When electrical energy is used what type of energy is also produced and considered to be waste energy?

Answers

Answer 1

Answer:

electrical energy is energy derived from electric potential energy or kinetic energy. when loosely electrical energy refers to energy that has been converted from electric potential energy

Answer 2

Answer:

electrical energy is energy derived from electric potential energy or kinetic energy. when loosely electrical energy refers to energy that has been converted from electric potential energy

Explanation:


Related Questions

Danny lowers the sails on his boat. He paddles upstream at 19 km/hr. The current is still running downstream at 15 km/hr. What is the actual velocity of the boat?

Answers

Answer:

4 km/hr

Explanation:

The computation of the actual velocity is shown below:

Because the path of its paddles is opposed to the current direction, the real velocity can be determined by deducting the current velocity to its velocity while paddling

So, the actual velocity is

= Upstream - downstream

= 19 km/hr - 15 km/hr

= 4 km/hr

As we can see it is in positive, so it is an upstream direction  

Answer:

since the direction of his paddles is opposite of the the direction of the current, so the actual velocity can be calculated by subtracting the velocity of current to to his velocity when paddling

v = 19 - 15

v = 4 since the answer is positive, then the direction is upstream

Explanation:

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las personas junto a él respiran para calentarlo con su aliento y aumentar su temperatura 1 grado Celsius. El tubo se hace más largo. También ya no queda ajustado. ¿A qué distancia sube sobre sobre el nivel del suelo? (solo tomar en cuenta la expansión radial al centro de la tierra, y aplicar la fórmula geométrica que relaciona la circunferencia C con el radio r: C= 2πr).

Answers

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

[tex]C=2\pi r[/tex]    (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

[tex]r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m[/tex]

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

[tex]L=Lo[1+\alpha \Delta T][/tex]         (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

[tex]L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m[/tex]

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

[tex]r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m[/tex]

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

Explanation:

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las personas junto a él respiran para calentarlo con su aliento y aumentar su temperatura 1 grado Celsius. El tubo se hace más largo. También ya no queda ajustado. ¿A qué distancia sube sobre sobre el nivel del suelo? (solo tomar en cuenta la expansión radial al centro de la tierra, y aplicar la fórmula geométrica que relaciona la circunferencia C con el radio r: C= 2πr).

During or after exercise,it is normal for a student to comment that his/her heart feels like it is "beating out of my chest"?

Answers

Figuratively? yes it is.

Answer:

Yes it is normal

Explanation:

When you exercise, your heart beat goes up, resulting in people saying that their heart feels like it is "beating out of their chests".

Answer part (d) please

Answers

Answer:

MARK me brainliest please and follow my page

Explanation:

All you have to do to get the average speed is to calculate the total distance covered and divide it by the total time taken

= 16/18 = 0.88m/s

Average speed = (distance covered) / (time to cover the distance)

For the full 18 seconds described by the graph . . .

Average speed = (16 meters) / (18 seconds)

Average speed = (16 / 18) m/s

Average speed =  0.89 m/s

A student throws a 120 g snowball at 7.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is 0.15 s

Answers

Answer:

The magnitude of the average force on the wall during the collision is 6 N.

Explanation:

Given;

mass of snowball, m = 120 g = 0.12 kg

velocity of the snowball, v = 7.5 m/s

duration of the collision between the snowball and the wall, t = 0.15 s

Magnitude of the average force can be calculated by applying Newton's second law of motion;

F = ma

where;

a is acceleration = v / t

a = 7.5 / 0.15

a = 50 m/s²

F = ma

F = 0.12 x 50

F = 6 N

Therefore, the magnitude of the average force on the wall during the collision is 6 N.

can all alpha beta and gamma radiation treat cancer

Answers

Answer:

No

Explanation:

Only a few of alpha beta and gamma radiation can treat cancer

Which measurement is a potential difference?
O A. 115 N
O B. 115 C
O C. 115 J
O D. 115 V​

Answers

Answer:

option d is answer because pd is measured in volt.

