Three 20-o resistors are connected in series across a 120-V generator. Find the equivalent resistance of the circuit.
a. 300
C. 200
b. 600
d. 800

Answer:

When light traveling through air enters water at an angle, it moves in a curve in air and changes direction when it enters water. This bending of light is known as refraction. The angle at which the light bends depends on the difference in the refractive indices of the two media, which is why the light bends when it enters the denser medium of water.

The magnitude of the applied force and the mass of the object, together determine how an object's motion will change in response to the applied force.

When describing the effect of an applied force on an object's motion, two important factors to consider are:

Magnitude of the Force: The magnitude or strength of the applied force determines the amount of acceleration or deceleration experienced by the object. According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. A greater force will result in a greater acceleration, while a smaller force will result in a smaller acceleration. Additionally, the direction of the force relative to the object's initial motion will determine if it speeds up, slows down, or changes direction.

Mass of the Object: The mass of the object being acted upon is another crucial factor. As mentioned earlier, according to Newton's second law, the acceleration of an object is inversely proportional to its mass. This means that for a given force, an object with a larger mass will experience a smaller acceleration compared to an object with a smaller mass. In simpler terms, it requires more force to accelerate a heavier object compared to a lighter object.

These two factors, the magnitude of the applied force and the mass of the object, together determine how an object's motion will change in response to the applied force.

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The most likely situation to occur if the climate changed and caused glaciers and ice caps to melt is:
D. Land masses would become smaller.

When glaciers and ice caps melt due to climate change, the water released from the melting ice flows into rivers, lakes, and eventually the oceans. This increase in water volume contributes to a rise in sea levels. As sea levels rise, low-lying coastal areas and islands may become submerged, leading to a reduction in the size of land masses.
In addition, the melting of glaciers and ice caps can cause other changes in the environment. While it might seem like the added freshwater could make the ocean water less salty, the influx of cold freshwater can actually disrupt ocean currents, which are crucial for regulating Earth's climate. As a result, certain regions may experience altered weather patterns and temperatures. However, these changes are complex and interconnected, making it difficult to predict specific outcomes, such as air temperature decrease or increased land fertility, solely based on melting ice caps and glaciers.
Overall, the most direct and likely impact of melting glaciers and ice caps on a global scale would be the reduction in land mass size due to rising sea levels.

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Answer:

The  velocity  is \(v_2 = 0.96 \ ft/s\)

Explanation:

From the question we are told that

   The initial speed is  \(v_1 = 4 \ ft/s\)

   The  cross -sectional area of the first pipe is  \(A_1 = 15.7 \ ft\)

   The  cross -sectional area of the second pipe is \(A_2 = 65.4 \ ft\)

Generally from continuity equation we have that

     \(A_1 * v_1 = A_2 * v_2\)

So  

     \(v_2 = \frac{A_1 * v_1 }{A_2 }\)

=>   \(v_2 = \frac{15.7 * 4 }{65.4 }\)

=>   \(v_2 = 0.96 \ ft/s\)

Explanation:

thermodynamics is the study of heat.

Answer The study of heat and its relationship to useful work is called thermodynamics and involves macroscopic quantities such as pressure, temperature, and volume without regard for the molecular basis of these quantitie

Explanation:

your answer is b to be presided

RLLY?

Explanation:

ALSO

MY DAD ALWAYS TELLS ME THAT IF YOUR FEET ARE COLD THAT YOU NEED TO PUT A HAT ON?!?!?!?!?!

A current of a 6 flows through a light bulb for 12 s. The total charge that passes through the light bulb during the given time is 72 coulombs.

To calculate the total charge that passes through the light bulb, we need to use the formula Q = I * t, where Q represents the charge in coulombs, I represents the current in amperes, and t represents the time in seconds.

Step 1: Identify the known values:

Current (I) = 6 amperes

Time (t) = 12 seconds

Step 2: Calculate the charge using the formula:

Q = I * t

Step 3: Substitute the known values into the formula:

Q = 6 amperes * 12 seconds

Q = 72 coulombs

Therefore, the total charge that passes through the light bulb during the given time is 72 coulombs.

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Answer: Pressure of gases.

Explanation:

In a heat engine, the air-fuel mixture is ignited, which causes an increase in pressure of the gases inside the cylinder. This pressure pushes the piston, which is connected to a crankshaft, causing it to rotate and do work. Therefore, pressure is an essential property of gases for the ability of a heat engine to do work.

Answer:

The parent cell

Explanation:

Answer

Mitosis phase - The DNA in the parent cell nucleus makes a copy of itself and is then split between the two daughter cells during mitosis

Explanation

This is biology, not really physics.

0.00455 moles of air must be released from the tire to restore the pressure to 210 kPa at 50 degrees C.

