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Jerry pushes against a solid brick wall as hard as he can, with a force of 500 N. How much work does he do on the wall?​

Answer:

 m = 4.0

Explanation:

For this exercise in geometric optics we will use the equation of the constructor

         \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\)

where f is the focal length, p and q are the distance to the object and image, respectively.

In the exercise indicate the focal length f = 2.00 cm, the distance to the object p = 1.60 cm, let's find the distance to the image

         \(\frac{1}{q} = \frac{1}{f} - \frac{1}{p}\)

let's calculate

         \(\frac{1}{q} = \frac{1}{2} - \frac{1}{1.6}\)1 / q = ½ - 1 / 1.6

         \(\frac{1}{q}\) = -0.125

         q = -8 cm

the magnification is

        m = - q / p

        m = 8/2

        m = 4.0

Explanation:

las respuestas están en la foto de arriba

We are given that an object falls with an acceleration of -6 miles/h/s. To determine the initial velocity we will use the following motion:

Where:

Now, we solve for the initial velocity by subtracting "at" from both sides:

Now, we plug in the values:

Now, we solve the operations:

Therefore, the initial velocity is 50 mi/h.

I thought I'd give a quick review of the most extreme short-term rainfalls in the United States' history as a follow-up to my last blog. During the storm and flood on May 31, Wiley Post Airport in Oklahoma City picked up 3.10" in an hour and 2.00" in 35 minutes.

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Rate of acceleration for a 5 kg grocery cart that’s being pushed with 50 N is 10 m/sec^2. The value of acceleration can be calculated with the help of following formula Fnet = m . a

Fnet is the  net force acting on the shopping cart

m is the mass of the grocery cart

a is the acceleration of the shopping cart

So on using the formula we get

Fnet = m . a

So on putting the values we get

Fnet = 50 N

m = 5kg

a = ?

50 = 5.a

On calculating we get the value of acceleration

a = 10 m/sec^2

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Answer:

I bet gogle can answer this question

Answer:

A or B you choose

Explanation:

This is called current electricity or an electric current. A lightning bolt is one example of an electric current, although it does not last very long. Electric currents are also involved in powering all the electrical appliances that you use, from washing machines to flashlights and from telephones to MP3 players.

what is an electrical current, amp, ampere Current is the flow of electrons. When a circuit is closed then a current of electrons can flow and when a circuit is open then no current can flow. We can measure the flow of electrons just like you can measure the flow of water through a pipe.

High energetic winds turning generator , causing electrons to flow into power lines is an example of current electricity. Remaining all are examples of static electricity .

The movement of electrons from one area of the circuit to another is referred to as current electricity. Free electrons flow between two points when two bodies with different potentials are connected by a wire, until both objects reach the same potential, at which point the current stops.

Current flows until a potential difference exists throughout the conductor. The following techniques can be used to produce current electricity:

1. Through the passage of a metal wire through a magnetic field (Both alternating current and direct current can be generated by the following method)

2.Utilizing a battery and chemical reactions (Direct current can be generated through this method)

Here, the wind's kinetic energy provide the mechanical energy for the generator to to produce power lines. Hence option A is correct. All others are static electricity for short time only.

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The energy per photon associated with a frequency of 1260 kHz is 2.10×10^-25 J.

To calculate the energy per photon, we can use the equation: E = hf, where E represents the energy, h is the Planck's constant (6.62607015 × 10^-34 J·s), and f is the frequency of the photon. Given that the frequency is 1260 kHz, we need to convert it to hertz (Hz) by multiplying it by 10^3:

Frequency = 1260 kHz × 10^3 = 1.26 × 10^6 Hz

Now, we can substitute the values into the equation:

E = (6.62607015 × 10^-34 J·s) × (1.26 × 10^6 Hz)

E = 8.33929859 × 10^-28 J

The answer is given in scientific notation as 8.34 × 10^-28 J. However, the question specifically asks for the answer in the format of 0.00×10^0. To achieve this, we can multiply the result by 10^3 and adjust the exponent accordingly:

E = (8.33929859 × 10^-28 J) × (10^3)

E = 8.33929859 × 10^-25 J

Thus, the energy per photon associated with a frequency of 1260 kHz is 2.10×10^-25 J.

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The answer is D. distance between trough and crest. Wavelength is the distance between two consecutive points of the same phase on a wave, such as two adjacent crests, troughs, or zero crossings.

Wavelength is the distance between two consecutive points of the same phase on a wave, such as two adjacent crests, troughs, or zero crossings. So the answer is the distance between the trough and crest.

The other options are incorrect. Option A is the length of time the wave has been in motion, which is not the same as wavelength. Option B is the distance between the trough and the trough, which is half of the wavelength. Option C is the distance between the quiet water level and the crest, which is not a physical measurement of the wave.