Guys I'm in kind of a PICKLE!!!!!! I know people say it a lot but I will give Brainiest to the best explained answer. Determine the net force charge acting at q1 (+ 2.0 × 10^-5C), caused by q2 (-4.0 × 10-5 C) and q3 (-4.0 × 10^-5C). They create a right angles triangle, where q1 is at the 90° corner

Determine the net electric field acting at q1

Answers

Answer:

E≅1.2×10^7 N/C

Explanation:

First off I'd like to say that I'm taking "net electric field" to mean that they don't want this answer to be put into vector component form and instead want magnitudes. Sometimes the wording of these questions throws me off, so sorry ahead of time if that's what they want from you!

Edit: I ended up adding it anyways ;P

Since we are observing the net electric field acting at q1, we need to use the formula:  [tex]E=k\frac{q}{r^{2} }[/tex]

And since we are observing the effects of multiple charges at once...

E=ΣE, which just means wee need to add all the observed electric fields together:

ΣE= [tex]k\frac{q2}{r^{2} } +k\frac{q3}{r^{2} }[/tex]

Since we are observing [static] electric fields here, we don't actually need q1's charge. (Though if you wanted to find the net force you would.) Now, before we start plugging values in, let's acknowledge what we know. We know that:

q2=q3they are the same distance from q1

These are actually really nice to have, because now we can simplify our expression to:

[tex]E=k\frac{2q}{r^{2} }[/tex]

Now let's plug in our values and get an answer out.

E= 2(8.99×10^9)(4×10^-5)/(0.24)

Plugging all that in, I get:

E≅1.2×10^7 N/C

If you end up needing the net force, F=(q1)(E). That is, you just multiply the electric field by the value of q1. And again, if your teacher wants the answer in vector component form, then the answer will look different.

Let me know what doesn't make sense, or if I got something wrong. Good luck with AP Phy.!

Edit: I put the component form for my answer in the attachment. I also noticed a small calculator related error in my original answer. I updated that to match the new one.

10 advantages of friction

Answers

Answer:

1. it helps to change the direction.

2. it helps us to walk on ground.

3. it helps the vechils to break while moving.

4. helps in changing one form of enegry to another form. eg when we rub our hands we feel heat energy.

5. it opposites the force.

6. it helps us to change shape of objects.eg we roll the dough to make it roti.

7. it changes the state of body from rest motion.eg when we push any obj from inclined plane it moves.

i all know is just 7..

coma Narrows Bridge across the Puget Sound collapsed. The bridge was a suspension bridge. The wind blowing through the narrows matched the natural frequency of the bridge. This resulted in a large movement of roadway, which eventually caused the bridge to fail. What characteristic of waves caused the bridge to collapse

Answers

Answer:

amplitude

Explanation:

Amplitude is the characteristic of waves which caused the bridge to collapse.   Amplitude of a wave is the maximum amount of displacement of a particle occurs in the medium from its rest position. When the frequency of a wave reaches the natural frequency of the bridge, the oscillation of the bridge produce an amplitude where it causing the destruction of the bridge which is called Resonance. So we can say that amplitude is the characteristic of waves which is responsible for the collapse of the bridge.

Answer: C.interference, because constructive interference occurred when the wind frequency matched the natural frequency of the bridge

Explanation:

Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de sección por el que sale el agua con un chorro continuo. A) ¿Qué cantidad de líquido saldrá del depósito expresada en m3/s?

Answers

Answer:

a) El caudal de salida del chorro es [tex]1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex].

Explanation:

a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:

[tex]\Delta z = \frac{v_{out}^{2}}{2\cdot g}[/tex]

Donde:

[tex]\Delta z[/tex] - Diferencia de altura, medida en metros.

[tex]g[/tex] - Constante gravitacional, medida en metros por segundo al cuadrado.

[tex]v_{out}[/tex] - Rapidez de salida del chorro, medida en metros por segundo.

Se despeja la rapidez de salida del chorro:

[tex]v_{out} = \sqrt{2\cdot g \cdot \Delta z}[/tex]

Si [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] y [tex]\Delta z = 0.3\,m[/tex], entonces la rapidez de salida del chorro es:

[tex]v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}[/tex]

[tex]v_{out} \approx 2.426\,\frac{m}{s}[/tex]

Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:

[tex]\dot V_{out} = v_{out}\cdot A_{t}[/tex]

Donde:

[tex]v_{out}[/tex] - Rapidez de salida del chorro, medida en metros por segundo.

