9. An object of mass 88.444 kg is sliding on a horizontalsurface. Calculate the normal force exerted bythe surface on the object. (1 point)A. O 270.156 NB. O 659.934 NC. O1289.839 ND. O1165.906 NE. O866.751 N[

Answer:

The wavelength of a wave with the frequency of 330hz and a speed of 343m/s would be 1.04m

Explanation:

You can get the wavelength of a wave by dividing the speed of the wave by its frequency, which in this case would be:

343/300, which as a decimal number, it'd be 1.04.

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Answer:

Probeware and computer

Explanation:

Computers are more powerful and better than a graphing calculator for this situation.

are the tools he must use.

The magnitude of normal force (N) on motorcycle is 704.537 N

In this type of scenario, we have been told that the mass, denoted by "m," is 350 kilograms, that the speed of the motorcycle, denoted by "v," is 29 meters per second, and that the radius of curvature, denoted by "R," is 105 meters. We can utilize the second law of motion that Newton proposed to compute the magnitude of the normal force, which is as follows:

\(m . g - N = \frac{mv^{2} }{R}\)

By applying the formula above we have:

\(350 x 9.8 - N= \frac{350x29^{2} }{108}\)

3430 - N = 2725.46296

N = 3430 - 2725.46296

N = 704.537 N

Thus, the magnitude of normal force on motorcycle would be 704.537 N

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Taking into account the definition of kinetic energy, the velocity of the object with a kinetic energy of 250 J and a mass of 32 kg is 3.95 m/s².

Kinetic energy is the energy possessed by a body or system due to its movement.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a certain mass and in a position of rest, until it reaches a certain speed. Once the final speed is reached, the amount of kinetic energy accumulated will remain constant, that is, it will not vary, unless another force acts on the body again.

Kinetic energy is represented by the following expression:

Ec = 1/2×m×v²

Where:

In this case, you know:

Replacing in the definition of kinetic energy:

250 J = 1/2× 32 kg×v²

Solving:

250 J÷ (1/2× 32 kg) = v²

15.625 J÷kg = v²

√15.625 J÷kg = v

3.95 m/s² = v

Finally, the velocity of the object is 3.95 m/s².

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The total initial mechanical energy is 590,650 J.

Work done on the projectile is 155,436 J.

Speed of the projectile is 115.4 m/s.

(a) The initial total mechanical energy of the system of the projectile and the earth can be found using the equation:

E = KE + PE

where E is the total mechanical energy, KE is the kinetic energy, and PE is the potential energy. Initially, the projectile is at a height of 116 m, so the potential energy is:

PE = mgh

PE = (54.0 kg)(9.81 m/s²)(116 m)

PE = 61,450 J

The initial speed of the projectile is 1.40 x 10² m/s, so the initial kinetic energy is:

KE = (1/2)mv²

KE = (1/2)(54.0 kg)(1.40 x 10² m/s)²

KE = 529,200 J

Therefore, the initial total mechanical energy is:

E = KE + PE

E = 529,200 J + 61,450 J

E = 590,650 J

The initial total mechanical energy of the system of the projectile and the earth is 63,014 J.

(b) At the maximum height, the potential energy of the projectile is:

PE = mgy

PE = (54.0 kg)(9.81 m/s²)(320 m)

PE = 169,517 J

At the maximum height, the kinetic energy of the projectile is:

KE = (1/2)mv²

KE = (1/2)(54.0 kg)(99.2 m/s)²

KE = 265,697 J

The total mechanical energy of the projectile at the maximum height is:

E = KE + PE

E = 169,517 J + 265,697 J

E = 435,214 J

The work done by air resistance is the difference between the initial total mechanical energy and the total mechanical energy at the maximum height:

W = Ei - Ef

W = 590,650 J - 435,214 J

W = 155,436 J

The amount of work done on the projectile by air resistance is 155,436 J.

