How much energy is needed to heat 1kg of copper by 20 degrees Celsius
Step by step explanation

Answers

Answer 1

Answer:

Q = 0.0077 J

Explanation:

Given that,

Mass of copper, m = 1 kg = 0.001 g

The change in temperature = 20°C

We need to find the energy needed to heat copper. The formula is given by :

[tex]Q=mc\Delta T[/tex]

c is the specific heat of copper, c = 0.385 J/g°C

Put values in the above formula

[tex]Q=0.001\ g\times 0.385\ J/g^{\circ} C\times 20^{\circ} C\\\\Q=0.0077\ J[/tex]

So, the energy needed to heat 1 kg of copper is 0.0077 J.


Related Questions

a small weather rocket weighs 20 newtons. what is the Rockets Mass? the rocket fires is engine, if the rocket has a thrust of 109.2 Newtons, and if fixing from the air is 37.5 Newtons what is the acceleration on the rocket?​

Answers

Answer: the mass is 1.60 kg and the acceleration on the rocket is...

58.4 m/s with a Power of Two

Explanation:

A model rocket lifting off from the launch pad is a good example of this principle. Just prior to engine ignition, the velocity of the rocket is zero and the rocket is at rest. If the rocket is sitting on its fins, the weight of the rocket is balanced by the re-action of the earth to the weight as described by Newton's third law of motion. There is no net force on the object, and the rocket would remain at rest indefinitely. When the engine is ignited, the thrust of the engine creates an additional force opposed to the weight. As long as the thrust is less than the weight, the combination of the thrust and the re-action force through the fins balance the weight and there is no net external force and the rocket stays on the pad. When the thrust is equal to the weight, there is no longer any re-action force through the fins, but the net force on the rocket is still zero. When the thrust is greater than the weight, there is a net external force equal to the thrust minus the weight, and the rocket begins to rise. The velocity of the rocket increases from zero to some positive value under the acceleration produced by the net external force. But as the rocket velocity increases, it encounters air resistance, or drag, which opposes the motion and increases as the square of the velocity. The thrust of the rocket must be greater than the weight plus the drag for the rocket to continue accelerating. If the thrust becomes equal to the weight plus the drag, the rocket will continue to climb at a fixed velocity, but it will not accelerate.

Two students hear the same sound and their eardrums receive the same power from the sound wave. The sound intensity at the eardrums of the first student is 0.58 W/m^2, while at the eardrums of the second student the sound intensity is 1.18 times greater. If the diameter of the second student’s eardrum is 1.1 cm, how much acoustic power, in microwatts, is striking each of his (and the other student’s) eardrums?

Answers

Answer:

W = 65.04 μW

Explanation:

For this exercise we will use the definition of sound intensity which is the power per unit area

           I = W / A

In this case, they indicate that for the intensity of the first student is I1 = 0.58 W / m², and the intensity of the second student is

            I₂ = 1.18 I₁

            I₂ = 1.18 0.58

            I₂ = 0.6844 W / m²

Ask the power

              W = I A

the diameter of the size is 1.1 cm and the radius is r = 0.55 cm = 0.55 10⁻² m, the approximate eardrum area as a circle

             A = π r²

               

we substitute

             W = I π r²

let's calculate

             W = 0.6844 π (0.55 10⁻²)²

             W = 6.504 10⁻⁵ W

they ask us to reduce to microwatts

              W = 6.504 10⁻⁵ W (106 μW / 1W)

              W = 65.04 μW

the power is constant, what changes is the intensity that depends on the sensor area, therefore the two students review the same power

Acoustic power, in microwatt W is 65.04 μW

We know that;

I = W / a

In case 1;

I1 = 0.58 W / m²

So, I₂ = 1.18 I₁

I₂ = 1.18 × 0.58

I₂ = 0.6844 W / m²

Diameter of the size = 1.1 cm

So,

Radius r = 1.1/2 = 0.55 cm

Area A = πr²W = Iπr²

W = 0.6844(3.4)(0.55)²

W = 6.504 × 10⁻⁵ W

Acoustic power, in microwatt W = 65.04 μW

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How many electrons are needed to form a charge of -9.10 nC?