[tex]A_{t}[/tex] - Área transversal del orificio, medido en metros cuadrados.

[tex]\dot V_{out}[/tex] - Caudal de salida del chorro, medido en metros cúbicos por segundo.

Dado que [tex]v_{out} = 2.426\,\frac{m}{s}[/tex] y [tex]A_{t} = 5\,cm^{2}[/tex], el caudal de salida del chorro es:

[tex]\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}} \right)[/tex]

[tex]\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex]

El caudal de salida del chorro es [tex]1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex].

Una bola de 1 kg gira alrededor de un circulovrtical en el extremo de un cuerda. El otro extremo de la cuerda esta fijo en el centro del circulo. Calcular la diferencia entre las tensiones (de la cuerda) maxima y minima

Answers

Answer:

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

Explanation:

Puesto que la bola gira en un círculo vertical, existe claramente una diferencia entre las tensiones debido a la influencia de la gravedad y la tensión que resulta de la aceleración centrípeta experimentada por la masa. La máxima tensión ocurre cuando la bola se encuentra en el nadir (o la sima) del trayecto circular, la cual se describe por la Segunda Ley de Newton:

[tex]T_{max} - m\cdot g = m\cdot \frac{v^{2}}{L}[/tex]

En cambio, la mínima tensión aparece cuando la bola se encuentra en el cénit (o la cima) del trayecto circular, descrita por la misma ley de Newton:

[tex]T_{min} + m\cdot g = m\cdot \frac{v^{2}}{L}[/tex]

Donde:

[tex]T_{min}[/tex], [tex]T_{max}[/tex] - Tensiones mínima y máxima, medidas en newtons.

[tex]m[/tex] - Masa de la bola, medida en kilogramos.

[tex]g[/tex] - Constante gravitacional, medida en metros por segundo al cuadrado.

[tex]L[/tex] - Distancia con respecto al eje de rotación, medida en metros.

[tex]v[/tex] - Rapidez tangencial, medido en metros por segundo.

Se elimina la aceleración centrípeta de ambas expresiones por igualación:

[tex]T_{min} + m\cdot g = T_{max} - m\cdot g[/tex]

Ahora, la diferencia entre las tensiones máxima y mínima es:

[tex]T_{max} - T_{min} = 2\cdot m \cdot g[/tex]

Si [tex]m = 1\,kg[/tex] y [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], entonces:

[tex]T_{max} - T_{min} = 2\cdot (1\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]T_{max}-T_{min} = 19.614\,N[/tex]

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

When hitting the golf ball the golfer swings the golf club to keep it in contact with the golf ball for as long as possible. The force acting on the golf ball is constant during this time. Explain the effect that the time of contact between the golf club and the golf ball has on the distance the golf ball travels

Answers

Answer:

Explanation:

We shall apply the concept of impulse which is given as follows .

Impulse = force x time

Impulse = change in momentum

If u be the initial velocity of golf ball and v be the final velocity , m be the mass

change in momentum

= mu - ( - mv )

= mu+ mv

If F be the force applied and t be the duration of touch with the ball

Impulse = F x t

F x t = mu + mv

mv = Ft - mu

For given mu , greater the value of t , greater will be the value of v

so v is increased when t is increased .

Increased value of v will help in achieving greater distance attained by

golf ball

A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of Ï3 ft/s. Find the times at which the mass is heading downward at a velocity of 3 ft/s.

Answers

Answer:

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

Explanation:

To solve the problem/ we first write the differential equation governing the motion. So,

[tex]m\frac{d^{2} x}{dt^{2} } = -kx \\ m\frac{d^{2} x}{dt^{2} } + kx = 0\\\frac{d^{2} x}{dt^{2} } + \frac{k}{m} x = 0[/tex]

with m = 1 slug and k = 9 lb/ft, the equation becomes

[tex]\frac{d^{2} x}{dt^{2} } + \frac{9}{1} x = 0\\\frac{d^{2} x}{dt^{2} } + 9 x = 0[/tex]

The characteristic equation is

D² + 9 = 0

D = ±√-9 = ±3i

The general solution of the above equation is thus

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity  v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ    where k is an integer

3t  = ±π/6 + 2kπ - 4π/3

t  = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3  or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6. So,

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

The time interval at which the mass will head downward at the velocity of 3 ft/s is t = 5π/18 + 2nπ/3.

