When heat is transferred by direct
contact, it is called...

Answers

Answer 1
C. Conductionnnnn :))

Related Questions

Gravitational potential energy is always calculated with
kinetic energy
mechanical energy
total potential energy
mass gravity and height

Answers

Answer:

kinetic energy as it involves dispersal of energy

A pendulum makes 50 complete swings in 2 min 40 second . what is time period for 1 complete swing?

Answers

Answer:

3.2 seconds

Explanation:

50 swings = 2 mins 40 secs

= 50 swings = 2 x 60 + 40 secs

= 50 swings = 160 secs

= 1 swing = 160 / 50 secs

= 1 swing -= 3.2 secs

A body at rest or moving with uniform velocity will have acceleration equals
to:
A.
1
B.
0
C.
Negligible
D.
Infinity

Answers

Answer:

B:O

Explanation:

As acceleration is the rate of change of velocity with time, if velocity is uniform, there will be no change in it and acceleration will be zero.

What the answer to the three of them ill mark brainlest

Answers

lhormmia  masng ies igentetel

Explanation:

PLSS HELPPP

A ball is at rest on a trapdoor. State the forces acting on the ball, including the direction they act it.

Answers

Explanation:

Weight/gravity - Downwards

Normal force - Upwards

keep in mind that the length of the arrows of both the weight and the normal forces have to be equal since these are vectors and are equal to each other to cancel each other out

the refractive index of rock salt with respect to water is 1.17. What will be the refractive index of water with respect to rock salt

Answers

Answer:

1.17

Explanation:

Given that,

The refractive index of rock salt with respect to water is 1.17.

We need to find the refractive index of water with respect to rock salt.

We know that,

Refractive index = (speed of light in air or vaccum)/( speed of light in that medium)

Refractive index of rock salt with respect to ice = (speed of light in ice)/(speed of light in rock salt)

The speed of light in ice = c/1.31

The speed of light in rock salt = c/1.54

Required refractive index,

[tex]\mu=\dfrac{1.54}{1.31}\\\\=1.17[/tex]

So, the refractive index of water with respect to rock salt is equal to 1.17.

Which two pairs of labeled stars (A-G) in the diagram have the same temperature?

Answers

Answer:

Please find the graph file in the question:

Explanation:

The D and E will be having the same temperature as they lie on the same line drown from the temp axis if it was drawn.

Which equation can correctly express the relation between force, acceleration, and mass?
a) F = m × a
b) F = m ÷ a
c) F = m − a
d) F = m + a

Answers

Hi there!

[tex]\large\boxed{\text{a) F = m * a}}[/tex]

According to Newton's Second Law, force is calculated with the formula:

F = m × a, where:

F = force (N)

m = mass (kg)

a = acceleration (m/s²)

Answer:

The Answer Is A . F= ma

Rerember Newtons second law of motion.

Describe the parts of the wheel and axle

Answers

Answer:

A simple machine that may be used the most often is called the wheel and axle. The wheel and axle has two basic parts: wheel and axle.  It has two circular objects which includes a larger disc and a small cylinder both joined at the center.

Explanation:

Explanation:

A simple machine that may be used the most often is called the wheel and axle. The wheel and axle has two basic parts: wheel and axle. They can be found everywhere. It has two circular objects which includes a larger disc and a small cylinder both joined at the center.The large disc is the wheel, and the small cylinder or rod is the axle. There may be two wheels attached to the axle.

There are two basic ways a wheel and axle can work together to help move things.

1. The Force is applied to the Wheel

For example, a screwdriver is an example of a wheel and axle. The handle is the wheel where the force is applied. It turns or spins and increases the force of the shaft or axle, which helps turn the screw.

Another example of force being applied to the wheel is when a doorknob is turned. The wheel (doorknob) is turned and the locking mechanism connected to the shaft turns and the door can then be opened.

2. The Force is applied to the Axle

A Ferris wheel is an example of force being applied to the axle. When the axle turns it results in the giant wheel turning. The wheel is much larger than the axle and covers more distance and area. A ceiling fan works the same way.

(02.06)plz help me 20 points ​

Answers

Answer:

the answer is 80 degrees

Explanation:

Figure 1.18 shows an oscillating pendulum. If the time taken for the pendulum to swing from A to C to B is 3 s, what is the period of the pendulum?​

Answers

Answer:

a

Explanation:

Does it take more energy to heat up water or less than
most other substances?

