How is temperature related to heat

The electric potential energy of the two-sphere system will be converted into kinetic energy as the spheres move away from each other. The direction of the spheres' movement will be away from each other, in opposite directions.

This is because the two positively charged spheres repel each other due to their like charges. As they move away from each other, the electric potential energy of the system decreases, while the kinetic energy of the spheres increases.

The law of conservation of energy dictates that the total energy of the system remains constant, but the energy is converted from potential to kinetic energy as the spheres move away from each other.

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Answer:

Eliza’s suggestion is the most promising

Explanation:

Newton’s law of universal gravitation is \(F = G\frac{m_1m_2}{r^2}\) where G is a constant and both masses are constant in this experiment as well. So it only depends on r, the distance between the center of mass of both objects.

Bringing it high in the sky, as Flo suggests, is definitely not a good idea because it only increases the distance from the center of the Earth mass.

A mountain contains significant mass, so the center of the Earth mass is somewhat shifted to regions with the largest mountains. However, standing on top of a mountain, as Lorenzo suggests, doesn’t help since the shift of the center of mass, if any, is far smaller than the height of the mountain.

Standing near sea level, as Eliza suggests, is a good way to minimize the distance to the center of the Earth mass.

Answer:

because I dont know

Explanation:

first you add the multiply

The instantaneous velocity at 60 m is zero, at - 40 m is - 1 m / s, at 15 s is zero and at 25 s is 0.8 m / s if Displacement (m) 0 20 40 40 80 80 60 400 Time (s) 10 10 20 30 40 50 60 70 80.

Instantaneous velocity = Position with respect to time / Time

1 ) At d = 60 m, the instantaneous velocity is zero because the car is at rest.

2 ) At d = - 40 m, t = 40 s

Instantaneous velocity = - 40 / 40

Instantaneous velocity = - 1 m / s

3 ) At t = 15 s, d = 60 m

At 60 m, the car is at rest, so the Instantaneous velocity is zero

4 ) At t = 25 s, d = 20 m

Instantaneous velocity = 20 / 25

Instantaneous velocity = 0.8 m / s

Instantaneous velocity of a given curve in a position-time graph can be found by drawing a tangent to the curve and finding the slope of the tangent. Instantaneous velocity = 0

Instantaneous velocity = - 1 m / s

Instantaneous velocity = 0

Instantaneous velocity = 0.8 m / s

Therefore, The instantaneous velocity at 60 m is zero, at - 40 m is - 1 m / s, at 15 s is zero and at 25 s is 0.8 m / s if Displacement (m) 0 20 40 40 80 80 60 400 Time (s) 10 10 20 30 40 50 60 70 80.

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Explanation:

a) Gamma rays are high-energy photons and are the most energetic part of the electromagnetic spectrum.

b) Beta particles are high-energy, high-speed electrons emitted by certain radioactive nuclei.

c) There is no radiation called delta radiation.

d) Alpha particles are particles that are composed of two protons and two neutrons.

Final answer:

Thus, the correct option is (B) Beta

Answer: 0.390625 grams

Explanation:

A half-life of an element is the amount of time that it takes for half the mass of the element to decay.

Gold-198 having a half-life of 2.5 days therefore means that every 2.5 days, the mass is cut in half.

If there are 20 days, find out how many half-life periods there are:

= 20 / 2.5

= 8 periods

The half life is:

= Original mass * 0.5^number of half-life periods

= 100 * 0.5⁸

= 0.390625 grams

Answer:

To correct the defects of vision by measuring the radius of curvature and thus the power of the lenses.

Explanation:

A spherometer is an instrument used to measure the curvature of objects such as lenses and curved mirrors.

Generally it consists of a fine screw which is moving in a nut carried on the center of a 3 small legged table or frame. The feet forms the vertices of an equilateral triangle. The lower end of the screw and those of the table legs are finely tapered and terminate in hemispheres.

If the screw has two turns of the thread to the milli meter the head is generally divided into 50 equal parts, so that differences of 0.01 millimeter may be measured without using a vernier scale.

