A car accelerates from 13-m/s to 24.1-m/s in 5.4-s. what was its acceleration in m/s2?

Answer: The efficiency of the lever is 80%.

Explanation:

An efficiency is the measure of how much wok or energy is conserved in a given process. It is defined as the ratio of output work and input work. If the energy is totally conserved in a process, the the percentage efficiency will be 100%.

Mathematically,

[tex]\%\text{ efficiency}=\frac{W_{out}}{W_{in}}\times 100[/tex]

Where,

[tex]W_{out}[/tex] = Work done by the lever = 20 J

[tex]W_{in}[/tex] = Work put in the lever = 25 J

Putting values in above equation, we get:

[tex]\%\text{ efficiency}=\frac{20}{25}\times 100\\\\\%\text{ efficiency}=80\%[/tex]

Hence, the efficiency of the lever is 80%.

We must always take note that mass is an inherent property of an object. It never change unless there is change in size of the object. While weight is the product of mass and gravity, so it changes depending on the gravitational pull.

If the gravitational pull decreases as he moves from the poles to the equator, therefore its weight will decrease.

Answer:

D.Its weight will decrease

Answer:

Energy  the electron have initially [tex]E_4=-0.85 \rm ev[/tex],

The change in energy if the electron from part a now drops to the ground state is [tex]12.75\rm ev[/tex]

Explanation:

Given information:

Electron have initially in the n=4 excited state,

We know,

Energy of an electron is given by,

[tex]E_n=-13.6\rm ev\times\frac{z^2}{n^2}[/tex]

On substituting n=4 for excited state energy,

[tex]E=-13.6\rm ev\times\frac{1^2}{4^2}=-0.85\rm ev[/tex]

Change in energy from excited state to ground state,

For ground state n=1,

[tex]E=-13.6\rm ev\times z^2\times( \frac{1}{(n_2)^2}-\frac{1}{(n_1)^2})[/tex]

On substituting [tex]n_1=1[/tex],[tex]n_2=4[/tex],

[tex]\Delta E=E_4-E_1=-13.6\rm ev\times 1^2\times( \frac{1}{(4)^2}-\frac{1}{(1)^2})=12.75ev[/tex]

Hence, energy  the electron have initially [tex]E_4=-0.85 \rm ev[/tex],

NOTE:- [tex]1\rm ev=-1.602\times 10^{-19}joule[/tex]

Energy  the electron have initially [tex]E_4=-0.85 \rm ev[/tex],

The change in energy if the electron from part a now drops to the ground state is [tex]12.75\rm ev[/tex]

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The percent accuracy is 10⁻⁵%. It is determined by the formula of percent accuracy.

Given information:

Distance of satellite = 20,000km

= 2 x 10⁷ m

Uncertainty = 2 meter

Uncertainty refers to a lack of exactness or precision in measurement, calculation, or prediction. It represents the degree of doubt or error associated with a particular value or result. Uncertainty is an essential concept in various fields, including science, engineering, statistics, and decision-making.

The formula to determine percent accuracy is:

[tex]\rm Percent \ accuracy= \frac{Uncertainty}{Measured d\ distance} \times 100[/tex]

Substituting the values in the formula:

[tex]\rm Percent \ accuracy = \frac{2}{2\times 10^7}\times 100 \\\rm Percent \ accuracy = \frac{1}{10^5} \\[/tex]

Simplifying:

Percent accuracy = 10⁻⁵%.

Therefore, the percent accuracy is 10⁻⁵%.

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Answer:

a)Mechanical to heat energy.

b)potential to kinetic energy

c)kinetic to potential energy

d)kinetic to heat energy.

According to the calculations, the water balloon does not hit the professor. It comes closest to her when it is approximately 0.316 meters away.

In order to determine if the water balloon hits the professor, we need to calculate the time it takes for the balloon to reach the ground. The time can be found using the equation:

t = \sqrt{\frac{2h}{g}}

Where t is the time, h is the height (18.0 meters), and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, we find that it takes approximately 1.52 seconds for the balloon to reach the ground.

We also need to calculate the horizontal distance the professor walks in that time. The distance can be found using the equation:

d = vt

Where d is the distance, v is the velocity (0.450 m/s), and t is the time (1.52 seconds). Plugging in the values, we find that the professor walks approximately 0.684 meters during that time.

Therefore, since the professor is 1.00 meter from the point directly beneath the window, the balloon does not hit her. It comes closest to her when it is 0.316 meters away from her.

Final answer:

To calculate the electric field magnitude at a distance from a charged rod, determine the charge per unit length and use the electric field formula.