Given data:

The mass suspended from spring is, m = 1 slug.

The spring constant is, k = 9 lb/ft.

The magnitude of upward velocity is, v = 3 ft/s.

The magnitude of downward velocity is, v' = 3 ft/s.

The given problem can be resolved by framing a differential equation that governs the motion of spring. The differential equation governing the motion of spring is,

[tex]m \dfrac{d^{2}x}{dt^{2}}=-kx\\\\\\\dfrac{d^{2}x}{dt^{2}}+\dfrac{k}{m}x=0[/tex]

With m = 1 slug and k = 9 lb/ft, the equation becomes

[tex]\dfrac{d^{2}x}{dt^{2}}+\dfrac{9}{1}x=0\\\\\\\dfrac{d^{2}x}{dt^{2}}+9x=0[/tex]  

Now, the characteristic equation is,

D² + 9 = 0

D = ±√-9 = ±3i

And the general solution of the above equation is,

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂²

              = √[(-1)² + (-√3/3)²]

              = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3

and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ    

where k is an integer

3t  = ±π/6 + 2kπ - 4π/3

t  = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3  or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6.

t = 5π/18 + 2nπ/3

here, n is a natural number.

Thus, we can conclude that the time interval at which the mass will head downward at the velocity of 3 ft/s is t = 5π/18 + 2nπ/3.

Learn more about the differential equation here:

https://brainly.com/question/14620493  

How many atoms of oxygen in the chemical formula 2 Ca(ClO2)2?

Answers

Answer:

8

Explanation:

Ca(ClO2)2  -    2*2 = 4 Oxygen atoms

2 Ca(ClO2)2  - 2*4 = 8 Oxygen atoms

What type of image does a concave lens form? A. real image B. magnified image C. virtual image D. reverse image

Answers

the answer to this question is C. virtual image

state the observations made in this experiment ​

Answers

Answer:

the water is going to boil and the mercury ill melt and shoot the cork out the bottom of the tube

Explanation:

1. Find the energy required to melt 255g of ice at 0°C into water at 0°C

Answers

Answer:

E = 85170 J (/ 85.2 kJ)

Explanation:

Take the latent heat of fusion of water be 334J / g.

From the equation E = ml,

E = energy required (unknown),

mass m = 255g,

latent heat of fusion l = 334J / g,

E = 255 x 334

E = 85170 J (/ 85.2 kJ)

A ball has a mass of 0.25 kg and is moving to the right at 1.0 m/s. It hits a ball of mass 0.15 kg that is initially at rest. After the collision, the 0.15 kg ball moves off to the right with a velocity of 0.75 m/s. What is the final velocity of the 0.25 kg ball? 0.42 m/s to the right 0.42 m/s to the left 0.55 m/s to the right 0.55 m/s to the left

Answers

Answer:

C-0.55 m/s to the right

Explanation:

edg

Answer:

0.55 to the right

Explanation:

Un contenedor de 1800 N está en reposo sobre un plano inclinado a 28°, el coeficiente de fricción entre el contenedor y el plano es de 0.4. Calcule la fuerza P paralela al plano y dirigida hacia arriba de éste que hará que el contenedor se mueva hacia arriba con una velocidad constante.

Answers

Answer:

F = 1480.77N

Explanation:

In order to calculate the required force to push the container with a constant velocity, you take into account the the sum of force on the container is equal to zero. Furthermore, you have for an incline the following sum of forces:

[tex]F-Wsin\alpha-F_r=0\\\\F-Wsin\alpha-N\mu cos\alpha=0\\\\F-Wsin\alpha-W\mu cos\alpha=0[/tex]     (1)

F: required force = ?