The spherometer is used to measure the radius of curvature of the lenses so that the opthalmologist find the focal length of the lens and then give the  power to the lens to correct the defects of vision.  

The thermosphere has the highest temperature of any layer in Earth’s atmosphere.

The term troposphere is the region that is found  12 to 50 kilometers from Earth’s surface. This region is found to be the region where you can find a lot of gases.

Both the Troposphere and the stratosphere  get colder as altitude increases. However, the ozone in the stratosphere  protects people from ultraviolet (UV) radiation.

The thermosphere has the highest temperature of any layer in Earth’s atmosphere.

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Answer:

2.1x10^6m/s

Explanation:

One electron has a charge of –1.602e-19 C

mass of electron is 9.1e-31 kg

mass of proton is 1.6726e−27 kg

mass ratio is 1.6726e−27 / 9.1e-31 = 1838

The force is constant, F

distance is constant, d

a = F/m

a increases by a factor 1838, as m decreases by that factor

a = a₀1838

v₀² = 2a₀d

v² = 2a₀d1838

v²/v₀² = 2a₀d1838 / 2a₀d = 1838

v² = 1838v₀² = 1838(45000)²

v = 45000√1838 = 2.1e6 m/s

Answer:

a) 1.1418 kg

b) 0.9135 kg/s

Explanation:

Given data :

spring constant ( k ) = 3 kg/s^2

amplitude of vibrations will decay like :  

The mass of the object will be 1.418 kg and the value of the friction coefficient, b is 0.9135 kg/s.

In physics, a periodic motion is a movement that repeats at regular intervals. A swing in motion, a rocking chair, a bouncing ball, a tuning fork vibrating, the Earth orbiting the Sun, and a water wave are all examples of periodic motion. The time interval for repetition, or cycle, of the motion, is referred to in each case as a period, and the frequency is the number of periods per unit of time.

According to the question, the given values are :

Spring constant, k = 3 kg/s²,

Time, T₁ = 4 sec

r = 0.4

The formula for damped oscillation is : \(e^-^r^t\)

r = b / 2 m

w₁ = \(\sqrt{w_0^2-r^2}\)

T₁ = 2π / w₁

= 2 π /        \(\sqrt{w_0^2-r^2}\)

w₀ = \(\sqrt{4 \pi^2/T_1 +r^2}\)

\(\sqrt{k/m}\) = \(\sqrt{4 \pi^2/T_1 +r^2}\)

M = k / \({4 \pi^2/T_1 +r^2}\)

M = 3/ (4π/4² +(0.4)²)

M = 1.1418 kg

b = 2 Mr = 2 × 1.1418 × 0.4

b = 0.9135 kg/s

Hence, the mass of the object is 1.1418 kg and the friction coefficient is 0.9135 kg/s.

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The total distance traveled by the object is 3h.

The total distance traveled by the object can be determined by applying the principle of conservation of energy as follows;

Since the object loses half its energy each time it hits the ground.

Hence it will rise half of the previous height each time, as potential energy at height h is given by;

P.E =  mgh

Thus, the object will rise h/2 after the first hit, then h/4 and so on.

Hence the total distance traveled by the object is calculated as follows;

s = h + (h/2 + h/2 + h/4 + h/4 + ---)

s = h + (h + h/2 + ---)

s = h + h(1 + 1/2 + 1/4 + ---)

s = h + h(1 +  (1/1 - 1/2) )

s = 3h

Thus, the total distance traveled by the object is 3h.

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The complete question is below:

An object is dropped from h height. graph v vs h? Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as t→∞ is :

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

a 150-dB sound is approximately 10 million times louder than an 80-dB sound.

Provide instances of the differences between the audible, ultrasonic, and infrasonic frequency ranges. Three sorts of sound waves, each covering a distinct frequency range, are used. These are what they are:

The difference in decibels between two sounds is related to the ratio of their intensities (or power) by the following formula:

dB₂ - dB₁ = 10 log10(I₂ / I₁)

where dB₁ and dB₂ are the decibel levels of the two sounds, and I₁ and I₂ are their intensities (or power).