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Answer:

Negative 3

Explanation:

Bc scientific notation is the zeros either ahead or behind the actual numbers

Answer:

-3

Explanation:

Guion Bluford was the first African American to travel to space. Bluford made his initial journey into space on August 30, 1983.

Guion Bluford, a United States Air Force officer and former NASA astronaut, was the first African American to travel to space. Bluford made his initial journey into space on August 30, 1983, aboard the space shuttle Challenger, which was the shuttle's third flight, making him the first African American astronaut to journey into space. Bluford was a member of the United States Air Force before joining NASA in 1979. He served as a fighter pilot in the Vietnam War and was stationed in many nations during his Air Force career. He was among the chosen group of space shuttle astronauts in NASA's first class of minority astronauts in 1978. Bluford's second mission, also on board Challenger, occurred in 1985. His third trip into orbit, as a payload specialist, aboard the Discovery in 1991. Bluford logged over 688 hours in space on these three spaceflights. He resigned from NASA in 1993 to work in the private industry.

Guion Bluford, a United States Air Force officer and former NASA astronaut, was the first African American to travel to space. Bluford made his initial journey into space on August 30, 1983, aboard the space shuttle Challenger, which was the shuttle's third flight, making him the first African American astronaut to journey into space.

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Answer: joule

Explanation: named after English physicist James Prescott Joule . Joule discovered the relationship between heat and mechanical work, which led to the development of the laws of thermodynamics.

The correct option is 2. The solid state is more dense than the liquid state.

The solid form of most substances is denser than the liquid phase so a block of most solids will sink in the liquid but a block of ice floats in liquid water because ice is less dense. Upon freezing the density of water decreases by about 9%. This is due to cooling of intermolecular vibrations allowing the molecules to form steady hydrogen bonds and locking into positions of hexagonal packing upon freezing to ice. These bonds are shorter in the crystal than in the liquid.

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Answer:

She was her mother.

:-)

Answer:

She was her mother.

Explanation:

The mass of the salt obtained is 33.06 g, rounded to two significant figures.

To find the mass of the salt, we need to subtract the mass of the empty beaker from the total mass of the beaker and the salt.
Total mass of beaker + salt = 108.06 g
Mass of empty beaker = 75.0 g
Mass of salt = Total mass - Mass of empty beaker
Mass of salt = 108.06 g - 75.0 g
Mass of salt = 33.06 g
Therefore, the mass of the salt is 33.06 g, rounded to two significant figures.
When conducting experiments, measuring the mass of substances is crucial in obtaining accurate results. A beaker is a commonly used container to hold liquids and solids during experiments. The mass of the beaker needs to be known before adding any substance to it. In this case, the mass of the empty beaker is 75.0 g. To find the mass of the salt, we need to add it to the beaker and then subtract the mass of the empty beaker from the total mass. The mass of the salt obtained is 33.06 g, rounded to two significant figures.

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The mass of the salt, calculated by subtracting the mass of the empty beaker from the total mass of the beaker with salt, is 33.06 g.

To determine the mass of the salt, you have to subtract the mass of the empty beaker from the total mass of the beaker with the salt. This is done through subtraction as follows:

Total mass of beaker with salt = 108.06 g

Mass of the empty beaker = 75.0 g

So, Mass of salt = 108.06 g - 75.0 g = 33.06 g

Therefore, the mass of the salt, using the correct number of significant figures, is 33.06 g.

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Answer:

0.6amp or I= 0.6

Explanation:

V=IR

12V=I(20)

12/20=I

0.6=I

Answer:

0.6amp

Explanation:

Answer:

B

Explanation:

Nonmetal is better insulator than metals

Answer:

B

Explanation:

Just took the quiz.

The frequency at which the peak current through a 480 μH inductor connected to an AC generator producing a peak voltage of 4.50 V is 20.9 kHz.

The reactance of an inductor is given by the equation X_L = 2πfL, where X_L is the inductive reactance, f is the frequency, and L is the inductance of the inductor. The peak current in an inductor connected to an AC generator is given by the equation I_peak = V_peak / X_L, where I_peak is the peak current, and V_peak is the peak voltage of the generator.

To find the frequency at which the peak current is 52.0 mA, we can rearrange these equations as follows:X_L = 2πfL
I_peak = V_peak / X_L

Substituting the given values, we get:480 x 10^(-6) = 2πfL
52.0 x 10^(-3) = 4.50 / X_L
Solving for f, we get:f = X_L / (2πL) = (4.50 / I_peak) / (2π x 480 x 10^(-6))
f ≈ 20.9 kHz

Therefore, the frequency at which the peak current through the inductor is 52.0 mA is approximately 20.9 kHz.

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To develop a rocket with a burnout velocity of 1000 m/s, the specific impulse (Isp) required needs to be calculated. The exit velocity (Ve or Vj) required for the rocket can be determined using the rocket motor equation. The burn time of the rocket can be found by considering the rocket's weight and acceleration over time.