Answers

Answer:

Water takes more energy to heat up

Two loudspeakers are about 10 m apart in the front of a large classroom. If either speaker plays a pure tone at a single frequency of 400 Hz, the loudness seems pretty even as you wander around the room, and gradually decreases in volume as you move farther from the speaker. If both speakers then play the same tone together, what do you hear as you wander around the room?
A. The sound is louder but maintains the same relative spatial pattern of gradually decreasing volume as you move away from the speakers.
B. The pitch of the sound increases to 800 Hz, and the sound is louder but not twice as loud. It is louder closer to the speakers and gradually decreases as you move away from the speakers−except near the back wall, where a slight echo makes the sound louder.
C. As you move around the room, some areas seem to be dead spots with very little sound, whereas other spots seem to be louder than with only one speaker.
D. The sound is twice as loud−so loud that you cannot hear any difference as you move around the room.
E. At points equidistant from both speakers, the sound is twice as loud. In the rest of the room, the sound is the same as if a single speaker were playing.

Answers

Answer:

Chupapi moñaño mi bola mcnfjdjdkdjdkek

Explanation:

Two loudspeakers are about 10 m apart in the front of a large classroom. If either speaker plays a pure tone at a single frequency of 400 Hz, the loudness seems pretty even as you wander around the room, and gradually decreases in volume as you move farther from the speaker. If both speakers then play the same tone together, what do you hear as you wander around the room?

A. The sound is louder but maintains the same relative spatial pattern of gradually decreasing volume as you move away from the speakers.

B. The pitch of the sound increases to 800 Hz, and the sound is louder but not twice as loud. It is louder closer to the speakers and gradually decreases as you move away from the speakers−except near the back wall, where a slight echo makes the sound louder.

C. As you move around the room, some areas seem to be dead spots with very little sound, whereas other spots seem to be louder than with only one speaker.

D. The sound is twice as loud−so loud that you cannot hear any difference as you move around the room.

E. At points equidistant from both speakers, the sound is twice as loud. In the rest of the room, the sound is the same as if a single speaker were playing.sanro de la pinga Chupamelo

Calculate the momentum of a 4,800 kg car with a velocity of 25 m/s.

Answers

Answer:

p = 120000 kg · m/s

Explanation:

p = mv

p=momentum

m=mass

v=velocity

p = (4800)(25)

p= 120000 kg · m/s

Answer:

[tex]momentum = mass \times velocity \\ = 4800 \times 25 \\ = 120000 \: kgm {s}^{ - 1} [/tex]

A 8.6*10^ -6 C charge is places 1.00 m away from a 2.3 * 10 ^ - 4 * C charge. How large is the force of repulsion between the two charges?

Answers

Answer:

The force of repulsion is 17.802 N

Explanation:

Given;

magnitude of the first charge, q₁ = 8.6 x 10⁻⁶ C

magnitude of the second charge, q₂ = 2.3 x 10⁻⁴ C

distance between the two charges, r = 1 m

The force of repulsion is calculated using Coulomb's law;

[tex]F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{9\times 10^9 \times 8.6\times 10^{-6} \times 2.3\times 10^{-4}}{1^2} \\\\F = 17.802 \ N[/tex]

Therefore, the force of repulsion is 17.802 N

A capacitor with a capacitance of 50µf when connected to a battery of 400 V. The charge and energy stored on it is? a. 0.05 C and 5 J b. 0.05 C and 10 J c. 0.02 C and 4 J d. 0.08 C and 12 J

Answers

Answer:

c. 0.02 C and 4 J

Explanation:

Applying,

Q = CV................ Equation 1

Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.

From the question,

Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V

Substitute these values into equation 1

Q = (50×10⁻⁶)(400)

Q = 0.02 C.

Also Applying

E = CV²/2............. Equation 2

Where E = Energy stored.

Therefore,

E = (50×10⁻⁶ )(400²)/2

E = 4 J

Hence the right option is c. 0.02 C and 4 J

what is unit of pressure?why is it called a derived unit ? give reasons​

Answers

Answer:

Unit of pressure is pascal (Pa)

The SI-derived unit of measurement for pressure. The pascal is one newton (an SI-derived unit itself) per square meter.