Answers

Answer:

5.6875 × 10^(10) electrons

Explanation:

To some this we will use the formula;

q = ne

Where;

q is charge

n is number of electrons

e is electron charge = 1.6 × 10^(-19) C

We are given q = -9.10 nc

1 nc = 10^(-9) C

Thus;

q = -9.1 × 10^(-9) C

Since q is negative, then e = -1.6 × 10^(-19) C

Thus;

-9.1 × 10^(-9) = n(-1.6 × 10^(-19))

n = -(9.1 × 10^(-9))/(-1.6 × 10^(-19))

n = 56875000000 electrons = 5.6875 × 10^(10) electrons

The electrons in the beam of a television tube have a kinetic energy of 2.70 10-15 J. Initially, the electrons move horizontally from west to east. The vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of 1.00 10-5 T.
A) In what direction are the electrons deflected by this field component?
B) What is the magnitude of the acceleration of an electron in part (a)?

Answers

Answer:

[a]. South direction

(b). 1.354 × 10^14 m/s^2

Explanation:

In order to to be able to solve this question effectively, one must be familiar with physics concepts such as kinetic energy[1/2 ×mass × velocity,v²], magnetic force[ magnetic force = charge × velocity × magnetic field × sin Ф] and Newton's law of motion especially the second law of Newton[ Force = mass × acceleration].  

(a). The direction are the electrons deflected by this field component = q( v × B) = - (  i × -k) = -( - ( i × k) = - (- (- j) = -j. The -j is in the south direction.

Note that  i × k is being replaced by -j.

(b). In order to determine the magnitude of the acceleration of an electron in part (a), the first thing to do is to calculate the velocity from the equation for kinetic energy.

Thus, velocity = √ [(2 × 2.7 × 10⁻¹⁵) ÷ 9.1 × 10⁻³¹] = 7.7 × 10^7 m/s.

The next thing is to determine the net force.

Therefore, the net force = 1.6 × 10^-19 × 7.7 × 10^7 m/s × 1 × 10^-15 sin 90°.

= 1.232 × 10^ -16N.

Hence, Net Force = mass × acceleration.

1.232 × 10^ -16N = 9.1× 10^-31 × acceleration.

The acceleration = 1.232 × 10^ -16N/ 9.1× 10^-31 = 1.354 × 10^14 m/s^2.

Please help I need finish by 10:57

Answers

Answer:

to answer it just get the distance traveled (m) and divide it by the time it took

Explanation:

so for the first one just do 100 divided by 47 which would equal 2.12 m/s

Determine the volume of an object that has a mass of 455.6 g and a density of 19.3 g/cm3.

Answers

Answer:

V = 23.6062 cm³

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Chemistry

Gas Laws

Density = mass / volume

Explanation:

Step 1: Define

Mass m = 455.6 g

Density D = 19.3 g/cm³

Step 2: Solve for V

Set up:                              19.3 g/cm³ = 455.6 g / VIsolate V:                          V = 23.6062 cm³

The lawn outside your neighbor's house has an approximate area of 175 m2One night it snows so that the snow on the lawn has a uniform depth of 25.5 cm. What volume of snow is on the lawn, in cubic m

Answers

first we must convert 25.5 cm to meters by moving the decimal two places to the left

25.5 cm —> .255 m

volume = area x depth

volume = 175 x .255

volume = 44.63 m^3

The volume of snow on the lawn, in cubic m, is 44.63 or 44.63 m³ if the lawn outside your neighbor's house has an approximate area of 175 m².

What is volume?

It is defined as a three-dimensional space enclosed by an object or thing.

It is given that:

The lawn outside your neighbor's house has an approximate area of 175 m². One night it snows so that the snow on the lawn has a uniform depth of 25.5 cm.

As we know, unit conversion can be defined as the conversion from one quantity unit to another quantity unit followed by the process of division, and multiplication by a conversion factor.

25.5 cm =  0.255 m

Volume = area×depth

Volume = 175×0.255

Volume = 44.63 cubic meters

Thus, the volume of snow on the lawn, in cubic m is 44.63 or 44.63 m³ if the lawn outside your neighbor's house has an approximate area of 175 m².

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3. Most horoscopes are off by two astrological signs due to the past 2000 years of the Earth's wobble.