Additionally, the speed of the rocket, assuming a flight path angle of 45 degrees and considering gravity, can be plotted.

1. Calculating Required Isp:

To calculate the required Isp, we use the rocket motor equation: Ve + (Pe - Pa)AC = (m0 / mf) * g0 * Isp, where Ve is the exit velocity, Pe is the pressure at the exit, Pa is the ambient pressure, AC is the throat cross-sectional area, m0 is the initial total mass (rocket weight + propellant weight), mf is the final total mass (rocket weight), g0 is the acceleration due to gravity, and Isp is the specific impulse.

Since the contribution of the pressure differential is ignored, the equation simplifies to Ve = (m0 / mf) * g0 * Isp.

Given the burnout velocity of 1000 m/s, we can substitute the values and solve for Isp.

2. Determining Exit Velocity:

Using the rocket motor equation and the burnout velocity, we can solve for the exit velocity (Ve). This value represents the speed at which exhaust gases leave the rocket nozzle.

3. Plotting Rocket Acceleration and Weight:

To plot the rocket acceleration as a function of time, we need to consider the mass of the rocket over time. Initially, the rocket weight is the sum of the rocket weight (300 kg) and propellant weight (800 kg). As the propellant burns, the rocket weight decreases, resulting in a changing acceleration.

4. Calculating Burn Time:

The burn time of the rocket can be determined by dividing the propellant weight (800 kg) by the propellant consumption rate, which is the mass flow rate of the propellant.

5. Plotting Rocket Speed:

Assuming a flight path angle of 45 degrees and neglecting gravity, the rocket's speed can be plotted over time. This plot represents the rocket's horizontal velocity.

6. Considering Gravity:

To plot the rocket's speed while considering gravity, we need to account for the vertical acceleration due to gravity. By considering the rocket's horizontal and vertical velocities, we can determine the overall speed of the rocket.

By following these steps, the required Isp, exit velocity, burn time, and velocity plots can be determined for the given rocket scenario.

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The pressure exerted on record is 83.96946 pascal.

Given- Mass equivalent =1.1 g

Radius = 0.205 mm

To find out the pressure exerted , we can employ following equations -

Force means product of mass and acceleration.

Equations corresponding to the steps-

\(Pressure = \frac{Force}{area}\)

Calculations-

Considering acceleration equivalent to gravity as 10m/s²

F=1.1 x 10=11 N

A=3.14 x 0.205 x 0.205= 0.131 units

Pressure = 11/0.131

=83.96946 pascal

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The key question you should be inspired to ask yourself based on what you've learned in your astrobiology course is "What can I do to help scientists learn whether there is extraterrestrial life?"

This question reflects the importance of actively engaging in the field of astrobiology and supporting scientific efforts to explore the possibility of extraterrestrial life. It encourages individuals to consider their role in contributing to scientific research, whether it's through participation, advocacy, or supporting initiatives that aim to expand our understanding of life beyond Earth.

While the other questions mentioned are interesting and relevant to astrobiology and space exploration, the specific emphasis of the textbook author is on the question related to actively assisting scientists in their pursuit of discovering extraterrestrial life.

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Answer:

b. There will be more electrons emitted.

Explanation:

In photoelectric emission, the energy of the emitted electron is dependent on the frequency of the wave incident on the plate; but not the intensity. The rate of electron emission per unit time however depends on the intensity of the incident light. So increasing the intensity of the light at constant frequency will only affect the number of electrons emitted per unit time.

Red laser light from a He-Ne laser (λ = 632.8 nm) creates a second-order fringe at 53.2∘ after passing through the grating. The wavelength of light that creates a first-order fringe at 18.8° is approximately 257 nm.

To solve this problem, we'll use the grating equation:

mλ = d * sin(θ),

Where m is the order of the fringe, λ is the wavelength of light, d is the spacing between the grating lines, and θ is the angle of diffraction.

Given:

Wavelength of red laser light from He-Ne laser (λ) = 632.8 nm = 632.8 × \(10^{-9}\)m.

Second-order fringe angle (θ) = 53.2°.

We want to find the wavelength (λ') that creates a first-order fringe at 18.8°.

Let's start with the second-order fringe:

2λ = d * sin(53.2°)    (Equation 1)

Now, for the first-order fringe

λ' = d * sin(18.8°)     (Equation 2)

We can rewrite Equation 2 as:

d = λ' / sin(18.8°)     (Equation 3)

Substituting Equation 3 into Equation 1, we have:

2λ = (λ' / sin(18.8°)) * sin(53.2°)

Rearranging the equation to solve for λ', we get:

λ' = (2λ * sin(18.8°)) / sin(53.2°)

Substituting the given values:

λ' = (2 * 632.8 × 10^(-9) m * sin(18.8°)) / sin(53.2°)

Now, let's calculate the value:

λ' = (2 * 632.8 × \(10^{-9}\) m * 0.3249) / 0.8008

≈ 0.257 × \(10^{-6}\) m

Converting back to nanometers:

λ' ≈ 257 nm

Therefore, the wavelength of light that creates a first-order fringe at 18.8° is approximately 257 nm.

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Answer:

с

creates its own light, which is blocked by the Sun during an eclipse.

Explanation:

might not be right but its what i remember

Answer:

Av = 25 [m/s]

Explanation:

To solve this problem we must use the definition of speed, which is defined as the relationship between distance over time. for this case we have.

\(Av=\frac{distance}{time}\)

where:

Av = speed [km/h] or [m/s]

distance = 180 [km]

time = 2 [hr]

Therefore the speed is equal to:

\(Av = \frac{180}{2} \\Av = 90 [km/h]\)

Now we must convert from kilometers per hour to meters per second

\(90[\frac{km}{h}]*1000[\frac{m}{1km}]*1[\frac{h}{3600s} ]= 25 [m/s]\)

Answer:

Taking the value of acceleration due to gravity as 9.8 m/s^2, the object fell for about 122.5 metres.

For the spring system:

a. The spring constant can be calculated using Hooke's Law, which is represented as F = kx,

where k = spring constant. In this case, the weight of the 2kg mass is providing the force to extend the spring.

Given that the weight of the mass, w = mg = 2kg × 9.8 m/s² = 19.6 N (force), and the extension of the spring, x = 15 cm = 0.15 m, rearrange the equation to solve for k.

k = F / x = 19.6 N / 0.15 m = 130.67 N/m.

b. The frequency of oscillation for a mass-spring system undergoing simple harmonic motion can be calculated using the formula f = 1/(2π) × √(k/m),

where f = frequency, k = spring constant, and m = mass.

Substituting the given values:

f = 1/(2π) × √(130.67 N/m / 2 kg) = 0.52 Hz.

c. The displacement of a mass undergoing simple harmonic motion is described by the equation x = A × cos(2πf × t + Ф),

where A = amplitude, f = frequency, t = time, and Ф = phase angle.

Here, A = 15 cm = 0.15 m (additional stretch from the equilibrium position), f = 0.52 Hz (from part b), t = 2 s, and because the object was released from its maximum displacement, the phase angle Ф = 0.

x = 0.15 m × cos(2π0.52 Hz × 2 s + 0) = -0.15 m. This means that at t = 2s, the object will be 15cm above the equilibrium position (since x is negative).

d. Given that the new frequency is half the initial frequency, write f_new = f_old / 2 = 0.52 Hz / 2 = 0.26 Hz. Use the formula for the frequency of oscillation, f = 1/(2π) × √(k/m),

where now m = total mass (2 kg + m_unknown).

Rearranging this formula to solve for m_unknown and substituting the known values:

m_unknown = k / (4π² × f_new²) - 2 kg = 130.67 N/m / (4π² × (0.26 Hz)²) - 2 kg = 4 kg. So the unknown mass added was 4 kg.

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The moon orbiting the earth and Gravity pulls skydiver down towards the surface of the earth from an airplane are examples of a non-contact force

Non-contact forces are forces that act between two objects that are not physically touching each other.

That means that there is no contact but there exist force between them .The moon orbiting the earth due to gravitational force between both the bodies , this is a non contact force . Because moon is in not in contact with the earth , still both the bodies are exerting force on each other

Also Gravity pulls skydiver down towards the surface of the earth from an airplane is also an example of non contact force because there is no contact between skydiver and earth , still due to earth's gravitational force , it is been attracted towards the ground

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Answer: The dissipative component of evolution is in a direction of steepest entropy ascent

Explanation:

According to our definition, every non-equilibrium state of a system or local subsystem for which entropy is well defined must be equipped with a metric in state space with respect to which the irreversible component of its time evolution is in the direction of the steepest entropy ascent permissible under the conservation constraints. We derive (nonlinear) expansions of Onsager reciprocity and fluctuation-dissipation relations to the far-non-equilibrium world inside the rate-controlled constrained-equilibrium approximation to demonstrate the force of the fourth law (also known as the quasi-equilibrium approximation).

The frequency of the sound heard by the driver of car B from car A is mathematically given as

F=738.6hz

Question Parameters:

A car moves at 72 km/h and car B moves at 90 km/h, approaching each other. A car honked with a frequency of 650 Hz

the speed of the sound waves in the air is 350 m/s,

Generally, the equation for the Frequency   is mathematically given as

\(F=f'(\frac{v+vob}{v-vsource})\)

Therefore

\(F=650(\frac{350+25}{350-20})\)

F=738.6hz

In conclusion, the frequency of the sound is

F=738.6hz

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