Using this formula, we can find the ratio of the intensity of a 150-dB sound to that of an 80-dB sound:

150 dB - 80 dB = 10 log10(I₁₅₀ / I₈₀)

70 = 10 log10(I₁₅₀ / I₈₀)

7 = log10(I₁₅₀ / I₈₀)

10^7 = I₁₅₀ / I₈₀

I₁₅₀  = 10^7 * I₈₀

This shows that the intensity of a 150-dB sound is 10 million times greater than that of an 80-dB sound.

Therefore, a 150-dB sound is approximately 10 million times louder than an 80-dB sound.

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Answer: bismuth and nitrogen, because they have the same number of valence electrons

Explanation:

Elements are distributed in groups and periods in a periodic table.

Elements that belong to same groups will show similar chemical properties because they have same number of valence electrons.

The number of valence electrons in Bismuth and nitrogen are 5 and thus thus they will show similar chemical properties and thus belong to the same group.

The atomic masses of elements in a group will differ drastically.

The group number has got nothing to be the isolation year.

Thus bismuth and nitrogen belong to same group because they have the same number of valence electrons

Answer:

D

Explanation:

i got it right in my test

To encourage children to eat, create a pleasant and distraction-free mealtime environment, serve a variety of visually appealing and small portions of favorite foods, involve them in meal planning and preparation, provide a relaxed and comfortable atmosphere, and avoid using food as a reward or punishment. Appropriate snack foods from each food group for school-age children can be selected from fruits, vegetables, grains, protein, and dairy. Adults can help children learn to enjoy nutritious food and establish healthy eating habits by modeling healthy eating behaviors, involving children in grocery shopping and meal planning, making meals enjoyable and fun, providing healthy food choices, and avoiding using food as a reward or punishment.

1. To create an atmosphere that encourages children to eat, several features can be used, such as:

A pleasant and inviting mealtime environment that is free of distractions such as television, mobile phones, or tablets.

Serve a variety of colorful, visually appealing foods in small portions, making sure to include favorite foods of the child.

Involve children in meal planning and preparation, giving them the opportunity to select foods and participate in simple meal preparation tasks.

Provide a relaxed and comfortable atmosphere by allowing enough time for meals and offering positive reinforcement for good eating behaviors.

Avoid using food as a reward or punishment and ensure that mealtimes are stress-free and enjoyable for children.

2. MyPlate is a visual representation of the five food groups that form a healthy and balanced diet. Appropriate snack foods from each food group for school-age children could include:

Fruits: apple slices, berries, bananas, or orange wedges.

Vegetables: carrot sticks, cherry tomatoes, cucumber slices, or celery sticks.

Grains: whole-grain crackers, granola bars, rice cakes, or air-popped popcorn.

Protein: hard-boiled eggs, cheese sticks, nut butter, or hummus.

Dairy: yogurt, milk, or cheese cubes.

3. Adults can help children learn to enjoy nutritious food and establish healthy eating habits by:

Modeling healthy eating behaviors and food choices themselves.

Encouraging children to try new foods and flavors, involving them in grocery shopping, and meal planning.

Making meals enjoyable and fun by creating a positive mealtime environment and involving children in meal preparation tasks.

Providing healthy food choices and limiting the availability of unhealthy options at home and school.

Offering regular mealtimes and avoiding using food as a reward or punishment.

Hence, Encourage children to eat by providing a pleasant and distraction-free mealtime environment, serving a variety of visually appealing and small portions of favorite foods, involving them in meal planning and preparation, providing a relaxed and comfortable environment, and avoiding using food as a reward or punishment. Fruits, vegetables, grains, protein, and dairy are examples of appropriate snack foods from each food group for school-age children. Adults can assist children in learning to enjoy nutritious food and develop healthy eating habits by modeling healthy eating behaviors, involving children in grocery shopping and meal planning, making meals enjoyable and fun, providing healthy food options, and avoiding using food as a reward or punishment.