Henry mixed salt and water together in a cup until he observed a clear solution. He measured the mass of the solution. Then he placed the cup outside for several sunny days during the summer. After a week, he observed that only solid salt remained in the cup and the mass had decreased. Henry concluded that a physical and chemical change occurred in this investigation.

Which statements correctly defend or dispute his conclusion?

He is correct. Dissolving salt in water is a physical change, but evaporating the water is a chemical change. Formation of a solid is evidence that a chemical change occurred.
He is correct. Evaporation is a physical change, but dissolving salt in water is a chemical change. The change in mass is evidence that a chemical change occurred.
He is incorrect. Dissolving salt in water and evaporation of the water are both physical changes. The reappearance of salt is evidence that the change was reversible by a physical change, so it could not be a chemical change.
He is incorrect. Dissolving salt in water and evaporation of the water are both chemical

Answers

Answer:

He is incorrect. Dissolving salt in water and evaporation of the water are both physical changes. The reappearance of salt is evidence that the change was reversible by a physical change, so it could not be a chemical change.

Answer:c

Explanation:

Sue wanted to know the mass of her earrings. She put them on a balance and added two 50 g weights, one 10 g weight, and three 1 g weights so that the balance was even. What was the mass of her earrings?  A. 120 g  B. 100 g  C. 140 g  D. 113 g

Answers

Answer:

113 grams

Explanation:

She put them on a balance and added two 50 g weights, one 10 g weight, and three 1 g weights so that the balance was even.

Two 50 g weights = 100 g

One 10 g weight = 10 g

Three 1 g weights = 3 g

Total weight = 100 g + 10 g + 3 g

= 113 g

So, the mass of her earrings is 113 grams.

Question 18 of 25
Which type of wave occurs where two different mediums meet?
A. Transverse
B. Electromagnetic
C. Longitudinal
D. Surface

Answers

The type of wave that occurs where two different mediums meet is known as a transverse wave. Hence, option A is correct.

What are waves?

Wave is the regular, systematic spread of disturbances from one location to another. The ripples that travel on water's surface are the most well-known, but waves also exist in sound, sight, and the movement of subatomic particles.

The disturbance periodically oscillates with a set frequency and wavelength in the simplest waves. In contrast to electromagnetic waves, which can travel in a vacuum, mechanical waves, like sound, need a medium to move through. The characteristics of the medium affect how a wave travels across it.

A transverse wave's crest and trough are the high and low points, respectively. The shock absorbers and rarefactions of longitudinal waves are comparable to the troughs and crests of transverse waves. The wavelength is the separation between subsequent crests or troughs.

To know more about Waves:

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I will be so thankful if u answer correctly!!​

Answers

Answer:

(a) 10N

Explanation:

The sketch of the two cases has been attached to this response.

Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.

In this case, a frictional force [tex]F_{r}[/tex] is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;

∑F = ma    -------------------(i)

Where;

∑F = effective force acting on the object (box)

m = mass of the object

a = acceleration of the object

∑F = F -  [tex]F_{r}[/tex]

m = 50kg

a = 0   [At constant velocity, acceleration is zero]

Substitute these values into equation (i) as follows;

F -  [tex]F_{r}[/tex] = m x a

F -  [tex]F_{r}[/tex] = 50 x 0

F -  [tex]F_{r}[/tex] = 0

F =  [tex]F_{r}[/tex]            -------------------(ii)

Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²

In this case, the same frictional force [tex]F_{r}[/tex] is opposing the movement of the box.

∑F = 1.5F -  [tex]F_{r}[/tex]

m = 50kg

a =  0.1m/s²

Substitute these values into equation (i) as follows;

1.5F -  [tex]F_{r}[/tex] = m x a

1.5F -  [tex]F_{r}[/tex] = 50 x 0.1

1.5F -  [tex]F_{r}[/tex] = 5            ---------------------(iii)

Substitute [tex]F_{r}[/tex] = F from equation (ii) into equation (iii) as follows;

1.5F - F = 5            

0.5F = 5            

F = 5 / 0.5

F = 10N

Therefore, the value of F is 10N

6. Which of the following graphs correctly demonstrates the relationship between the
electromagnetic force and distance between charges?

Answers

Answer:

But where are the graphs

Coulomb's Law describes the relationship between electromagnetic force and the distance between charges.

What is the relationship between electromagnetic force and the distance between charges?