(c) Let Wup be the work done by air resistance when the projectile is going up, and let Wdown be the work done by air resistance when the projectile is going down. According to the problem, Wdown = (3/2)Wup. Use the principle of conservation of energy to relate the speed of the projectile at its maximum height to its speed just before hitting the ground:

Ef = Ei

KEf + PEf = KEi + PEi

At the maximum height, the kinetic energy of the projectile is zero, so:

PEf = Ei - PEi

PEf = 590,650 J - (54.0 kg)(9.81 m/s²)(320 m)

PEf = 421,133 J

The total mechanical energy just before hitting the ground is the same as the total mechanical energy at the maximum height, so:

KEf + PEf = KEi + PEi

0 + 421,133 J = (1/2)mv² + 61,450 J

(1/2)(54.0 kg)v² = 359,683 J

v² = 13,321.59

v = 115.4 m/s

Therefore, the speed of the projectile immediately before it hits the ground is 115.4 m/s.

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The Kinetic energy of the car with a mass of 80 grams is 0.40 joules

From the graph given, we can see that the as the mass increase, the kinetic energy also increase. Thus, we can say that the kinetic energy and mass of the car are in direct proportionality.

Now, we shall obtain the velocity of the car. Details below:

KE = ½mv²

0.1 = ½ × 0.02 × v²

0.1 = 0.01 × v²

Divide both side by 0.01

v² = 0.1 / 0.01

Take the square root of both side

v = √(0.1 / 0.01)

v = 3.16 m/s

Finally, we shall determine the kinetic energy of the car of mass 80 grams. Details below:

KE = ½mv²

KE = ½ × 0.08 × 3.16²

Kinetic energy = 0.40 joules

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The wavelength of A2 with frequency (110hz) is 3.01 m.

Wavelength of a sound wave is the distance between successive similar points in the wave such as rarefactions or compressions.

A tuba is a musical instrument that produces sound waves.

Wavelength is related to frequency and velocity by the formula below:

The wavelength of A2 with frequency (110hz) is 3.01 m.

In conclusion, the wavelength of a wave is inversely proportional to frequency.

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Answer:

south

Explanation:

The pattern of lines of force around two positive charges separated from each other demonstrates the repulsive nature of like charges and the direction and strength of the electric field between them.

When two positive charges are separated from each other, the lines of force (also known as electric field lines) originate from one positive charge and terminate on the other positive charge. The lines of force follow a pattern that reflects the repulsion between the positive charges.

Here's a verbal description of the pattern:

From each positive charge, the lines of force radiate outward in all directions.

The lines of force are evenly spaced and radially symmetric around each charge, indicating that the electric field strength is the same at all points along a given line.

As the lines of force move away from each charge, they curve away from each other, reflecting the repulsion between the positive charges.

The density of lines of force is higher near the charges and decreases as they move further apart.

The lines of force never cross each other, maintaining their continuous and unbroken nature.

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Answer:

20 characters

Explanation:

The bottle is squeezed, when a bottle with a reservoir is squeezed, is the fluid will come up through the bottom of the reservoir. Thus, the correct option is D.

A reservoir is an artificial lake where the water is stored. Most of the reservoirs are formed through the construction of the dams across major rivers. A reservoir can also be formed from a natural lake whose outlet has been dammed to control the water level in that.

When we squeeze an open bottle which is filled with water, the water in the bottle will spill out. When we squeeze a bottle, the material collapses where we squeeze it, however it expands in other areas, resulting in a constant volume. If the volume is constant, then we would think that the water should not spill out.

Therefore, the correct option is D.

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The mass of the object is approximately 0.457 kg.

The mass of the object attached to the trolley can be calculated using the principle of conservation of momentum. Since the two trolleys couple together and move as a single system after the collision, the total momentum before and after the collision should be the same. Given the mass of one trolley is 0.80 kg and the initial speed is 1.1 m/s, the momentum before the collision is 0.80 kg * 1.1 m/s = 0.88 kg·m/s. After the collision, the total mass is the sum of the two trolleys, and the final speed is 0.70 m/s.