The pressure rise in a 0k tire when the air temperature rises from 25 degrees C to 50 degrees C can be calculated using the ideal gas law. The formula is

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature. In this case, the temperature is rising from 25 degrees C to 50 degrees C, so the pressure will increase with the temperature.

By plugging these values into the equation, we can calculate that the pressure increase is 45.5 kPa.

To restore the pressure to its original value of 210 kPa at 50 degrees C, we need to bleed off the amount of air equal to the pressure increase. This is equal to 45.5 kPa, or 0.00455 moles. This means that 0.00455 moles of air must be released from the tire to restore the pressure to 210 kPa at 50 degrees C.

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Based on the observations during the experiment, the car's velocity can be described as follows. .

The car had a   constant speed of approximately 60 km/h throughout the experiment,indicating a consistent rate of motion.

In terms of   direction, the car initially traveledin a straight line towards the east.

However, after a certain point,   it made a sharp turn towards the north, changing its direction but maintaining thesame speed.

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Energy required to heat the gas at constant pressure is greater than the energy needed to heat it at constant volume. The correct statement is option (a).

This is because, during heating, the gas undergoes an increase in temperature and, therefore, an increase in internal energy. The amount of energy required to increase the internal energy of the gas depends on the specific heat of the gas and the change in temperature. At constant pressure, the volume of the gas also changes, requiring additional energy to perform work against the surrounding pressure.

Hence, more energy is required to heat the gas at constant pressure compared to constant volume, where the volume remains the same and no work is done.

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--The complete question is, A gas is heated from temperature T1 to T2 (T2 > T1). In this context, which of the following statements is correct?

Multiple Choice

a. Energy required to heat the gas at constant pressure is greater than the energy needed to heat it at constant volume.

b. Energy required to heat the gas at constant pressure is lower than the energy needed to heat it at constant volume.

c. Energy required to heat the gas at constant pressure is non-zero and the energy required to heat it at constant volume is zero.

d. Energy required to heat the gas at constant pressure is zero and the energy required to heat it at constant volume is non-zero.--

The mass of the metal ball is approximately equal to 0.1 kg.

Freefall is a type of motion in which an object falls with gravitational acceleration without resistance from air, surface, or medium. In a freefall, the only force acting on an object is the force of gravity. In a freefall, the acceleration of an object is given by the formula:  `g = 9.81 \(m/s^2`\)

The final velocity of a freefalling object can be calculated using the formula: `v = g*t `Where v is the final velocity, g is the gravitational acceleration, and t is the time taken to reach the final velocity. The force acting on an object is equal to the product of the mass of the object and the acceleration acting on the object. This is known as Newton's Second Law of Motion.

The formula for calculating force is:  `F = ma`.Where F is the force acting on the object, m is the mass of the object, and a is the acceleration acting on the object.Given that the velocity of the metal ball at the moment of striking the ground was 30 m/s, and that it stopped after penetrating the sand for 0.01 s, we can calculate the mass of the metal ball using the formula for force.

We can assume that the force acting on the metal ball is equal to the average resistance force of the sand against the motion of the ball.

The formula for calculating force is: `F = ma`.

Rearranging the formula, we get: `m = F/a` .

Substituting the given values, we get:  `

m = -3010/10`

Thus, the mass of the metal ball is approximately equal to 0.1 kg.

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15.8m/s is the final velocity

Deceleration is calculated by dividing the final velocity minus the initial velocity by the time it takes for this velocity to drop. Here the acceleration formula can be used with a negative sign to determine the deceleration value.

The rate of change of the velocity of an object in the time is known as Acceleration . Velocity is defined as the speed of an object in motion by time.

here we know that the initial velocity is 32m/s.

the deceleration is -1.5m/s2 and the time taken is 10.8s

so from the formula

                                       V=U + at

  = 32+-1.5 x10.8

  =15.8m/s          

So the final velocity is 15.8m/s

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Answer:

Walking –We can walk only if we apply frictional force. Friction is what holds your shoe to the ground. The friction present on the ice is very little, this is the reason why it is hard to walk on the slippery surface of the ice.

Writing – A frictional force is created when the tip of the pen comes in contact with the surface of the paper. Rolling friction is what comes into play while writing with a ballpoint pen while sliding friction arises when one writes with a pencil.

Skating – A thin film of water under the blade is necessary to make the skate slide. The heat generated by the skate blade rubbing against the surface of ice causes some of the ice to melt right below the blade where the skater glides over the ice. This water acts as a lubricant reducing friction.

Lighting a matchstick – When the head of the matchstick is rubbed against a rough surface, heat is generated and this heat converts red phosphorous to white phosphorous. White phosphorous is highly inflammable and the match stick ignites. Sometimes, matchsticks fail to ignite due to the presence of water. Water lowers friction.