W: weight of the container = 1800N

N: normal force = weigth

α: angle of the incline = 28°

g: gravitational acceleration = 9.8m/s^2

μ: coefficient of friction = 0.4

You solve the equation (1) for F and replace the values of the other parameters:

[tex]F=W(sin\alpha+\mu cos\alpha)\\\\F=(1800N)(sin28\°+(0.4)cos28\°)=1480.77N[/tex]

The required force to push the container for the incline with a constant velocity is 1480.77N

What can happen if a body moves through speed of light

Answers

As waves travel into the denser medium, they slow down and wavelength decreases.

Explanation:

The denser the medium the slower the waves (speed of light) travels.

◦•●◉✿When the body approaches the speed of light, the body's length appears to contract in the direction of travel, and its mass appears to increase from the point of view of a stationary observer. Only photons move to light velocity. They don´t have length.✿◉●•◦

Can someone please help me with this ? Neon gas in a container was heated from 20°C to 120°C. It's new volume is 150ml. What was the original volume ?

Answers

Answer:

110 mL

Explanation:

Ideal gas law:

PV = nRT

Assuming the container isn't rigid, and the pressure is constant, then:

V/T = V/T

Plug in values (remember to use absolute temperature).

V / 293 K = 150 mL / 393 K

V = 110 mL

a child rides her bicycle in her level drive way. what form of energy does she possess A. Elastic B. Nuclear C. Potential D. Kinetic

Answers

Answer:

Kinetic Energy

Explanation:

Kinetic energy is energy due to motion.

Answer:

D

Explanation:

Elastic energy is energy stored in a object when there is a strain or compression on the object.

Nuclear energy is the energy found in the nucleus of an atom.

Potential energy is energy that an object stores because of its position to other objects.

Kinetic energy is the energy that an object has due to motion.

The child is riding her bicycle, therefore the child is in motion. So, the correct answer must be D. Kinetic energy

Two identical resistors are connected in parallel across a 26-V battery, which supplies them with a total power of 7.1 W. While the battery is still connected, one of the resistors is heated so that its resistance doubles. The resistance of the other resistor remains unchanged. Find (a) the initial resistance of each resistor, and (b) the total power delivered to the resistors after one resistor has been heated.

Answers

Answer:

A) R = 190.42 Ω

B) P = 5.325 W

Explanation:

We are given;

Total power;P_tot = 7.1 W

Voltage;V = 26 V

A)We are told that while the battery is still connected, one of the resistors is heated, so that its resistance doubles.

Thus, the power is doubled.

Now, formula for power is;

P = IV

Thus, since power is doubled, we have;

P = 2(IV)

Now, formula for current is; I = V/R

So, P = 2V²/R

Making R the subject, we have;

R = 2V²/P

In this question, P is p_total = 7.1 W

Thus;

R = (2 × 26²)/7.1

R = 190.42 Ω

B) Now, the resistance of the resistors are R and 2R.

Formula for power in this context is;

P = V²/R

Thus,

Total power delivered to the resistors is;

P = V²/R + V²/2R

P = 3V²/2R

P = (3 × 26²)/(2 × 190.42)

P = 5.325 W

The skier starts from rest. The total distance travelled by the skier during the descent is 2800 m. The average resistive force on the skier is 220 N. Calculate the work done against the resistive force

Answers

Answer:

Explanation:

Force equal to resistive force will be applied for movement . So force applied

F = 220 N .

displacement = 2800 m

work done against resistive force

= force x displacement

= 220 x 2800 J

= 6.16 x 10⁵ J .

Q.) Miscellaneous conversations. a) mass=120*10^8 g (Convert this value in mg and kg Write in standard form after converting) b) length=200000*10^3 Convert this value in micrometres cm and km Write in standard form after converting

Answers

Answer:

a. Convert  120 × 10⁸ g to i mg = 1.2 × 10¹³ mg ii. to g = 1.2 × 10⁷ kg

b. Convert 200000 × 10³ m to i. micrometers = 0.2 × 10³ μm  ii. to cm = 2 × 10⁶ cm iii. to km = 2 × 10⁵ km