Using the momentum equation, the mass of the object can be calculated as follows:

Total momentum before collision = Total momentum after collision

0.88 kg·m/s = (0.80 kg + mass of the object) * 0.70 m/s

Solving for the mass of the object, we get:

0.88 kg·m/s = (0.80 kg + mass of the object) * 0.70 m/s

0.88 kg·m/s = 0.56 kg + 0.70 kg * mass of the object

0.88 kg·m/s - 0.56 kg = 0.70 kg * mass of the object

0.32 kg = 0.70 kg * mass of the object

Dividing both sides by 0.70 kg, we find:

mass of the object = 0.32 kg / 0.70 kg = 0.457 kg

The two trolleys collide and couple together, the total momentum before the collision is equal to the total momentum after the collision according to the principle of conservation of momentum.

The momentum of an object is defined as the product of its mass and velocity. In this case, the mass of one trolley is known (0.80 kg) and the initial speed is given (1.1 m/s), allowing us to calculate the momentum before the collision.

After the collision, the two trolleys move together at a new speed (0.70 m/s). By setting the initial momentum equal to the final momentum and solving for the unknown mass of the object, we can find its value.

In the calculation, we subtract the masses of the two trolleys from the total mass in order to isolate the mass of the object.

Dividing the difference in momentum by the product of the known mass and the new speed, we obtain the mass of the object. In this case, the mass of the object is approximately 0.457 kg.

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The acceleration due to gravity near the surface of the hypothetical planet is 3.02 m/s².

The formula for acceleration due to gravity is:

g = GM/r² Where, g = acceleration due to gravity G = universal gravitational constant M = mass of the planet r = radius of the planet

In this case, since the mass of the hypothetical planet is the same as that of Earth, we can use the mass of Earth instead of M.

Therefore, g is proportional to 1/r².

So, using the ratio of radii given (1.8), we can write:

r = 1.8 x r Earth, where r Earth is the radius of Earth.

Substituting this value of r in the formula for acceleration due to gravity, we get:

g = GM/(1.8 x r Earth)² = GM/(3.24 x rEarth²) = (1/3.24)GM/rEarth²

We know that the acceleration due to gravity on Earth (g Earth) is 9.8 m/s².

Therefore, we can calculate the acceleration due to gravity on the hypothetical planet (gh) as follows:

gh = (1/3.24) x g Earth = 3.02 m/s²

Thus, the acceleration due to gravity near the surface of the hypothetical planet is 3.02 m/s².

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Answer:

Hope I helped you.

Answer: The answer [b] or also known as k

Explanation:

gear C and Z will run at the same RPM

RPM is not a unit, according to the International System of Units (SI). The word "revolution" is a semantic annotation rather than a unit, which explains this. RPM formula sign must be f for (rotational) frequency and or for angular velocity due to the measured physical quantity. s−1  or Hz is the basic SI-derived unit that corresponds. The radians per the second unit are used to express angular speed.

As a result, a disc rotating at 60 RPM  is considered to be rotating at either 2 rad/s or 1 Hz, where the former refers to angular velocity and the latter to the rate of revolutions per second.

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Question :-

Answer :-

Explanation :-

As per the provided information in the given question, we have been given that the Velocity of the ball is 12 m/s . Time is given as 36 sec . And, we have been asked to calculate the Acceleration .

For calculating the Acceleration , we will use the Formula :-

\( \bigstar \: \: \boxed{ \sf{ \: Acceleration \: = \: \dfrac{v \: - \: u}{t} \: }} \)

Where ,

Therefore , by Substituting the given values in the above Formula :-

\( \dag \: \: \: \sf { Acceleration \: = \: \dfrac{Final \: Velocity \: - \: Initial \: Velocity}{Time} } \)

\( \longmapsto \: \: \sf { Acceleration \: = \: \dfrac{0 \: - \: 12}{36}} \)

\(\longmapsto \: \: \sf {Acceleration \: = \: \dfrac{ \: 12 \: }{36}}\)

\(\longmapsto \: \: \sf {Acceleration \: = \: \dfrac{ \: 1 \: }{3}}\)

\( \longmapsto \: \bf {Acceleration \: = \: 0.33 \: m/s^{2}} \)

Hence :-

\( \underline {\rule {212pt} {4pt}} \)

(1) The acceleration of the car is determined as 17.42 m/s².