Driving of the vehicle on a surface- While driving a vehicle, the engine generates a force on the driving wheels. This force initiates the vehicle to move forwards. Friction is the force that opposes the tyre rubber from sliding on the road surface. This friction avoids skidding of vehicles.

Applications of breaks in the vehicle to stop it- Friction braking is the most widely used braking method in vehicles. This process involves the conversion of kinetic energy to thermal energy by applying friction to the moving parts of a vehicle. The friction force resists the motion and in turn, generates heat. This conversion of energy eventually bringing the velocity to zero.

Flight of aeroplanes- Drag is the force that opposes the forward motion of the aeroplane. The friction which resists the motion of an object moving through a fluid or immobile in a moving fluid, as occurs when we fly a kite. The friction of the air is created as it meets and passes over an aeroplane and its components. Drag is generated by air impact force, skin friction and displacement of the air.

Drilling a nail into the wall- Friction is responsible for fixing of nails in a wall. As the nail is driven into the wall, the nearby material to the nail of the wall gets compressed. This exerts a force on the nail. This force is the friction that converts the normal force exerted by the compressed layers of the wall into the resisting shear force. In this manner the friction cause nails and screws to hold on to walls.

The dusting of the carpet by beating it with a stick- When the carpet is beaten with the stick, the dust comes out. The dust is carried off by the wind or falls on the floor. The carpet exhibits a little static friction that holds the dust to the carpet.  When the carpet is beaten, it will overcome the friction and the carpet will move away from the dust making the carpet free from dust.

Sliding on a garden slide- We know that friction is a force that is present whenever two objects rub against each other. In case of a slide in the garden such as a slide and a person’s backside rub each other’s surface. Without friction, a slide would accelerate the rider too quickly, resulting in possible injury due to the fall. The friction reduces the velocity of the sliding person and makes him stop.

Hope that helps! :D Sorry if it's too lengthy...

Explanation:

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

20 seconds

Explanation:

We are given 2 givens in the first statement

v0=0 and a=5

And we are trying to find time needed to cover 1km or 1000m.

So we use

x-x0=v0t+1/2at²

Plug in givens

1000=0+2.5t²

solve for t

t²=400

t=20s

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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Answer:

24 cm/s

Explanation:

Applying

Pythagoras theorem,

a² = b²+c²............. Equation 1

Where a = resultant, b = vertical component, c = horizontal component

From the question,

Given: a = 26 cm/s, c = 10 cm/s

Substitute these values into equation 1

26² = b²+10²

676 = b²+100

b² = 676-100

b² = 576

b = √576

b = 24 cm/s

Answer:

2

Explanation:

pulling force because of it force

Answer:

5.9 cm

Explanation:

f: frequency of oscillation

frequency of oscillationk: spring constant

frequency of oscillationk: spring constantm: the mass

\(f = \frac{1}{2\pi} \sqrt{ \frac{k}{m} } \)

in this problem we know,

F= 1.4 Hz

m= 0.26 kg

By re-arranging the formula we get

\(k = {(2\pi \: f )}^{2} m = {(2\pi(1.4hz))}^{2} 0.26kg = 20.1 \frac{n}{m} \)

The restoring force of the spring is:

F= kx

where

F= 1.2 N

k= 20.1 N/m

x: the displacement of the block

\(x = \frac{f}{k} = \frac{1.2 \: n}{20.1 \frac{n}{m} } = 0.059m \: = 5.9 \: cm\)

A procedure known as magnetic separation can be used to extract magnetic elements from coal.

This method makes use of the magnetic characteristics of some materials to distinguish them from non-magnetic materials like coal. A description of how magnetic separation can be used to remove magnetic components from coal is given below:

Putting a magnetic field around the coal and magnetic material mixture is the first step in the magnetization process. This can be achieved by creating an electromagnetic field or by putting the mixture close to a powerful magnet.

Magnetism: Magnetic materials, such as iron atoms or magnetite that are frequently found in coal, will be drawn to the magnetic field and become magnetized. They line up their magnetic moments with the magnetic field's direction.

Separation: The magnetic coal components can be physically separated from the non-magnetic coal once they have been magnetized. To create this separation, there are numerous techniques:

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Answer:

a) v(f) = -4i - 5j

b) 4.18 m

Explanation:

The equation to be used for this question is

v(c)m(c) + v(t)m(t) = [m(c) + m(t)] v(f)

if we rearrange and make v(f) subject of formula, then

v(f) = v(c)m(c) + v(t)m(t) / [m(c) + m(t)]

One vehicle is headed towards south and the other vehicle, west when they collide they will travel together in a southwestern direction. This means that both vehicles are traveling in the negative direction taking a standard frame of reference. Thus, we can write the equation in component form by substituting the values as

v(f) = 1225(-9.5i) + 1654(-8.6j) / 1225 + 1654

v(f) = -11637.5i - 14224.4j / 2879

v(f) = -4i - 5j m/s

From the answer,

v(f) = √(4² + 5²)

v(f) = √41

v(f) = 6.4 m/s

And we know that

KE = ½mv²

Fd = umgd

And, KE = Fd, so

½mv² = umgd

½v² = ugd

Making d the subject of formula,

d = v²/2ug

d = 6.4² / 2 * 0.5 * 9.8

d = 41 / 9.8

d = 4.18 m

(a) The velocity of the system after collision is 4.04 i + 4.9 j.