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Answer: tennis ball

Explanation:

Answer:

For example, speed is a length divided by time. Force is mass times acceleration, and is therefore a mass times a distance divided by the square of a time. We therefore say that [Force] = MLT−2.

Grace Scholars Program is a scholarship program that selects the best and brightest students for the program.

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The total distance covered by the car is 300 miles.

The total displacement covered by the car is zero.

The average speed of the car is 17.88 m/s.

The average velocity of the car is also zero.

Distance between the points A and B, d = 150 miles

Time taken by the car to travel from A to B, t₁ = 3 hours

Time taken by the car to travel from B to A, t₂ = 5 hours

a) Given that the car travelled from A to B and then back to A.

Therefore, the total distance covered by the car is,

Distance = 2 x d

Distance = 2 x 150

Distance = 300 miles

Since the car is travelling from A to B and then returning back to the initial point A, the total displacement covered by the car is zero.

b) The speed with which the car travelled from A to B is,

v₁ = d/t₁

v₁ = 150/3

v₁ = 50 miles/hr

v₁ = 22.35 m/s

The speed with which the car travelled from B to A is,

v₂ = d/t₂

v₂ = 150/5

v₂ = 30 miles/hr

v₂ = 13.41 m/s

Therefore, the average speed of the car is,

v = (v₁ + v₂)/2

v = (22.35 + 13.41)/2

v = 17.88 m/s

As, the total displacement of the car is zero, the average velocity of the car is also zero.

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The change in surface area of Gaussian surface with radius (r) is 8πr.

The electric field experienced by a charge is calculated as follows;

\(E = \frac{Q}{4\pi \varepsilon_o r^2}\)

where;

The electric field reduces by a factor of \(\frac{1}{r^2}\)

The surface area of a sphere is given as;

\(A = 4\pi r^2\)

\(\frac{dA}{dr} = 8\pi r\)

Thus, the change in surface area of Gaussian surface with radius (r) is 8πr.

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Answer:

471392.4 N

Explanation:

From the question,

Just before contact with the beam,

mgh = Fd.................... Equation 1

Where m = mass of the beam, g = acceleration due to gravity, h = height. F =  average Force on the beam, d = distance.

make f the subject of the equation

F = mgh/d................ Equation 2

Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m

Constant: g = 9.8 m/s²

Substitute into equation 2

F = 1900(4)(9.8)/0.158

F = 471392.4 N

The volume of the gas at 101.3 kPa would be approximately 651.25 cm³.

To calculate the change in volume of a gas from an initial pressure to a final pressure, we can use Boyle's law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.

Boyle's law can be expressed as:

P1 * V1 = P2 * V2

Where:

P1 = Initial pressure (80.8 kPa)

V1 = Initial volume (817 cm³)

P2 = Final pressure (101.3 kPa)

V2 = Final volume (to be calculated)

Let's plug in the values into the equation and solve for V2:

80.8 kPa * 817 cm³ = 101.3 kPa * V2

V2 = (80.8 kPa * 817 cm³) / 101.3 kPa

V2 ≈ 651.25 cm³

Therefore, the volume of the gas at 101.3 kPa would be approximately 651.25 cm³.

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Answer:

A piece of copper at 100°C is transferred to a copper calorimeter with a liquid at 30°C. The final temperature is 50°C. By applying the principle of conservation of energy, the specific heat capacity of the liquid is calculated to be approximately 2100 J/kg/°C.

To calculate the specific heat capacity of the liquid, we can apply the principle of conservation of energy. The heat lost by the copper piece will be equal to the heat gained by the liquid and calorimeter.