Explanation:

Electric Field and Charged Rod:

To find the magnitude of the electric field at a distance from a charged rod, you can use the formula for the electric field of a long, charged rod. Given the total charge on the rod and its length, you can calculate the charge per unit length. Then, apply the formula to find the electric field at the specified location.

Let the rod be on the x-axes with endpoints -L/2 and L/2 and uniform charge density lambda (2.6nC/0.4m = 7.25 nC/m).

The point then lies on the y-axes at d = 0.03 m.

from symmetry, the field at that point will be ascending along the y-axes.

A charge element at position x on the rod has distance sqrt(x^2 + d^2) to the point.

Also, from the geometry, the component in the y-direction is d/sqrt(x^2+d^2) times the field strength.

All in all, the infinitesimal field strength from the charge between x and x+dx is:

dE = k lambda dx * 1/(x^2+d^2) * d/sqrt(x^2+d^2)

Therefore, upon integration,

E = k lambda d INTEGRAL{dx / (x^2 + d^2)^(3/2) } where x goes from -L/2 to L/2.

This gives:

E = k lambda L / (d sqrt((L/2)^2 + d^2) )

But lambda L = Q, the total charge on the rod, so

E = k Q / ( d * sqrt((L/2)^2 + d^2) )

By six times, a person should jump farther on the moon as compared to the earth if the takeoff speed and angle are the same.

The height of the earth can be computed using the third equation of motion. The third equation of motion, referred to as one of the kinematic equations, relates the final velocity, initial velocity, acceleration, and displacement of an object.

The height of the earth is given as:

[tex]v^2-u^2= 2gH_e\\v^2 = 2\times9.8\timesH_e\\H_e = v^2/2g[/tex]

The height of the moon is given as:

[tex]v^2-u^2 = 2(g/6)H_m\\v^2 = g/3H_m\\H_m = 3v^2/g[/tex]

The ratio of both heights is given as:

[tex]H_m/H_e = 3v^2/g/(v^2/2g)\\H_m/H_e = 6\\H_m = 6H_e[/tex]

Hence, by six times, a person should jump farther on the moon as compared to the Earth if the takeoff speed and angle are the same.

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A person can jump about six times further on the Moon compared to Earth given the same initial angle and speed. This is because the acceleration due to gravity on the moon is one-sixth that of Earth, resulting in a weaker gravitational pull.

The topic in question involves the concept of gravitational forces and its influence on the jump length of a person on Earth versus the Moon. The acceleration due to gravity on the Moon is about one-sixth that of Earth, meaning the gravitational force is much weaker. Therefore, if a person jumps with the same speed and angle on the moon as on Earth, they would be able to jump approximately six times farther due to the weaker gravitational pull.

For better understanding, consider this example. If a person jumps and covers a certain distance on Earth in a given time, the same person would be airborne six times as long on the Moon if they jumped with the same velocity. This is due to the inverse relationship between the time of flight (airborne time) and acceleration due to gravity (g). Since g for the Moon is one-sixth that of Earth, a person can jump about six times further.

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Answer:

Sulfuric acid.

Explanation:

A battery is a device which contains one or more electrical chemical cells which are connected in a electrical circuit to power the objects like flashlight, speaker, etc.

When a battery is supply electric power then its positive terminal classified as cathode and negative as anode.

And as we know that the electrolyte in a typical wet storage battery or cell contains 67% water and 33% of sulfuric acid.

Actually we could create a triangle in this case. The hypotenuse is the radius of Earth plus the height above the earth, while the two sides are the radius of Earth and the scope of vision.

That is:

(6,400,000 m + 100 m)^2 = (6,400,000 m)^2 + b^2

b^2 = 1,280,010,000

b = 35,777.23 m

You can see 35,777.23 m far.

Final answer:

The magnitude of the average acceleration the football player experiences is 38.02 m/s², and the direction of the acceleration is backward, towards the line of scrimmage.

Explanation:

To calculate the magnitude and direction of the average acceleration experienced by the football player, we use the following kinematic equation:

a = (v_f - v_i) / t

Where:

Given:

Now we can plug these values into our equation to find a:

a = (0 - 7.68 m/s) / 0.202 s

a = -7.68 m/s / 0.202 s

a = -38.02 m/s2

The negative sign indicates that the acceleration is in the opposite direction of the player's initial motion. Since the player was moving forward and came to a stop, the acceleration is directed backward, towards the line of scrimmage.

The puma, cougar, or mountain lion leaves the ground to make a leap with a speed of approximately 8.32 m/s.