(b)The distance traveled by the vehicles after collision is 1.73 m.

The given parameters;

Apply the principle of conservation of linear momentum to determine the velocity of the system after collision;

\(m_1u_x_1 + m_2 u_y_2 = V(m_1 + m_2)\\\\V = \frac{(1225\times 9.5)_x \ + \ (1654 \times 8.6)_y}{m_1 + m_2} \\\\V = \frac{(1225\times 9.5)_x \ + \ (1654 \times 8.6)_y}{1225+ 1654} \\\\V= \frac{(11,637.5)_x \ + \ (14,224.4)_y}{2879} \\\\V = 4.04x \ + 4.94y\\\\V = 4.04i \ + 4.9 j\)

The magnitude of the final velocity of the system is calculated as;

\(V = \sqrt{v_x^2 + v_y^2} \\\\V = \sqrt{(4.04)^2 + (4.9)^2} \\\\V = 6.35 \ m/s\)

The change in the mechanical energy of the system;

\(\Delta K.E = K.E_f - K.E_i\\\\\)

The initial kinetic energy of the cars before collision is calculated as;

\(K.E_i = \frac{1}{2} m_1u_1_x^2 \ + \frac{1}{2} m_1u_2_y^2 \\\\K.E_i = \frac{1}{2} (1225)(9.5)^2\ + \frac{1}{2} (1654)(8.6)^2\\\\K.E_i = 55,278.13_x \ + \ 61,164.92_y\\\\K.E_i = \sqrt{55,278.13^2 \ + \ 61,164.92^2} \\\\K.E_i = \sqrt{6,796,819,094.9} \\\\K.E_i = 82,442.82 \ J\)

The final kinetic energy of the system;

\(K.E_f = \frac{1}{2} (m_1 + m_2)V^2\\\\K.E_f = \frac{1}{2} (1225 + 1654)(6.35)^2\\\\K.E_f = 58,044.24 \ J\)

The change in kinetic energy is calculated as;

\(\Delta K.E = K.E_f -K.E_i\\\\\Delta K.E= (58,044.24) - (82,442.82)\\\\\Delta K.E = -24,398.58 \ J\)

Apply the principle of work-energy theorem, to determine the distance traveled by the vehicles after collision;

\(W = \Delta K.E\\\\- \mu Fd = - 24,398.58\\\\\mu mgd= 24,398.58\\\\d = \frac{24,398.58}{\mu mg} \\\\d = \frac{24,398.58}{0.5 \times 9.8(1225 + 1654)} \\\\d = 1.73 \ m\)

Thus, the distance traveled by the vehicles after collision is 1.73 m.

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The energy of the capacitor at this larger plate separation will be= D) U/4

The energy stored in a parallel-plate capacitor with plate area A, plate separation d, and charge Q is given by the formula:

U = (1/2) * (Q^2 / (ε_0 * A * d))

where ε_0 is the permittivity of free space.

If the separation between the plates is increased from d to 2d without changing the charge on the plates, the capacitance of the capacitor will be reduced by a factor of 2. This is because the capacitance of a parallel-plate capacitor is given by the formula:

C = ε_0 * A / d

So, when the separation is doubled, the capacitance is halved.

Since the charge on the plates remains constant, the energy stored in the capacitor is proportional to the square of the charge and inversely proportional to the capacitance. Thus, the new energy of the capacitor when the plates are separated by 2d is given by:

U' = (1/2) * (Q^2 / (ε_0 * A * (2d))) * (1/2)

where the factor of 1/2 is included because the capacitance is halved.

Simplifying this expression, we get:

U' = U / 4

Therefore, the energy of the capacitor at the larger plate separation is (d) U/4.

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Answer:

Specific heat capacity is an intensive property and does not depend on sample size.

Explanation:

Answer:

10

Explanation:

i dont know this is right answer

0.35

0.91

1.26

1.57

Explanation:

I did it on ed2020 and got it correct

Answer:

c = 115.92 q/g C

Explanation:

0.016 kg = 16 grams

q = Joules

m = mass (g)

t = (final temp. - initial temp)

c = specific heat

c = (1086.75 joules) / (175 C - 25 C) (16 g)

c = 115.92 q/g C