experienced yet but want to find out more on.