To find the speed at which the animal leaves the ground, we can use the kinematic equation for projectile motion:

[tex]\[v = \sqrt{2gh},\][/tex]

where [tex]\(v\)[/tex] is the speed, [tex]\(g\)[/tex] is the acceleration due to gravity (approximately [tex]\(9.81 \, \text{m/s}^2\))[/tex], and [tex]\(h\)[/tex] is the vertical height.

Given that the height  [tex]\(h\)[/tex] is 13.7 ft, we first convert it to meters:

[tex]\[h = 13.7 \, \text{ft} \times 0.3048 \, \text{m/ft} = 4.1756 \, \text{m}.\][/tex]

Substituting [tex]\(h\)[/tex] into the equation, we have:

[tex]\[v = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 4.1756 \, \text{m}} \approx 8.32 \, \text{m/s}.\][/tex]

In order to maintain equilibrium and have a downslope acceleration of 0.64 m/s² with a 2.3 kg mass, the right-hand mass should be approximately 0.15 kg.

In physics, the question is referring to the concept of equilibrium through the force of gravity. When a downtrend pushes left-hand mass with a certain acceleration, it implies that any other force (potentially from another mass on the opposite direction) counteracts. This counteraction or the right-hand mass is what we are now to calculate.

Assuming there is no friction, the force acting on the left-hand mass (F1) can be calculated using Newton's second law of motion, F = ma where m is the mass and a is the acceleration, yielding F1 = 2.3 kg * 0.64 m/s² = 1.472 N

For equilibrium to be maintained, the force due to the right-hand mass (F2) must equal F1. Therefore, if g is the acceleration due to gravity (9.81 m/s²), the right-hand mass (m2) can be found using the formula m2 = F1 / g, simplifying to m2 = 1.472 N / 9.81 m/s² = 0.15 kg.

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Answer: B. standard deviation

Explanation:

In statistics , Standard deviation is a term which is used to to represent the measure of dispersion of data from the mean-value.

It is used to determine how closely the data values are with the mean of the entire data.

It is the average distance from each data value to the mean.

Hence, the average distance between the variable scores and the mean in a set of data is the standard deviation.

If the mass of the crate is doubled but the initial velocity is not changed, the crate slides distance d before stopping.

[tex]\texttt{ }[/tex]

Let's recall the formula of Kinetic Energy as follows:

[tex]\large {\boxed {E_k = \frac{1}{2}mv^2 }[/tex]

Ek = Kinetic Energy ( Newton )

m = Object's Mass ( kg )

v = Speed of Object ( m/s )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

initial height = h₁ = 26 m

final height = h₂ = 16 m

initial speed = v₁ = 0 m/s

coefficient of friction = μ

gravitational acceleration = g

distance = d

Asked:

final speed = v₂ = ?

Solution:

We will use Work and Energy formula to solve this problem as follows:

[tex]W = \Delta Ek[/tex]

[tex]-f d = Ek_{final} - Ek_{initial}[/tex]

[tex]-\mu N d = \frac{1}{2}m (v_f)^2 - \frac{1}{2} m (v_i)^2[/tex]

[tex]-\mu mg d = \frac{1}{2}m (v_f)^2 - \frac{1}{2} m (v_i)^2[/tex]

[tex]-\mu mg d = \frac{1}{2}m (0)^2 - \frac{1}{2} m (v_i)^2[/tex]

[tex]-\mu mg d = \frac{1}{2} m (v_i)^2[/tex]

[tex]\mu g d = \frac{1}{2} (v_i)^2[/tex]

[tex]d = \frac{1}{2} (v_i)^2 \div ( \mu g )[/tex]

[tex]\boxed {d = \frac { (v_i)^2 } { 2 \mu g } }[/tex]

[tex]\texttt{ }[/tex]

From information above we can conclude that the distance is independent to the mass of the crate.

If the mass of the crate is doubled but the initial velocity is not changed, the crate slides the same distance d before stopping.

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

If the mass of the crate is doubled while maintaining the same initial velocity, the stopping distance will be halved. This is due to the doubled frictional force from the increased mass. Therefore, the crate will slide a distance of d/2 before coming to a stop.

Thus, the crate will slide half the distance before stopping if the mass is doubled but the initial velocity is unchanged.

Answer:

[tex]108\ N[/tex]

Explanation:

Mass of object, [tex]m=1.50\ kg.[/tex]

Length of string, [tex]r=0.50\ m.[/tex]

Tangential speed, [tex]v_t=6.0\ m/s.[/tex]

Now, centripetal force F of a object moving in given radius r and mass m moving with velocity v.