(2) The acceleration of the crate is determined as 8.89 m/s².

The acceleration of the car is calculated from the net force acting on the car.

∑F = ma

F - Ff = ma

F - μW = ma

where;

m = W/g

m = (9320)/(9.8)

m = 951.02 kg

2(9320) - 0.222(9320) = 951.02a

16,570.96 = 951.02a

a = 17.42 m/s²

F - μW = ma

F - μmg = ma

302 - 0.75(18.6 x 9.8) = 18.6a

165.29 = 18.6a

a = 8.89 m/s²

Thus, the acceleration of the car is determined as 17.42 m/s².

The acceleration of the crate is determined as 8.89 m/s².

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The momentum of the car is 15030kgm/s

The momentum of a body is the product of the mass of the body and it's velocity. It is a vector quantity and measured in kgm/s.

Momentum is expressed as :

momentum = mass× velocity

using the relation v² = u²+2as to get the velocity of the car.

v² = 0²+2×3.5× 40

v² = 280

v = √280

v = 16.7m/s

from momentum = mass× velocity

= 900× 16.7

= 15030kgm/s

therefore the momentum of the car is 15030kgm/s

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Answer:

Hey, I'm not sure I understand exactly what your assignment is but hopefully this helps . . .

put the other end against the side of the bubls base to light up the light bulb.

This will act as the bulbs conductor, lighting up bc you are making a complete electrical circuit between positive and negative terminals of the battery.

Ok hopefully this helps you out

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.  

4.44×\(10^{9}\) kg/s is the rate at which the sun mass is decreasing.

The Sun radiates energy through a process called nuclear fusion, where hydrogen atoms combine to form helium, releasing a tremendous amount of energy in the process. According to Einstein's mass-energy equivalence principle (E=mc²), this energy release corresponds to a decrease in mass.

To calculate the rate at which the Sun's mass is decreasing, we can use the formula ΔE = Δmc², where ΔE is the change in energy, Δm is the change in mass, and c is the speed of light.

Given that the Sun radiates energy at a rate of 4×10^26 W, we can substitute this value into the equation as ΔE and solve for Δm.

ΔE = 4×10^26 W

c = 3×10^8 m/s (speed of light)

Using the equation ΔE = Δmc² and rearranging it, we get Δm = ΔE / c².

Substituting the values, we have:

Δm = (4×10^26 W) / (3×10^8 m/s)²

Evaluating this expression, we find that the rate at which the Sun's mass is decreasing is approximately 4.44×10^9 kg/s.

This calculation demonstrates that the Sun's mass is gradually decreasing as it continuously radiates energy into space, primarily through the process of nuclear fusion in its core.

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Answer:

A , power

Explanation:

Hope this is useful. Have a lovely rest of your day! God bless you.

Answer:

They have the same acceleration.

Explanation:

50km/h to 60km/h = 10km/h

0km/h to 10 km/h = 10km/h

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Answer:

they have a same acceleration

Explanation:

50km/h to 60 km/h=10km/h

0km/h to 60 km/h=10km/h

Newton's laws of motion describe the behaviour of objects when they are at rest or in motion.

The 5 lbs bag of sugar is at rest in this situation, and you are exerting a 5 lbs upward force on it. The sugar bag will continue to be at rest unless an outside force acts upon it, in accordance with the first law of motion. It is being subjected to an external force that you are applying.

According to the second law of motion, theforce exerted on the bag of sugar will have a direct proportional effect on how fast it accelerates. The mass of the bag of sugar is 5 pounds, and the force you are exerting on it is 5 lbs. Because\(\textsf{$\frac{\textsf{5 lbs (force)}}{\textsf{ 5 lbs (mass)}}$ = 1 (unit of accleration)}\), the bag of sugar will accelerate by 1 (unit of acceleration).

The third law of motion states that there is an equal and opposite reaction to every action. In this case, the sugar bag pushes down on your hand with a 5 lb force when you push up on it with a 1 lb force. The force you are providing to the bag of sugar will have an equal and opposite reaction here.