Explanation:

a. i. To convert the mass = 120 × 10⁸ g to mg, We know that 1000 mg = 10³ mg = 1 g, Since we are converting to mg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10³ mg = 120 × 10¹¹ mg = 1.2 × 10² × 10¹¹ mg = 1.2 × 10¹³ mg

ii. To convert the mass = 120 × 10⁸ g to kg, We know that 1000 g = 10³ g = 1 kg, 1 g = 10⁻³ kg. Since we are converting to kg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10⁻³ kg = 120 × 10⁵ kg = 1.2 × 10² × 10⁵ kg = 1.2 × 10⁷ kg

b. i.To convert the length = 200000 × 10³ m to micrometers, We know that 1/1000000 μm = 10⁻⁶ mg = 1 m, Since we are converting to micrometers, μm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ ×  1/1000000 μm = 200000/1000000 × 10³ μm = 0.2 × 10³ μm

ii. To convert the length = 200000 × 10³ m to cm, We know that 100 cm = 10² cm = 1 m, 1 m = 10⁻² cm = 1/100 cm. Since we are converting to cm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ × 1/100 cm = 200000/100 × 10³ cm = 2000 × 10³ cm  = 2 × 10³ × 10³ cm = 2 × 10⁶ cm

iii. To convert the length = 200000 × 10³ m to km, We know that 1000 m = 10³ m = 1 km, 1 m = 10⁻³ km = 1/1000 km Since we are converting to km, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ ×  1/1000 km = 200000/1000 × 10³ km = 200 × 10³ km = 2 × 10² × 10³ km = 2 × 10⁵ km

What kind of substance can you pour from one container into another without a change in volume

Answers

Liquid is the answer
Please mark brainliest please

Answer:

Liquids

Explanation:

Liquids take up the shape of the container it is poured into but will never change its volume.

An air-track glider of mass 0.25 kg moving at 0.60 m/s collides with and sticks to a glider of mass 0.50 kg at rest. How much kinetic energy is converted into other forms of energy as a direct result of this collision?

Answers

Answer:

0.03 Joules have been converted into other forms of energy as the direct result of the collision.

Explanation:

Let's start studying the conservation of momentum for the system:

[tex]P_i=P_f\\(0.25\,kg)\,{0.6\,m/s)+(0.5\,kg)\,(0\,m/s)=(0.25\,kg+0.5\,kg)\, v_f \\\\\\ 0.15\,kg\,m/s=0.75\,kg\,\,v_f\\v_f=0.15/0.75\,\,m/s\\v_f=0.2\,\,m/s[/tex]

Now that we know the speed of the newly created object, we can calculate how the final kinetic energy differs from the initial one:

[tex]K_i=\frac{1}{2} (0.25)\,(0.6)^2+\frac{1}{2} (0.5)\,(0)^2=0.045\,\,J\\ \\K_f=\frac{1}{2} (0.75)\,(0.2)^2=0.015\,\,J\\[/tex]

Then, when we subtract one from the other, we can estimate how much kinetic energy has been converted into other forms of energy in the collision:

0.045 J - 0.015 J = 0.03 J

A pendulum at position A is released and swings through position B to position Con the other side.
B
1. Describe the total mechanical energy at each of the following positions. (3)
A.
B.
C

Answers

Explanation:

Given the conditions A,B and C when the pendulum is released, at point A the initial velocity of the pendulum is zero(0), the potential energy stored is maximum(P.E= max),

the conditions can be summarized bellow

point A

initial velocity= 0

final velocity=0

P.E= Max

K.E= 0

point B

initial velocity= maximum

final velocity=maximum

P.E=K.E

point C

initial velocity= min

final velocity=min

P.E= 0

K.E= max

Magnetic field lines exit out of the . Magnetic field lines enter into the . Magnetic field lines travel around a bar magnet in

Answers

Answer:

Magnetic field lines exit out of the North pole . Magnetic field lines enter into the South pole. Magnetic field lines travel around a bar magnet in closed loops.

Explanation:

Magnetic field lines shows the direction of a magnetic force and how it acts, it gives the direction of the magnetic field at that point in time.

For a bar magnetic, the magnetic field lines runs from the north pole to the south pole, i.e. it exits the north pole and enters into the south pole. This magnetic field lines also go through the magnet forming closed loops without ends.

Answer:

Magnetic field lines exit out of the

✔ north pole

.

Magnetic field lines enter into the

✔ south pole

.

Magnetic field lines travel around a bar magnet in

✔ a closed loop

.

Explanation:

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