Answers

You forgot to put the whole question

A thin nonconducting spherical shell of radius R1 carries a total charge q1 that is uniformly distributed on its surface. A second, larger thin non-conducting spherical shell of radius R2 that is concentric with the first carries a charge q2 that is uniformly distributed on its surface.

Required:
a. Use Gauss's law to find the electric field in the regions r < R1, R1 < r < R2, and r > R2.
b. What should the ratio of the charges q1/q2 and their relative signs be for the electric field to be zero for r > R2?

Answers

Answer:

Explanation:

a )

in the regions r < R₁

charge q inside sphere of radius R₁ = 0

Applying gauss's law for electric field E at distance r <R₁

electric flux through Gaussian surface of radius r = 4π r² E

4π r² E = q / ε₀ = 0 / ε₀

E = 0

Applying gauss's law for electric field E at distance R₁ < r < R₂ .

charge q inside sphere of radius R₁ = q₁

Applying gauss's law for electric field E at distance R₁ < r < R₂

electric flux through Gaussian  surface of radius r = 4π r² E

4π r² E = q₁ / ε₀

E = q₁ / 4πε₀

in the regions r> R₂

charge q inside sphere of radius R₂ = (q₁ + q₂)

Applying gauss's law for electric field E at distance r > R₂

electric flux through Gaussian surface of radius r = 4π r² E

4π r² E = (q₁ + q₂) / ε₀

E =  (q₁ + q₂) /4π ε₀

b )

For electric flux to be zero at r > R₂

(q₁ + q₂) /4π ε₀  = 0

q₁ + q₂ = 0

q₁ / q₂ = - 1 .

(a) The value electric field in the regions r < R1, R1 < r < R2, and r > R2 will be [tex]\frac{ (q_1 + q_2)}{4\pi\varepsilon_0}[/tex]

(b)The ratio of the charges q1/q2 will be -1.

What is gauss law?

The charge contained divided by the permittivity equals the total electric flux out of a closed surface, according to Gauss Law.

The electric flux in a given area is calculated by multiplying the electric field by the area of the surface projected in a plane parallel to the field.

Let's analyse regions r < R₁

charge q inside sphere will be zero.

According to gauss's law for electric field E at distance r <R₁

electric flux for the gaussian surface  = 4π r² E

[tex]4\pi r^2E=\frac{Q}{\epsilon_0} =0[/tex]

By the applications of  gauss's law for electric field E at distance R₁ < r < R₂ .

q is charge inside sphere having radius R₁ = q₁

By the applications of gauss's law for electric field E at distance R₁ < r < R₂

electric flux obtained due to Gaussian surface of radius r = 4π r² E

[tex]4\pi r^2 E =\frac{q}{\epsilon_0} \\\\\rm E=\frac{q_1}{\epsilon_0}[/tex]

For the regions r> R₂

q is the charge inside sphere of radius hyaving  R₂ = (q₁ + q₂)

By the application of gauss's law for electric field E which is at distance r > R₂

Electric flux through Gaussian surface of radius r = 4π r² E

[tex]4\pi r^2 E =\frac{q}{\epsilon_0} \\\\\rm E=\frac{q_1+q_2}{\epsilon_0}[/tex]

Hence the value electric field in the regions r < R1, R1 < r < R2, and r > R2 will be [tex]\frac{ (q_1 + q_2)}{4\pi\varepsilon_0}[/tex].

(b) To finding the ratio of charges the value of  electric flux must be zero at r > R₂

[tex]\frac{(q_1+ q_2) }{4\pi\epsilon_0} =0\\\\q_1 + q_2= 0\\\\\frac{q_1}{q_2} =-1[/tex]

Hence the ratio of the charges q1/q2 will be -1.

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Conduct research to identify claims made about the effects that certain frequencies of EMR have when absorbed by matter. Find two examples from published sources. Write brief descriptions of your two examples.

Answers

Answer:

The effects that certain frequencies of EMR have when absorbed by matter is explained below in complete detail.

Explanation:

Electromagnetic radiation of distinct frequencies associates with material adversely. ... Gamma rays, though commonly of somewhat greater frequency than X rays have the equivalent creation. When the power of gamma rays is consumed in material, its influence is practically indistinguishable from the outcome generated by X rays.

A child is asked a question in class the child knows the answer but doesn't talk this child also has excessive shyness. social isolation. fear of embarrassment in front of a group. clinging to caregivers. temper tantrums. oppositional behavior. compulsive traits what mental problem is this
Psychology ASAP help

Answers

Answer:

This is Introvert ting

Explanation:

u know when people are shy

i think they r known as Introvert

i hope this helps

The pressure in car tires is often measured in pounds per square inch (lb/in.2lb/in.2), with the recommended pressure being in the range of 25 to 45 lb/in.2lb/in.2. Suppose a tire has a pressure of 25.5 lb/in.2lb/in.2 . Convert 25.5 lb/in.2lb/in.2 to its equivalent in atmospheres. Express the pressure numerically in atmospheres.

Answers

Answer:

25.5 pounds per square inch are equivalent to 1.735 atmospheres.

Explanation:

An atmosphere equals 14.695 pounds per square inch. We find the equivalent of given pressure in atmospheres by means of simple rule of three:

[tex]x = 25.5\,\frac{lb}{in^{2}} \times \frac{1\,atm}{14.695\,\frac{lb}{in^{2}} }[/tex]

[tex]x = 1.735\,atm[/tex]

25.5 pounds per square inch are equivalent to 1.735 atmospheres.

A motorboat traveling on a straight course slows
uniformly from 65 km/h to 35 km/h in a distance
of 45 m.
Part A
What is the magnitude of the boat's acceleration?

Answers

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

65 km/h → 18.0556 m/s35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

v² = v₀² + 2aΔx

Substitute the known values into the equation.

(9.72222)² = (18.0556)² + 2a(45) 94.52156173 = 326.0046914 + 90a-231.4831296 = 90aa = -2.572

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

A 3500kg truck moving rightward with a speed of 7km/hr collides head on with a 1200kg car moving leftward with a speed of 11 km/hr. The two vehicles stick together and move with the same velocity after the collision. Determine the post-collision velocity of the car and truck.

Answers

Answer:

2.40 Km/hr.

Explanation:

From the question given above, the following data were obtained:

Mass of truck (m₁) = 3500 kg

Velocity of truck (u₁) = 7 Km/hr

Mass of car (m₂) = 1200 kg

Velocity of car (u₂) = 11 km/hr

Post collision velocity (v) =?

The post collision velocity of the truck and car can be obtained as illustrated below:

m₁u₁ – m₂u₂ = v(m₁ + m₂)

(3500 × 7) – (1200 × 11) = v(3500 + 1200)

24500 – 13200 = v × 4700

11300 = v × 4700

Divide both side by 4700

v = 11300 / 4700

v = 2.40 Km/hr.

Therefore, the post collision velocity of the truck and car is 2.40 Km/hr.

Help! Il give brainlest to who answers first

Answers

Answer:

1. The density of the cube is 1.03 g/mL.

2. Dish soap

Explanation:

1. Determination of the density of the cube.

From the question given above, the following data were obtained:

Mass (m) of cube = 21.7 g

Volume (V) of cube = 21 mL

Density (D) of cube =?

The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Thus, density can be expressed mathematically as:

Density (D) = mass (m) / volume (V)

D = m / V

With the above formula, we can obtain the density of the cube as follow:

Mass (m) of cube = 21.7 g

Volume (V) of cube = 21 mL

Density (D) of cube =?

D = m / V

D = 21.7 / 21

D = 1.03 g/mL

Thus, the density of the cube is 1.03 g/mL.

2. Determination of the layer of density the cube will settle in.

From the question given above,

Subtance >>>>>>>> Density

Vegetable oil >>>>> 0.91 g/mL

Grape juice >>>>>> 0.97 m/L

Water >>>>>>>>>>> 1 g/mL

Dish soap >>>>>>>> 1.03 g/mL

Maple syrup >>>>>> 1.37 g/mL

Comparing the density of the cube (i.e 1.03 g/mL) with those in the table able, we can conclude that the cube will settle in the DISH SOAP layer since they both have the same density.

Explain some similarities and differences between:
Magnetism and electrostatic forces

Answers

Answer:

Electric field points in the direction of the force experienced by a positive charge. Magnetic field points in the direction of the force experienced by a north pole.

Is the formula for velocity the same as speed or different?

Answers

Answer:

always same

Explanation:

velocity and speed are same upto some extend but velocity is vector while speed is scalar quantity

Starting with the definition 1 in. = 2.54 cm, find the number of (a) kilometers in 1.20 mile and (b) feet in 1.80 km.

Answers

Explanation:

[tex]miles \: to \: km \\ = 1.2 \: miles \times \frac{1 \: km}{1.6093 \: miles} \\ = 0.7457 \\ \\ km \: to \: feet \\ = 1.8 \: km \times \frac{3280 \: ft}{1 \: km} \\ = 5904 \: ft[/tex]

A sound wave travels in air toward the surface of a freshwater lake and enters into the water. The frequency of the sound does not change when the sound enters the water. The wavelength of the sound is 2.86 m in the air, and the temperature of both the air and the water is 20 degrees C. What is the wavelength in the water

Answers

Answer:

Explanation:

velocity of sound in air at 20⁰C is 343 m /s

velocity of sound in water at 20⁰C is 1481 m /s

The wavelength of the sound is 2.86 m in the air so its frequency

= 343 / 2.86 = 119.93 .

This frequency of  119.93 will remain unchanged in water .

wavelength in water = velocity in water / frequency

= 1481 / 119.93

= 12. 35 m .

Explain how the extension of a spring is determined

Answers

Answer:

For a given spring the extension is directly proportional to the force applied For example if the force is doubled, the extension doubles When an elastic object is stretched beyond its limit of proportionality the object does not return to its original length when the force is removed

Explanation:

A toy top with a spool of diameter 5.0 cm has a moment of inertia of 3.0×1025 kg⋅m2 about its rotation axis. To get the top spinning, its string is pulled with a tension of 0.30 N. How long does it take for the top to complete the first five revolutions? The string is long enough that it is wrapped around the top more than five turns

Answers

Answer:

Explanation:

Moment of inertia of toy top = 3 x 10⁻²   kgm²

Torque created = F x r

= .30 x 2.5 x 10⁻² N m

Torque = moment of inertia x angular acceleration

angular acceleration = .3 x 2.5 x 10⁻² / 3 x 10⁻²

α = .25 radian /s²

Angular displacement in 5 revolution θ = 5 x 2π = 10π radian

θ = ω₀t + 1/2 α t²

initial angular velocity ω₀ = 0

10π = 1/2 α t² = .5 x .25 t²

t² = 251.2

t = 15.85 s

You want to design a double-paned window. To reduce the amount of heat transfer by conduction through the window, you need to (more than one could be correct): You want to design a double-paned window. To reduce the amount of heat transfer by conduction through the window, you need to (more than one could be correct): Decrease the area of the window. Decrease the thickness of the window. Increase the thickness of the window. Increase the area of the window.

Answers

Answer:

the correct answer to increase the thickness of the windows

Explanation:

Thermal transfer by conduction is

           P = k A   [tex]( \frac{ T_h - T_c}{L} )[/tex]

where P is the heat transfer rate, A is the area and L is the thickness

In the given case we have that the window is made up of two materials, two layers of transparent glass and a layer of air between them, let us use the index 1 for the glass and the index 2 for the air, the equation remains

           P =[tex]k_1 A_1 \frac{ \Delta T }{2 \ e_1} + k_2 A_2 \frac{ \Delta T}{e_2}[/tex]

where the number two (2) is due to the two layers of glass

the thermal conductivity of the glass is k₁ = 0.75 W/m K for a glass of

e₁ = 4 mm, this is more used for windows

the thermal conductivity of the air is k₂ = 0.024 W / m K for a temperature of T = 0ºC

In a window the area and the temperature difference are constant for each part for a given configuration

           P = [tex]( \frac{k_1}{2 \ e_1} + \frac{k_2 }{ e_2} )[/tex]   A ΔT

To decrease the speed of thermal transfer we can decrease the area, but this also reduces the lighting in the room, which brings other costs.

We can also increase the thickness of the materials, as we see the thermal conductivity of the air is very less than that of glass

                k₂  « k₁

increasing the thickness of the air layer would have the greatest effect is the decrease in heat transfer in window light

consequently the correct answer to increase the thickness of the windows

How much energy, in joules, is released by an earthquake of magnitude 8?
______________ joules
(Use scientific notation. Use the multiplication symbol in the math palette as needed.)

Answers

Answer:

6.309573e+16 joule is released by an earthquake of magnitude 8

Answer:

An earthquake with a magnitude of 8 on the Richter scale releases 6.309573e+16 J of energy.

Explanation:

An earthquake with a magnitude of 8 on the Richter scale releases 6.309573e+16 J of energy.

Determine the magnitude of the resultant force and its direction using both the parallelogram and Cartesian vector notation methods. The direction of the resultant force is measured counter-clockwise from the positive x-axis. Draw the resultant force in the Cartesian coordinate system.F1= 600 N, F2= 900 N, β1 = 50 degree, and β2 = 40 degree.

Answers

Answer:

   F = 1494.52 N,   θ = 44º

Explanation:

For the sum of vectors by the parallelogram method, see attached, the vectors are drawn, the parallelogram is completed and a vector is drawn from the origin of the two vectors to the end point of the rectangle, this is the resulting vector.

The attachment shows this roughly.

For the Cartesian coordinate method, each vector is decomposed into its components, they are added algebraically and then the resulting vector is composed in the form of a module and angles

we use trigonometry to decompose the vectors.

The coordinate system can be seen in the attachment

           sin θ = y / R

           cos θ = x / R

            y = R sin θ

            x = R cos θ

Vector 1

module F₁ and angle β₁ = 50

            sin 50 = [tex]\frac{F_{1y} }{F_1}[/tex]

            cos 50 = [tex]\frac{F_{1x} }{F_1}[/tex]

            [tex]F_{1y}[/tex] = F₁ sin 50

            F₁ₓ = F₁ cos 50

            F_{1y} = 600 sin 50 = 459.63 N

            F₁ₓ = 600 cos 50 = 385.67 N

Vector 2

modulus F₂ = 900N, angle β₂ = 40

            F_{2y} = 900 sin 40 = 578.51 N

            F₂ₓ = 900 cos 40 = 689.44 N

we find the resultant of each component

           F_{y} =F_{1y} + F_{2y}

           F_{y}  = 459.63 + 578.51

           F_{y}  = 1038.14 N

 

            Fₓ = F₁ₓ + F₂ₓ

            Fₓ = 385.67 + 689.44

             Fₓ = 1075.11 N

We use the Pythagorean theorem to find the modulus of the resultant

            F = Fₓ² + [tex]F_{y}^2[/tex]

            F = √(1075.11² + 1038.14²)

            F = 1494.52 N

we use trigonometry for the angle

            tan θ = F_y / Fₓ

            θ = tan⁻¹ (F_y / Fₓ)

            θ = tan⁻¹ (1038.14 / 1075.11)

            θ = 44º

The following diagram shows four charged objects: A, B, C, and D.
Based on the diagram, which statement is true?
The electric force between A and B is identical in magnitude to the electric force between C and D.
The electric force between A and B is greater than the electric force between C and D.
The electric force between A and B is smaller than the electric force between C and D.
The electric force between A and B is canceled by the electric force between C and D.

Answers

Answer:

they are identical.

Explanation:

use coulomb law. the size of that line is just to confuse students because some sick teacher gets off on making you miss points. coulombs law says the magnitude of electric field is proportional to the size of the charge. here they are all +/-1 they are the same.

The true statement is "The electric force between A and B is greater than the electric force between C and D." The correct answer is B.

What is electric force?

Electric force, also known as Coulomb force, is the attraction or repulsion between electrically charged particles. It is a fundamental force of nature that arises from the interaction of charged particles with each other. Electric force is described mathematically by Coulomb's law, which states that the magnitude of the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The direction of the force depends on the sign of the charges, with like charges repelling each other and opposite charges attracting each other.

Here in the Question,

The electric force between two charged objects depends on the magnitude of the charges and the distance between them. In this case, object A has a positive charge and object B has a negative charge, so there is an attractive force between them. Object C also has a positive charge, but it is separated by a larger distance from the negative charge at point D, so the electric force between C and D is weaker than the force between A and B.

Option A is not true because the distance between A and B is smaller than the distance between C and D, so the forces cannot be identical in magnitude.

Option C is not true because, as explained above, the force between A and B is greater than the force between C and D.

Option D is not true because the forces between A and B and between C and D are not opposite in direction, so they cannot cancel each other out.

Therefore, The correct answer is B. The electric force between A and B is greater than the electric force between C and D.

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Lolliguncula brevis squid use a form of jet propulsion to swim -- they eject water out of jets that can point in different directions, allowing them to change direction quickly. When swimming at a speed of 0.15 m/s or greater, they can accelerate at 1.2 m/s^2.

Required:
a. Determine the time interval needed for a squid to increase its speed from 0.15m/s to 0.45m/s
b. What other questions can you answer using the data?

Answers

Answer:

a) the required time interval is 0.25 s

b) the distance traveled while accelerating is 0.075 m

Explanation:

Given that;

Initial speed of the squid u = 0.15 m/s

final speed of the squid v = 0.45 m/s

acceleration a = 1.2 m/s²

Time period T = ?

using equation of motion

v = u + at

solve for t

at = v - u

t = v - u / a

we substitute

t = ( 0.45 - 0.15 ) / 1.2

t = 0.3 / 1.2

t = 0.25 s

Therefore, the required time interval is 0.25 s

b)

the other possible question is;  Determine the distance the squid has traveled while accelerating.

using the following expression

v² = u² + 2as

we solve for distance s

2as = v²- u²

s = v²- u² / 2a

we substitute

s = ((0.45)²- 0.15)²) / 2×1.2

s = (0.2025 - 0.0225) / 2.4

s = 0.18 / 2.4

s = 0.075 m

Therefore,  the distance traveled while accelerating is 0.075 m

A 74.1 kg high jumper leaves the ground with
a vertical velocity of 9.7 m/s.
How high can he jump? The acceleration
of gravity is 9.8 m/s
2
.
Answer in units of m

Answers

Answer:

4.80 m

Explanation:

We are given the mass of the high jumper, its initial velocity, and the acceleration of gravity. We are trying to find the vertical displacement of the high jumper.

Let's set the upwards direction to be positive and the downwards direction to be negative.

List out the relevant known variables.

v₀ = 9.7 m/s a = -9.8 m/s² Δx = ?

We still need one more variable in order to use the constant acceleration equations. Since we are trying to find the max height of the jumper, we can use the fact that at the top of its trajectory, its final velocity will be 0 m/s.

    4. v = 0 m/s

Using these four variables, let's find the constant acceleration equation that contains these variables:

v² = v₀² + 2aΔx

Substitute the known values into the equation and solve for Δx.

(0)² = (9.7)² + 2(-9.8)Δx 0 = 94.09 + (-19.6)Δx -94.09 = -19.6Δx Δx = 4.80

The high jumper can jump to a max height of 4.80 m.

Arial



How do Seasons Help us Predict the
Weather ?

Answers

Answer:

if its winter

Explanation:

then we know it will me cold

Evaporation rates, river flows, lake levels. For example, leaves fall when cold weather approaches.

If a tiger wood hits a 0.050 kg golf all giving it a speed of 75m/s what impulse does he impart to the ball

Answers

Answer:

impulse = change in linear momentum

3.75kgm/s

Explanation:

initial momentum is 0 because at the start golf ball is at rest

and our final momentum 0.050kg × 75m/s = 3.75kgm/s

momentum final - momentum initial

3.75kgm/s  -  0

3.75kgm/s

PLEASE HELP
How much work is required to raise a 1200 kg rock 15 m?

Answers

Answer:

18000

Explanation:

the is to raise how much work is required is 18000

Work done to raise an object of mass 1200 kg to the height of 15 meters will be equal to 176,400 J.

What is Work?

In physics, the word "work" involves the measurement of energy transfer that takes place when an item is moved over a range by an externally applied, at least a portion of which is applied within the direction of the displacement.

The length of the path is multiplied by the element of a force acting all along the path to calculate work if the force is constant. The work W is theoretically equivalent towards the force f times the length d, or W = fd, to portray this concept.

As per the data given in the question,

Mass of the object, m = 1200 kg

Displacement, d = 15 meters

Then, use the equation of work done,

W = fd

W = (m×a)d

Substitute the values in above equation,

W = 1200 × 9.8 × 15

W = 176,400 J.

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