Therefore, this scenario illustrates all three of Newton's laws of motion

----------------------------------------------------------------------------------------------------------

Yes, this is an example of Newton's Third Law of Motion. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. In this example, when you push upwards on the 5 lb bag of sugar with a force of 5 lb, the bag of sugar also pushes downwards on you with a force of 5 lb.In other words, the action is you pushing upwards on the bag of sugar and the reaction is the bag of sugar pushing downwards on you with the same amount of force. So, this is an example of Newton's Third Law of Motion.

Knowing the exits are exactly 7 km apart, the cars pass each other at 9:29 and 15 seconds a.m.

The relative velocity of the cars is 30 m/s - 26 m/s = 4 m/s.

The distance between the cars is 7 km = 7000 m.

The time it takes for the cars to pass each other is 7000 m / 4 m/s = 1750 seconds.

1750 seconds is 29 minutes and 15 seconds.

To calculate the time in minutes;

Let:

v_p = the speed of car P (m/s)

v_q = the speed of car Q (m/s)

d = the distance between the cars (m)

t = the time it takes for the cars to pass each other (s)

Given that:

v_p = 30 m/s

v_q = 26 m/s

d = 7000 m

Use the equation for relative velocity to find the velocity of the cars relative to each other:

v_r = v_p - v_q

v_r = 30 m/s - 26 m/s = 4 m/s

Use the equation for distance to find the time it takes for the cars to pass each other:

d = v_r × t

7000 m = 4 m/s × t

t = 7000 m / 4 m/s = 1750 s

Convert 1750 seconds to minutes and seconds:

1750 s = 29 minutes and 15 seconds

Therefore, the cars pass each other at 9:29 and 15 seconds a.m.

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Answer:

Conduction

Explanation:

Conduction is when heat or electricity is transmitted throughout a substance.

The energy transfer taking place through our body is called conduction. Our body is conductive and which make the electrical shock.

There are three different modes of energy transfer namely, conduction, convection and radiation. Conduction is the energy transfer mode on solids and convection is taking place in gases and liquids.

Radiation is the energy transfer mode through vacuum. In conduction the energy transfers easily through the chain of  closely packed molecules. Whereas in convection, the molecule which obtained an energy travels across the space and transfers to other molecules.

Our body transfers energy inside through conduction process. Skin is very conductive and the electrons from the wire can be passed through our body that's why we gets electrically shocked.

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If a child pulls a waxed wooden sled across dry snow at a constant speed with a 2.6 N force, the mass of the sled is 6.63 kg

Since the waxed wooden sled is pulled across dry snow at a constant speed,

a = 0

∑ Fx = m ax

F - f = 0

2.6 - f = 0

f = 2.6 N

f = μ N

N = m g

f = Frictional force

μ = Co-efficient of friction

N = Normal force

m = Mass

g = Acceleration due to gravity

μ = 0.04

f = μ m g

2.6 = 0.04 * m * 9.8

0.392 m = 2.6

m = 6.63 kg

Therefore, the mass of the sled is 6.63 kg

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Answer:

\(c=1\ J/g^\circ C\)

Explanation:

Given that,

Heat required, Q = 8000 J

Mass, m = 400 g

The change in temperature, \(\Delta T = 20^{\circ}\)

The heat required due to change in temperature is given by :

\(Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{8000 }{400\times 20}\\\\c=1\ J/g^\circ C\)

So, the specific heat of the substance is \(1\ J/g^\circ C\)

Answer:

Explanation:

Given parameters:

Electrical energy  = 18J

Quantity of charge Q = 12C

Unknown:

Voltage  = ?

Solution:

The expression of electrical potential energy is given as:

      Electrical potential energy  = \(\frac{1}{2}\) c v²

c is the quantity of charge

v is the voltage

           18  =  \(\frac{1}{2}\)  x 12 x v²  

           18 = 6v²  

              v² = 3  

               v = 1.7v

Answer: Height of object = 29.13475

Hope this helps!

Answer: 29.12 Meters.

Explanation: