A certain radio wave has a frequency of 2.0 x 10^6 Hz. What is its wavelength?

Can someone help me solve this??

Answers

Answer 1
Solution:  
Formula for finding wavelength from frequency is wavelength=wave velocity(c)/frequency(f). 
 General value of Wave velocity (c) = 3 x 10^8 m/s
 Given frequency = 2.0 x 10^6 Hz 
 Now,
 Wavelength = 3 x 10^8 m/s / 2.0 x 10^6 Hz
  = 150 m
 Therefore the value of wavelength is 150m.

Related Questions

As you add or remove neutrons from the nucleus, isotopes are formed

Answers

That's true: isotopes are chemical elements with same atomic number (i.e. same number of protons) but different mass number (i.e. different number of protons+neutrons, so if the number of protons is the same, the number of neutrons should be different). 
For instance, hydrogen has three different isotopes: 
- protium, with only one proton in the nucleus
- deuterium, with one proton and one neutron
- tritium, with one proton and two neutrons

Which has more thermal energy: lake or a cup of hot chocolate?

Answers

Though the hot cocoa would have a higher temperature, the lake would have more thermal energy because it has more molecules with a greater total internal energy.

Answer:

A cup of hot chocolate

Explanation:

A cup of hot chocolate will have more thermal energy has the hot cocoa would have a higher temperature, the lake would have more thermal energy because it has more molecules with a greater total internal energy.

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What is the relationship between atomic radius, ionization energy, and electronegativity? What is the relationship between atomic radius, ionization energy, and electronegativity? @Physics,

Answers

Ionization energy is the energy required to remove an electron from a given chemical species. Ionization energy is related to the atomic radiii; such that they are inversely related, therefore ionization increases, atomic radius decrease. Atoms with many electrons get bigger as the electrons become more shielded  from the nucleus. Electronegativity is the atom's chemical ability to attract and bind with electrons. It is the tendency of the atom to attract and form bonds with electrons. Electronegativity increases across the period of elements with decrease in atomic radii, while electronegativity decreases down the period with increase in atomic radii.

The relationship between electronegativity, ionization energy, and atomic radius is that larger atoms are less attracted by nuclear force, which impacts their ability to retain and attract electrons.

Atomic Radius:

It is generally defined as the distance from the nucleus to the outermost electron cloud. Atomic radius tends to increase as you move down a group on the periodic table due to the addition of electron shells.Conversely, it decreases as you move across a period from left to right because of the increasing positive charge in the nucleus, which pulls electrons closer.

Ionization Energy:

The energy needed to extract an electron from an atom in the gas phase is known as ionization energy.The first ionization energy refers to the energy needed to remove one electron. This energy increases across a period and decreases down a group. Higher ionization energy indicates that an atom holds its electrons more tightly.

Electronegativity:

An atom's capacity to draw in and form bonds with electrons is measured by its electronegativity.It goes higher periodically and down within a group.When an atom has a high electronegativity, it draws electrons toward itself in a chemical connection.

Relationship Between the Three Properties:

As the atomic radius increases, ionization energy and electronegativity decrease. This is because larger atoms have more electron shells, resulting in weaker attraction between the nucleus and the outermost electrons.Conversely, as the atomic radius decreases, ionization energy and electronegativity increase. This is due to stronger nuclear attraction in smaller atoms, making it harder to remove electrons and easier to attract electrons from other atoms.

Two Jupiter-size planets are released from rest 1.40×10^11m apart. What is their speed as they crash?,

Answers

To solve this problem, we derive Newton’s Law of Universal Gravitation as the basis of computation

Where: M₁ = mass of planet #1

M₂ = mass of planet #2

M = total mass

R₁ = radius of planet #1

R₂ = radius of planet #2

d₁ = initial distance between planet centers

d₂ = final distance between planet centers

a = semimajor axis of plunge orbit

v₁ = relative speed of approach at distance d₁

v₂ = relative speed of approach at distance d₂

 

To determine velocity during the impact of two heavenly bodies, the solution is as follows:

M₁ = M₂ = 1.8986e27 kilograms

M = M₁ + M₂ = 3.7972e27 kg

G = 6.6742e-11 m³ kg⁻¹ sec⁻²

GM = 2.5343e17 m³ sec⁻²

d₁ = 1.4e11 meters

a = d₁/2 = 7e10 meters

R₁ = R₂ = 7.1492e7 meters

d₂ = R₁ + R₂ = 1.42984e8 meters

v₁ = 0

v₂ = √[GM(2/d₂−1/a)]

v₂ = 59508.4 m/s

The final answer is 59508.4 m/s.

Answer:

this verified kid is way too smart for his own good

Explanation:

a body of mass 1.5kg, traveling along the positive x axis with speed 4.5m/s,collides with another body B of mass 3.2kg which,initially is at rest. A is deflected and moves with a speed of 2.1m/s in a direction which is 30 degrees below the x axis. B Is set in motion at angle b above the x axis. calculate the velocity of B after collision.,

Answers

Data:
m₁ = 1.5kg
m₂ = 3.2kg
α = -30° (negative because it is below the x-xis)
[tex] v_{1i} [/tex] = initial speed of object 1 = 4.5m/s
[tex] v_{2i} [/tex] = initial speed of object 2 = 0m/s
[tex] v_{1f} [/tex] = final speed of object 1 = 2.1m/s
[tex] v_{2f} [/tex] = ?
β = ?

Since the motion after the collision is in 2 dimentions, it is better to write the speeds with their components along the x and the y-axis:
[tex] v_{1ix} [/tex] = initial speed of object 1 along x-axis = 4.5m/s
[tex] v_{1iy} [/tex] = initial speed of object 1 along y-axis = 0m/s
[tex] v_{2ix} [/tex] = initial speed of object 2 along x-axis = 0m/s
[tex] v_{2iy} [/tex] = initial speed of object 2 along y-axis = 0m/s

[tex] v_{1fx} [/tex] = final speed of object 1 along x-axis = 2.1 cos(-30) = 1.82m/s
[tex] v_{1iy} [/tex] = final speed of object 1 along y-axis = 2.1 sin(-30) = -1.05m/s 

In this kind of collision, we have the conservation of momentum, therefore we can write the system:
[tex] \left \{ {{m_{1} v_{1ix} + m_{2} v_{2ix} = m_{1} v_{1fx} + m_{2} v_{2fx} } \atop { m_{1} v_{1iy} + m_{2} v_{2iy} = m_{1} v_{1fy} + m_{2} v_{2fy}}} \right. [/tex]

Considering the terms that are zero, it becomes:
[tex]\left \{ {{m_{1} v_{1ix} = m_{1} v_{1fx} + m_{2} v_{2fx} } \atop {0 = m_{1} v_{1fy} + m_{2} v_{2fy}}} \right.[/tex]

Let's face first the y-component:
[tex]m_{2} v_{2fy}[/tex] = [tex]-m_{1} v_{1fy}[/tex]

therefore:
[tex]v_{2fy}[/tex] = [tex]\frac{-m_{1} v_{1fy}}{m_{2}}[/tex]=[tex]\frac{-(1.5)(-1.05)}{3.2}[/tex] = 5.04m/s

Now, let's face the x-component:
[tex]v_{2fx}[/tex]=[tex]\frac{m_{1} v_{1ix} - m_{1} v_{1fx}}{m_{2}}[/tex] =
[tex]\frac{m_{1} (v_{1ix} - v_{1fx})}{m_{2}}[/tex] = [tex]\frac{(1.5)(4.5-1.82)}{3.2}[/tex] = 1.26m/s

Now that we have the two components, we can find:

[tex]v_{2f} [/tex] = [tex] \sqrt{ v_{2fx}^2 + v_{2fy}^2 } [/tex] = [tex] \sqrt{5.04^{2} + 1.26^{2} } [/tex] = 6.35m/s

Lastly, the angle can be found with trigonometry:

β = tan⁻¹([tex] \frac{ v_{2fy} }{ v_{2fx} } [/tex]) = tan⁻¹([tex] \frac{ 1.26} }{ 5.04} } [/tex]) = 14°



When red light and green light shine on the same place on a piece of white paper, the spot appears t obe?

Answers

When red light and green lights shine on the same place on a piece of white paper, the spot appears to be Yellow in color. Complementary colors are always located directly across from each other on the graphic. For example cyan is located across from red, magenta across from green, and yellow across from blue. When primary color mix the produce secondary colors , for example a red light and a green light will combine to form a yellow color. The production of various colors of light by the mixing of the three primary colors of light is known as color addition.

Answer:

yellow

Explanation:

red + green light makes yellow

Holding a book while walking means i am doing work on the book. true or false

Answers

False, while holding the book it means there is no work done on the book.

Which best explains what happens to the motion of the pendulum?
A.The mechanical energy transforms to thermal energy as the pendulum slows and eventually stops moving.
B.The mechanical energy disappears as the pendulum slows and eventually stops moving.
C.The pendulum will continue moving at the same speed until someone stops it because energy cannot be destroyed.
D.The pendulum will slow down but will never stop moving because energy cannot be destroyed.

Answers

The answer is A, I just learned about this a few days ago!
I think the answer is A

A force of 75 N is applied to a spring, causing it to stretch 0.3 m. What is the spring constant of the spring?

0.004 N/m
22.5 N/m
75.3 N/m
250 N/m

Answers

the answer is D. 250 N/m 

Answer:

D 250

Explanation:

EDGE

Calculate δe, if the system absorbs 7.24 kj of heat from the surroundings while its volume remains constant (assume that only p−v work can be done). express your answer using three significant figures.

Answers

I believe the answer is 7.24 kJ.
From the equation ΔE = dW + dQ; where W is the work done on/by the system and Q is the heat the system absorbs/loses.
Therefore; ΔE = 72.4 kJ since the system has bot done any p-v work (dV= zero) and has absorbed heat.

Answer:

[tex]\delta Q = \delta E = 7.24 kJ[/tex]

Explanation:

Heat absorbed by the system is given as

[tex]\delta Q = 7.24 kJ[/tex]

now from first law of thermodynamics we know that

[tex]\delta Q = \delta E + W[/tex]

here

W = work done by the system

[tex]\delta E[/tex] = change in internal energy

also we know that when volume of the system remains same then work done by the system must be zero

[tex]W = 0[/tex]

so from above equation

[tex]\delta Q = \delta E = 7.24 kJ[/tex]

Two very quick questions!!

1) If a pulley system with an actual mechanical advantage of 2 is used in lifting a 2,000 lb. car, what effort would be needed to move the car?


If a pulley system with an actual mechanical advantage of 2,000,000 is used in lifting a 2,000 lb. car, what effort would be needed to move the car?,

Answers

The weight lifted by a machine to the applied force on a machine is called mechanical advantage. This is written as Mechanical advantage, M. A, = load(weight)/effort. So for 1) M.A = 2 and load = 2, 000lb = 8896.446N. So 2 = 8896.446/ effort Effort = 8896.446/2 = 4448.48 Similarly for M.A of 2, 000, 000 we have Effort = 8896.446/ 2, 000, 000 = 0.004448

Explanation:

1. Mechanical advantage of a pulley system, m = 2

Mechanical advantage of a machine is defined as the ratio of load to the effort force i.e.

[tex]m=\dfrac{F_L}{F_E}[/tex]

Here, [tex]F_L=2000\ lb[/tex]

[tex]F_E=\dfrac{F_L}{m}[/tex]

[tex]F_E=\dfrac{2000\ lb}{2}[/tex]

[tex]F_E=1000\ lb[/tex]

Hence, 1000 lb effort will be needed to move the car.

2. Mechanical advantage of a pulley system, m = 2,000,000

Value of load, [tex]F_L=2000\ lb[/tex]

Mechanical advantage, [tex]m=\dfrac{F_L}{F_E}[/tex]

[tex]F_E=\dfrac{F_L}{m}[/tex]

[tex]F_E=\dfrac{2000\ lb}{2000000}[/tex]

[tex]F_E=0.001\ lb[/tex]

Hence, 0.001 lb effort would be needed to move the car.

Please help! A net force of 2.0 N acts on a 2.0-kg object for 10 seconds. What is the object’s kinetic energy after that 10 seconds (assuming the object starts from rest and there is no friction)?

Answers

For Newton's second law, the force F applied to the object of mass m will cause an acceleration a of the body:
[tex]F=ma[/tex]
So, the acceleration is 
[tex]a= \frac{F}{m}= \frac{2.0 N}{2.0 kg}=1 m/s^2 [/tex]

The object undergoes through this acceleration for 10 seconds, t=10 s. Since it is an accelerated motion, we can find its final velocity after 10 seconds:
[tex]v_f = v_i + at=0 m/s+(1m/s^2)(10 s)=10 m/s[/tex]
where [tex]v_i[/tex] is the initial velocity of the object, which is zero since it starts from rest.

Finally we can calculate the final kinetic energy of the object, which is given by
[tex]K= \frac{1}{2}mv^2= \frac{1}{2}(2.0 kg)(10 m/s)^2=100 J [/tex]

A positive charge of 6.0 × 10-4 C is in an electric field that exerts a force of 4.5 × 10-4 N on it. What is the strength of the electric field

Answers

The magnitude of the electric force exerted on a charge in an electric field is given by

[tex] F=qE [/tex]

where

q is the charge

E is the magnitude of the electric field


In this problem, we have a charge of [tex] q=6.0 \cdot 10^{-4} C [/tex], while the force exerted on it is [tex] F=4.5 \cdot 10^{-4}N [/tex], so we can rearrange the previous formula to calculate the magnitude of the electric field:

[tex] E=\frac{F}{q}=\frac{4.5 \cdot 10^{-4} N}{6.0 \cdot 10^{-4} C}=0.75 N/C [/tex]


Answer:

0.75

Explanation:

With a quarter-turn flick of the wrist, a student sets the frisbee rotating at 570 rpm . what is the magnitude of the torque, assumed constant, that the student applies?

Answers

Due to missing information such as the time of the wrist flick or the frisbee's moment of inertia, we cannot calculate the exact torque that the student applied to the frisbee. To find the torque, we would need to determine the angular acceleration based on the provided angular velocity and the additional data.

To answer the question about the magnitude of the torque applied by a student to a frisbee rotating at 570 revolutions per minute (rpm), we need to consider several factors, including the angular velocity, the moment of inertia, and the time during which the torque was applied. However, the problem statement does not provide all the necessary information (like the time duration and the moment of inertia of the frisbee).

Typically, to calculate the torque, you would use the formula: torque (\(\tau\)) = moment of inertia (I) \(\times\) angular acceleration (\(\alpha\)). The angular acceleration could be found if we had the time duration for the quarter-turn and the final angular velocity.

Since these key pieces of data are missing in the student's problem, we cannot calculate the exact torque without additional information. If the time duration of the wrist flick or the moment of inertia of the frisbee were provided, we could find the angular acceleration using \(\alpha = \Delta\omega / \Delta t\), where \(\Delta\omega\) is the change in angular velocity and \(\Delta t\) is the change in time. From there, the torque could be calculated.

The torque applied by the student on the frisbee can be calculated using the formula torque = moment of inertia x angular acceleration. The primary force acting on the student when flicking the frisbee is the force applied perpendicular to the frisbee's radius.

Calculating the torque: To find the torque applied by the student on the frisbee, we can use the formula for torque, which is given by torque = moment of inertia × angular acceleration. First, convert the frisbee's initial angular velocity of 570 rpm to radians per second by multiplying by 2π/60.

Forces acting on the student: The primary force acting on the student when flicking the frisbee is the force applied perpendicular to the frisbee's radius. This force creates a torque and initiates the rotational motion of the frisbee.

HELP PLZ!!!
Which describes the difference between a bowling ball sitting in the rack waiting to be used and the bowling ball knocking pins down? A. In the rack, the bowling ball has only potential energy. As it knocks pins down, its potential energy has decreased, while its kinetic energy has increased. B. In the rack, the bowling ball has no energy at all. As it knocks down the pins the energy is 75% kinetic and 25% potential. C. In the rack and when knocking down pins, the potential energy and the kinetic energy of the bowling ball are equal. D. In the rack, the bowling ball has only kinetic energy. As it knocks pins down, its potential energy has increased, while its kinetic energy has decreased.

Answers

The answer would be A. In the rack, the bowling ball has only potential energy. As it knocks down pins, it’s potential energy has decreased while it’s kinetic increased.

Kinetic energy is the energy of motion and potential energy is stored energy when and object is not in motion

I hope this helped! :-)

What is the frequency of a microwave that has a wavelength of 0.050 m?

Answers

Answer:

[tex]6x10^9Hz[/tex]

Explanation:

we use a formula that relates the frequency and the wavelength:

[tex]f=\frac{c}{\lambda}[/tex]

where [tex]f[/tex] is the frequency of the wave, [tex]c[/tex] is the speed of light [tex]c=3x10^8m/s[/tex], and [tex]\lambda[/tex] is the wavelength.

We know that the wavelength is: [tex]\lambda=0.050m[/tex], so substituting the known values in the equation for the frequency we get:

[tex]f=\frac{3x10^8m/s}{0.05m} \\f=6x10^9s^{-1}=6x10^9Hz[/tex]

The frequency is:

[tex]6x10^9Hz[/tex]

Final answer:

The frequency of a microwave with a wavelength of 0.050 m is [tex]6*10^9[/tex] Hz, or 6000 MHz, calculated using the speed of light and the wavelength-to-frequency formula.

Explanation:

To calculate the frequency of a microwave that has a wavelength of 0.050 m, you can use the formula

c = λf, where c is the speed of light in a vacuum (approximately [tex]3*10^8[/tex] m/s), λ is the wavelength, and f is the frequency. Solving for f, the equation becomes

f = c / λ.

Let’s plug in the values:

f = ([tex]3*10^8[/tex] m/s) / (0.050 m)

After calculating, the frequency f is [tex]6*10^9[/tex] Hz, or 6000 MHz.

How much does it cost to operate a 25-w soldering iron for 8.0 hours if energy costs 8.0¢/kwh?

Answers

The power P is defined as the energy E per unit of time t:
[tex]P= \frac{E}{t} [/tex]
In our problem, the power is P=25 W and the total time is t=8.0h, so we can calculate the total energy used during this time:
[tex]E=Pt=(25 W)(8.0 h)=200 Wh=0.2 kWh[/tex]
The energy costs 8€ per each kwh, so we can write the following proportion:
[tex]8€: 1kwh = x:0.2 kWh[/tex]
where x is the price corresponding to 0.2 kWh. Solving the proportion we find x:
[tex]x= \frac{(8€)(0.2 kWh)}{1kWh}=1.6 [/tex]€
Final answer:

To find the cost of operating a 25-watt soldering iron for 8 hours at an energy cost of 8.0 cents per kWh, first convert the iron's wattage to kilowatts, then multiply by the hours to get kWh, and finally multiply by the cost per kWh. The total cost comes to 1.6 cents.

Explanation:

To calculate the cost of operating a 25-watt soldering iron for 8.0 hours with electricity costing 8.0 cents per kilowatt-hour, we first need to convert the power usage into kilowatts and then multiply by the number of hours used to find the total energy consumed in kilowatt-hours (kWh). Finally, we multiply the energy consumed by the cost per kWh to find the total cost.

First, we convert watts to kilowatts:
25 watts = 0.025 kilowatts (kW).

Next, we multiply the power consumption by the number of hours used:
0.025 kW × 8.0 hours = 0.2 kWh.

Lastly, we calculate the total cost:
0.2 kWh × $0.08 per kWh = $0.016.

So, operating a 25-watt soldering iron for 8 hours would cost 1.6 cents.

What is the velocity of the object at 20.0s

Answers

So notice that at the beginning of the motion, the object moves to 16m in a span of about 10 seconds. You should know that the velocity in a pvt graph is the slope, so here the v = 1.6 m/s more or less. At 20 seconds, the object doesn't seem to be moving and you see it stays still from 10-26ish seconds. From observing you can tell that there is no velocity, but the slope of this segment is zero so you can be sure of it. 

Highway curves are marked with a suggested speed. If this speed is based on what would be safe in wet weather, estimate the radius of curvaure for a curve marked 50km/h?The coefficient of static friction of rubber on wet concrete is .7, the coefficient of kinetic friction of rubber on wet concrete is .5 .

Answers

Radial (aka centripetal) acceleration keeps cars on an unbanked road in a curve. This acceleration is given by

[tex]a_{rad}= \frac{v^2}{r} [/tex]

The force required to keep in going in a circle is given by Newton's second law, F=ma.  Here a is the radial acceleration from the above equation, so we get a radial force:

[tex]F_{rad}=m \frac{v^2}{r} [/tex]

This force is mediated by the friction force between the tires of the car and the road.  For friction force we use the coefficient of friction and the normal force of the road on the car (which will also be the force of the car on the road).  On an unbanked road this is just the weight of the car, mg.  

The question now is, which coefficient of friction to use?  It might surprise you to find that although the car is moving relative to the road, we should use the static coefficient.  To see why this is the case, think about what the wheel's surface is doing as it meets the road.  As a wheel rotates around and comes into contact with the road surface, it's bottom section is moving in a direction opposite to the car's forward movement.  The point that actually touches the road is moving backward at the same speed that the car moves forward and so it is at rest with the street.  If this were not the case, your tires would always be skidding on the road.  (If you're having trouble seeing this, there are plenty of animations of this phenomenon on the internet). Once a car is skidding it is no longer in control and cannot turn like one that is not skidding.

So now the friction force is just

[tex]F_{fric}= \mu_s F_n=\mu_smg[/tex]

We now have to set these equal and solve for r, the radius of the circle we are traveling in.  Note that mass will cancel and so we don't need to worry about the mass of the cars involved:

[tex]m \frac{v^2}{r}=\mu_smg \\ \\ \frac{v^2}{r}=\mu_sg \\ \\ \frac{v^2}{\mu_sg}=r \\ \\ \frac{(13.89m)^2}{0.7*9.8 \frac{m}{s^2} } =28.12m[/tex]

Where I have converted the speed to m/s for unit consistency. 

The radius of curvature for wet weather is required.

The radius of curvature would be 28.1 m.

[tex]\mu_s[/tex] = Coefficient of static friction for wet concrete = 0.7

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

m = Mass of car

v = Velocity of car = 50 km/h

The force balance is

[tex]\dfrac{mv^2}{r}=\mu_s mg\\\Rightarrow r=\dfrac{v^2}{\mu_s g}\\\Rightarrow r=\dfrac{(50\times \dfrac{1000}{3600})^2}{0.7\times 9.81}\\\Rightarrow r=28.1\ \text{m}[/tex]

The radius of curvature would be 28.1 m.

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Which platform did Mr. Ray’s marketing team advise him to use for the launch of a new product?
Mr. Ray is looking for a platform to launch his product. His marketing team advises him to launch the product via_____ as it is free advertising and improves_____.
1. A. Newspapers A. Product Quality
B.social Media B. Consumer Welfare
C.Billboards C. Brand Reconition

Answers

The answers are in answer B

Social Media and Brand Recognition

A 0.40 kg ball is suspended from a spring with spring constant 12 n/m . part a if the ball is pulled down 0.20 m from the equilibrium position and released, what is its maximum speed while it oscillates?

Answers

The total mechanical energy of the system at any time t is the sum of the kinetic energy of motion of the ball and the elastic potential energy stored in the spring:
[tex]E=K+U= \frac{1}{2}mv^2+ \frac{1}{2}kx^2 [/tex]
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.

Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.

When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:
[tex]E=U_{max}= \frac{1}{2}k(x_{max})^2 [/tex]

Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):
[tex]E=K_{max}= \frac{1}{2}m(v_{max})^2 [/tex]

Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:
[tex]U_{max}=K_{max}[/tex]
[tex] \frac{1}{2}k(x_{max})^2 = \frac{1}{2}m(v_{max})^2 [/tex]
[tex]v_{max}= \sqrt{ \frac{k x_{max}^2}{m} }= \sqrt{ \frac{(12 N/m)(0.20 m)^2}{0.4 kg} }=1.1 m/s [/tex]

The maximum speed of the 0.40 kg ball while it oscillates is approximately 1.10 m/s. This is found using the conversion of potential energy into kinetic energy at the equilibrium position. The mechanical energy principles of simple harmonic motion are applied.

To determine the maximum speed of a ball oscillating on a spring, we can use principles from simple harmonic motion. Given:

Mass of the ball, m = 0.40 kgSpring constant, k = 12 N/mDisplacement from equilibrium, x = 0.20 m

We start by calculating the total mechanical energy when the ball is at maximum displacement. The potential energy at maximum displacement is given by:

U = 0.5 * k * x2 = 0.5 * 12 N/m * (0.20 m)2 = 0.24 J

Since there is no damping, this total mechanical energy converts entirely into kinetic energy at the equilibrium position where the speed is maximum. The kinetic energy K at this point is:

K = 0.5 * m * [tex]v_{max}^2[/tex] = 0.24 J

Solving for [tex]v_{max}[/tex] :

[tex]v_{max}[/tex] = √((2 * K) / m) = √((2 * 0.24 J) / 0.40 kg) = √(1.2) ≈ 1.10 m/s

Therefore, the maximum speed of the ball while it oscillates is approximately 1.10 m/s.

Johnny was playing baseball with his friends and they noticed a bolt of lightning. They heard thunder seven seconds later. How far away is the storm?

Answers

To estimate the distance to a lightning strike, divide the number of seconds between seeing lightning and hearing thunder by five. With a 7-second delay, the storm is approximately 1.4 miles away.

The storm that Johnny and his friends experienced is associated with lightning and thunder. Light from lightning travels much faster than sound from thunder. This difference can be used to estimate the distance to a lightning strike. Since light travels at 3 × 108 meters per second, we see the lightning almost immediately. However, sound travels much slower, and for every 5 seconds we count after seeing lightning before hearing thunder, the storm is approximately 1 mile away.

In Johnny's case, they heard thunder 7 seconds after seeing lightning. Using the rule of thumb that every 5 seconds corresponds to about 1 mile, we can estimate the distance to the storm. 7 seconds divided by 5 gives us 1.4, so the storm is about 1.4 miles away from Johnny and his friends. This is a practical application of basic physics and the differences between light and sound velocity.

Which has a higher acceleration: a 10-kg object acted upon with a net force of 20 N or an 18-kg object acted on by a net force of 30 N?

A. Both accelerate at the same rate.

B. the 10-kg object

C. the 18-kg object

Answers

the first one accelerate at 2m/s
the second one accelerates at 1.6666m/s

therefore statement b. the 10kg object is your answer

hope this helps :)

Answer:

the answer is B, 10-kg object

Explanation:

Because of Newton we know that if a force is applied to an object, it will have an acceleration on the same direction of the force applied.

[tex]F=m*a\\\\where:\\m=mass\\a=acceleration[/tex]

in order to obtain the acceleration we have to reordenate the formula:

[tex]a=\frac{F}{m}[/tex]

The acceleration of the 10 kg object is:

[tex]a=\frac{20N}{10kg}=2\frac{m}{s^2}[/tex]

The acceleration of the object of 18 kg is:

[tex]a=\frac{30N}{18kg}=1.67\frac{m}{s^2}[/tex]

The biggest acceleration was for the object with less mass.

A laser beam of wavelength 439.4 nm is inci- dent on two slits 0.306 mm apart.
How far apart are the bright interference fringes on a screen 6.08 m away from the double slits?
Answer in units of cm.

Answers

Constructive interference happens when this condition is satisfied:
[tex]dsin(\theta)=m\lambda; m=1,2,3,4,5,...[/tex]
This means that the difference in the distance traveled by the waves, from the slit to the screen, is equal to the whole multiple of the wavelength.
If we say that the distance between two interference fringes is much smaller than the distance from the slit to the screen, we can use the following approximation:
[tex]sin(\theta)=\frac{y}{L}[/tex]
Finally for the bright spots we have:
[tex]y=\frac{m\lambda L}{d}[/tex]
The spacing between bright spots is:
[tex]y=\frac{\lambda L}{d}=\frac{6.08\cdot 439.4\cdot 10^{-9}}{0.306 \cdot 10^{-3}}=8.73mm[/tex]



in a lab investigation. one group of students (group A) measures the speed of a 0.1-kilogram car at 2.5 m/sec at the bottom of a hill. Another group of students (group B) measures the speed of the car at 3 m/sec at the bottom of the hill. the car's starting position at the top of the hill is one-meter high. what is the potential energy of the car at the beginning of the experiment before its speed is measured?

Answers

The potential energy of the car at the top of the hill, before the beginning of the experiment, is given by
[tex]U=mgh[/tex]
where [tex]m=0.1 kg[/tex] is the mass of the car, [tex]g=9.81 m/s^2[/tex] is the gravitational acceleration and [tex]h=1.0 m[/tex] is the height of the hill. Using these data, we have
[tex]U=(0.1 kg)(9.81 m/s^2)(1.0 m)=0.98 J[/tex]

The potential energy of the car at the beginning of the experiment, before its speed is measured, is approximately 0.98 Joules.

To calculate the potential energy of the car at the beginning of the experiment, we can use the formula for gravitational potential energy:

Potential energy (PE) = mass (m) × gravitational acceleration (g) × height (h)

Given:

Mass of the car (m) = 0.1 kg

Height of the hill (h) = 1 meter

Gravitational acceleration (g) = 9.8 m/s² (approximately)

The potential energy (PE) is as follows:

PE = m × g × h

= 0.1 × 9.8 × 1

= 0.98 Joules

Therefore, the potential energy of the car at the beginning of the experiment, before its speed is measured, is approximately 0.98 Joules.

To know more about the potential energy:

https://brainly.com/question/9349250

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A hydraulic jack is used to lift a car in order to change a tire. The car exerts a force of 9,800 N toward the ground due to the force of gravity.

1. To begin to lift the car, a hydraulic jack must produce ______ of force.

A. more than 9,800 N
B. less than 9,800 N
C. exactly 9,800 N

2. To keep the car in position once it is lifted, the jack must produce _____ of force.
A. more than 9,800 N
B. less than 9,800 N
C. exactly 9,800 N

Answers

To begin to lift the car, a hydraulic jack must produce
more than 9,800 N of force. (A)

To keep the car in position once it is lifted, the jack must
produce exactly 9,800 N of force. (C)

Force is defined as the product of mass and acceleration or gravity. Force is calculated in Newton.

The force required to lift the car should be more than 9,800 N, and, to keep the car in position, the force should be exactly same as 9,800 N.

The force is a vector quantity, calculated as the product of mass and acceleration (sometimes gravity.) In order to lift the car, Jack must apply a force greater than the weight of the car.

As given in the question,  the weight of the object is 9,800 N. The weight is equal to the force when the object is at rest.

Also, once the car is lifted, Jack must maintain the force exactly 9,800 N. The less force of the car will cause the gravity to pull the car downwards.

Therefore, the force should be exactly 9,800 N while lifting and should be more than 9,800 N, when beginning to lift the car.

To know more about force, refer to the following link:

https://brainly.com/question/12412965

Which best describes the current atomic theory?
A since it is only a theory it shouldn’t be used in practice
B it has not been tested enough to be useful for developing technology
C it is the most powerful explanation scientist have to offer at this time
D there is very lil change it will be changed in the future

Answers

Choice-C is a correct statement.

C. it is the most powerful explanation scientists have to offer at this time. I am 100% right, I just finish the test and got it right.

Calculate the total force on the Earth due to Venus, Jupiter, and Saturn, assuming all four planets are in a line. The masses are Mv=0.815ME, MJ=318ME, Msat=95.1ME, Msun=1.99x10^30kg, ME=5.98x10^24kg and the mean distances of the four planets from the Sun are 108, 150, 778, and 1430 million km. Apparently the answer is 9.56x10^17 N but I'm not sure how to get to that .-.,

Answers

You need to consider the following:
Me (mass of Earth) = 5.98 x 10^24 kg 
Ms (mass of Sun) = 1.99 x 10^30 kg 
G = 6.67 x 10^-11 N 

Formula:
 F = G * M1M2/r^2 
The ratio FT/F = 4.02x10^-4 / 14.8
 = 2.72x10^-5 

Since,
1/2.72x10^-5 = 36800
The fraction ratio is  1/36800
9.56x10^17 N

Answer: Total Force = [tex]9.56*10^{17}[/tex]

Explanation:

Line points are:  Sun - Venus - Earth - Jupiter - Saturn

Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This means:

[tex]F=G\frac{m_{1} m_{2}}{r^{2} }[/tex]

Where,

G is the gravitational constant,

m1 and m2 are the masses of the objects,

and r is the distance between the centers of their masses.

So, if G value is [tex]6.674*10^{-11}  [\frac{m^{3}}{kg*s^{2}}][/tex], then we replace the equation with the corresponding values:

[tex]F=6.674*10^{-11} (-\frac{0.815ME^{2}}{(4.2*10^{10})^{2}} + \frac{318ME^{2}}{(6.28*10^{11})^{2}} + \frac{95.1ME^{2}}{(1.28*10^{12})^{2}})[/tex]

To get the distances we subtract the distances between the sun and earth and the distances between the other planets and the sun.

The mass of a bucket full of water is 15 Kg. It is being pulled up from a 15m deep well. Due to a hole in bucket 6 Kg of water flows out. Find the work done in pulling it out of the well.,

Answers

Solution:
workdone = change in potential energy 
W = Force times distance

= ( m -dm)g • ( x)
= mgx -(dm(gx))
= (15 × 10 × 15) -(6/15 gx.dx)
=2250 - (6/15 x10x 225/2)
=2250 -  (3 x 15x 10)
= 2250 - 450
= 1800 Joule
The answer is  1800 Joules.

The work done in pulling a bucket full of water with a weight of 15 kg that lost 6 kg of it from the hole from a well 15 m deep is 1350 Joules

Further Explanation;Work done Work done refers to the measure of energy when a body is moved by a force through a certain distance.Work done is therefore equivalent to energy transferred.Work done is given by the product of force acting on an object and distance moved by the object.Therefore; Work done = Force × distanceForce refer to a pull or a push and is measured in Newtons, N.Distance is measured in meters, m.

Thus; Work done is measured in Joules, J, or Nm

In this case;

Mass of the bucket is 15 Kg

However, 6 kg of water flew out due to the whole, therefore the remaining mass is 9 kg.

Force or Weight = Mass (kg) × 10 N/kg

Therefore;

Force = 90 Newtons

Distance = 15 m

Therefore;

Work done = 90 N × 15 m

                  = 1350 Joules

Hence; Work done pulling out the bucket from the well is 1350 Joules.

Key words: Work done, Force, Distance

Learn more about;Work done; https://brainly.com/question/12029835Force: https://brainly.com/question/11368524Example of work done: https://brainly.com/question/12029835

Level: High school

Subject: Physics

Topic: Work, power and simple machines

High-voltage power lines are a familiar sight throughout the country. the aluminum (resistivity = 2.82 x 10-8Ï m) wire used for some of these lines has a cross-sectional area of 4.1 x 10-4 m2. what is the resistance of 8.2 kilometers of this wire?

Answers

The formula that relates the resistance R of a wire with the resistivity [tex]\rho[/tex] of the material of the wire is
[tex]R= \frac{\rho L}{A} [/tex]
where L is the length of the wire (in our problem, [tex]L=8.2 km=8200 m[/tex]) and A is the cross sectional area (in our problem, [tex]A=4.1 \cdot 10^{-4} m^2[/tex]).
For the aluminim wire ([tex]\rho =2.82 \cdot 10^{-8} \Omega \cdot m[/tex]) of the problem, the resistance is
[tex]R= \frac{(2.82 \cdot 10^{-8}\Omega m)(8200 m)}{4.1\cdot 10^{-4}m^2}=0.564 \Omega [/tex]
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