The heat lost by the copper piece can be calculated using the formula:

Heat lost = Mass of copper × Specific heat capacity of copper × Temperature change

Given:

Mass of copper = 400 g

Specific heat capacity of copper = 390 J/kg/°C (assuming it remains constant)

Temperature change of copper = 100°C - 50°C = 50°C

Heat lost = 400 g × 390 J/kg/°C × 50°C

Heat lost = 7,800,000 J

The heat gained by the liquid and calorimeter can be calculated using the formula:

Heat gained = (Mass of liquid + Mass of calorimeter) × Specific heat capacity of liquid × Temperature change

Given:

Mass of liquid = 100 g

Mass of calorimeter = 10 g

Temperature change of liquid = 50°C - 30°C = 20°C

Heat gained = (100 g + 10 g) × Specific heat capacity of liquid × 20°C

Now, by equating the heat lost and heat gained:

7,800,000 J = (110 g) × Specific heat capacity of liquid × 20°C

Specific heat capacity of liquid = 7,800,000 J / (110 g × 20°C)

Specific heat capacity of liquid ≈ 3545.45 J/kg/°C

Therefore, the specific heat capacity of the liquid is approximately 3545.45 J/kg/°C.

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Answer:

A

Explanation:

I just did the test

Answer:

B

Explanation:

Friction opposes the motion of a body

The two strings are taut. When the system is released from rest, it accelerates at 2 m/s2 then, Mass of A = 8m/5 kg.

Let the mass of the body A be ‘m’.

The two strings are taut so they exert a tension ‘T’ on body A.

Let ‘a’ be the acceleration produced in the system.

The free body diagram of body A is given below: mA + 2T = mA + ma = mA + m(2)mA + 10T = mA + ma = mA + m(2)

As the two strings are taut, we can say that tension in both strings is equal.

Therefore 2T = 10T or T = 5T As the body A is resting on a smooth horizontal table, there is no friction force acting on the body A.

The net force acting on body A is the force due to tension in the strings. ma = 2T – mg …(1)

As per the given problem, the system is released from rest.

Hence the initial velocity is zero.

Also, we are given that the system accelerates at 2 m/s2.

Therefore a = 2 m/s2 …(2)

From the equations (1) and (2), we get, m(2) = 2T – mg …(3)⇒ m(2) = 2×5m – mg⇒ 2m = 10m – g⇒ g = 8m/5

Thus, the mass of A is 8m/5 kg.

Answer: Mass of A = 8m/5 kg.

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Answer:

Explanation:

A television set is plugged into a 120 V outlet. The television circuit carries a current equal to 0.75 A. What is the overall resistance of the television set?​

V = IR

120 volt = 0.75A * R

R = 160 Ω

Given:

The magnetic field is B = 0.2 T

The angle is

The speed of the helical path is

To find the pitch.

Explanation:

The pitch can be calculated by the formula

Here, m is the mass of the electron whose value is 9.1 x 10^(-31) kg

q is the charge of the electron whose value is 1.6 x 10^(-19) C

On substituting the values, the pitch will be

Answer:

b and c

Explanation:

NASA might hire U.S. businesses and universities to do:  Research new ideas, build complex equipment. Hence, option (b) and (c) are correct.

Astronomy and space technology are used in space exploration to learn more about the universe. While astronomers use telescopes to explore the universe, both unmanned robotic space missions and human spaceflight also participate in the physical exploration of space. One of the main sources of space science, like its traditional form astronomy, is space exploration.

Although astronomy, or the study of celestial objects, has existed since the beginning of trustworthy written history, it wasn't until the mid-20th century that the possibility of physical space exploration became a reality.

Space exploration can lead to money going to U.S. businesses and universities. Hence,  NASA might hire U.S. businesses and universities to do:  Research new ideas, build complex equipment.

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According to the question the equivalent resistance is 37.0 Ω.

Equivalent resistance is the resistance of a circuit when its individual resistors are replaced with a single resistor that has the same overall effect on the circuit. It is a measure of resistance that is used to simplify calculations of electrical circuits. Equivalent resistance is calculated by taking the sum of the inverse of the individual resistances and then inverting the sum to find the equivalent resistance. This is useful for analyzing complex circuits as it allows for easier calculations.

The equivalent resistance of the three resistors joined in series is equal to the sum of the individual resistances.
Therefore, the equivalent resistance is 6.00 Ω + 11.0 Ω + 20.0 Ω = 37.0 Ω.

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