Here , object moves along the ends of string. Therefore, radius is equal to length of string.

[tex]F=\dfrac{m \times v_t^2}{r}.[/tex]

Putting values of m,v and r in above equation.

We get, [tex]F=\dfrac{1.50\times (6.0)^2}{0.50} \ N=108 \ N.[/tex]

Hence, this is the required solution.

The pitch of a note is determined by its frequency, with higher frequencies leading to higher pitches. The quality, or timbre, of a note depends on the shape of the waveform, influenced by various frequencies and phases of sound waves. These aspects combine to give each note its unique character.

The quality and pitch of a note depend on different aspects of sound waves. The pitch of a note is primarily determined by its fundamental frequency, which is measured in hertz (Hz). A higher frequency results in a higher pitch, making a note sound “sharper” or higher on the musical scale. For example, the piano note middle C has a frequency of 261.63 Hz. Musical intervals, like the octave, are based on the doubling of frequencies.

On the other hand, the quality or timbre of a note depends on the waveform's shape, which is influenced by the frequencies and phases of other sound waves produced alongside the fundamental frequency. This complexity allows us to distinguish between different instruments playing the same note due to the variety in waveforms. The timbre is what makes the same note played on a trumpet distinctly different from the same note played on a clarinet.

In summary, while pitch is a direct correlation to the frequency of sound, quality or timbre involves the intricate interplay of multiple frequencies and their waveform shapes, contributing to the unique character of each musical note.

Answer:

Newton's third law of motion

Explanation:

As per Newton's third law of motion, every action has its equal and opposite reaction.

When a jet aircraft takes off it exerts a force and gains upward acceleration, the same force is applied on your body as it is moving along the jet hence this force tries to push you back into the seat.

For instance, while rowing a boat you try to apply force on water with paddles in backward direction, the same amount of force is reverted by water thus moving the boat in forward direction.

Hence Newton's third law of motion explains the phenomena.

Answer:

Newton's third law of motion

Explanation:

As per Newton's third law of motion, When two bodies interact they apply force to one another that are equal in magnitude and opposite  in direction.

During the take off a jet aircraft it exerts a force and gain upward acceleration, the same amount of force is applied on our body because we are in contact with jet due to third law we pushed back into the seat..

[tex]F_1=-F_2[/tex]

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Michael Jordan's takeoff speed for his 47.2-inch vertical leap would be approximately 4.85 m/s, calculated using the kinematic equation for vertical motion.

To find Michael Jordan's takeoff speed, we can use the kinematic equation for vertical motion without air resistance: vf^2 = vi^2 + 2 * a * d, where vf is the final velocity (0 m/s at the peak of the jump), vi is the initial velocity (takeoff speed we want to find), a is the acceleration due to gravity (-9.81 m/s^2), and d is the vertical leap distance.

First, we convert the vertical leap from inches to meters: 47.2 inches = 1.2 meters (approximately).

Now, we can set vf to 0 m/s and solve for vi:

0 = vi^2 + 2 * (-9.81) * 1.2

vi^2 = 2 * 9.81 * 1.2

vi = sqrt(2 * 9.81 * 1.2)

vi ≈ 4.85 m/s (rounded to two decimal places)

Therefore, Michael Jordan's takeoff speed would be approximately 4.85 m/s.

Answer:

10 calories

Explanation:

The thermal energy needed to increase the temperature of a substance is given by

[tex]Q=m C \Delta T[/tex]

where

m is the mass of the substance

C is the specific heat of the substance

[tex]\Delta T[/tex] is the increase in temperature

In this problem, the mass of the ice is m=20 g, the specific heat is C=0.5 calories/gram°C, and the increase in temperature is [tex]\Delta T=1^{\circ}[/tex]. Therefore, the energy required is

[tex]Q=(20 g)(0.5 cal/g^{\circ}C)(1^{\circ}C)=10 cal[/tex]

Final answer:

Bernoulli's equation is a form of the conservation of energy principle in fluid flow. It states that the sum of pressure, kinetic energy, and potential energy is constant at any two points in an incompressible, frictionless fluid.

Explanation:

Bernoulli's equation is a form of the conservation of energy principle in fluid flow. It states that the sum of pressure, kinetic energy, and potential energy is constant at any two points in an incompressible, frictionless fluid.

For example, if a fluid flows through a pipe with varying diameters, according to Bernoulli's equation, the pressure decreases as the fluid velocity increases.

This concept is important in the study of fluid mechanics and is used to analyze and predict the behavior of fluids in various applications.