The relationship between electromagnetic force and the distance between charges is described by Coulomb's Law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, Coulomb's Law is expressed as:

F = k  (q₁x q₂) / r²

where F is the electromagnetic force, q1 and q2 are the charges, r is the distance between the charges, and k is the Coulomb constant.

This means that as the distance between two charges increases, the force between them decreases, and as the distance decreases, the force increases. The relationship between force and distance is inverse square, which means that the force decreases rapidly as the distance increases.

To know more about electric charges follow

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Which of the following measurements require the use of scientific notation?
A. The length of your hair
B. The distance to the sun
C. The width of your hand
D. Distance to Indianapolis in miles

Answers

Answer: B. The distance of the sun

Which of the following statements is false?
A. Weight is a vector quantity

B. Weight is measured in newtons, N

C. The weight of an object is the same on the Earth and the moon

Answers

Answer:

C

Explanation:

The weight of an object is 1/6 of the earth's weight in the moon

For example if your weight is 54 kg on earth,

Weight on moon=1/6 of 54=1/6 x 54 = 9kg

Answer:

C. The weight of an object is the same on the Earth and the moon

Explanation:

The mass of an object is ⅙ of earth's weight on moon. So, option (C) is the false statement.

Which of the motion variables is the same in both thex and y axis?
A) velocity
B) acceleration
C) time
D) displacement

Answers

Answer:

Acceleration (b) not sure tho

Explanation:

Magnitude of force F experienced by a certain object moving with speed v is given by F=Kv race to 2, where K is constant. Find the dimension of K

Answers

Answer:

MT⁻¹

Explanation:

Given equation:

F = Kv        --------------(i)

Where;

F = Magnitude of force experienced by the object

v = speed at which the object is moving

K = constant.

To get the dimension of K, follow the following steps:

i. Make K subject of the formula in equation(i)

K = F / v

ii. Get the dimension of force F and speed v

Dimension of force = MLT⁻²

Dimension of velocity = LT⁻¹

iii. Substitute these dimensions into the result of (i) above;

K = MLT⁻² / LT⁻¹

iv. Simplify the result in (iii)

First, the Ls in the numerator and denominator will cancel out

K = MT⁻² / T⁻¹

Next, the Ts will be expressed as follows

K = MT⁻² x T¹           [using the law of indices]

K = MT⁻²⁺¹

K = MT⁻¹

Therefore, the dimension of K is MT⁻¹

find out the speed of the car from home to school in meters per second.
Time - 15 mins
Distance - 8 km​

Answers

1.875

Explanation:

speed = distance moved ÷ time taken

The atomic mass of calcium is 40.08. What is the mass of 6.02 × 1023 atoms of calcium?

Answers

Answer:

It is 40.08 g

Explanation:

[tex]6.02 \times 10 {}^{23} \: atoms = 1 \: mole = 40.08 \: g[/tex]

IF YOU ANSWER I WILL MARK YOU BRAINLIEST
PPLLSSS

Answers

Answer:

the mushrooms

mushrooms are decomposers because they break down organic material around them.


A person riding a bike has a total mass of 95 kg, if the person bike system is moving at a velocity of 5 m/s, how much kinetic energy is there?

Answers

Answer:

Kinetic energy is 1187.5J

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Leighd. a concerned neighbor Enunciado del ejercicio n 1Se lanza un cuerpo verticalmente hacia abajo con una velocidad inicial de 7 m/s.a) Cul ser su velocidad luego de haber descendido 3 s?b) Qu distancia habr descendido en esos 3 s?c) Cul ser su velocidad despus de haber descendido 14 m?d) Si el cuerpo se lanz desde una altura de 200 m, en cunto tiempo alcanzar el suelo?e) Con qu velocidad lo har?Usar g = 10 m/sDesarrolloDatos:v0 = 7 m/st = 3 sy = 200 mh = 14 mFrmulas:(1) vf = v0 + gt(2) y = v0t + gt(3) vf - v0 = 2ghEnunciado del ejercicio n 2Se lanza un cuerpo verticalmente hacia arriba con una velocidad inicial de 100 m/s, luego de 4 s de efectuado el lanzamiento su velocidad es de 60 m/s.a) Cul es la altura mxima alcanzada?b) En qu tiempo recorre el mvil esa distancia?c) Cunto tarda en volver al punto de partida desde que se lo lanzo?d) Cunto tarda en alcanzar alturas de 300 m y 600 m?Usar g = 10 m/sDesarrolloDatos:v0 = 100 m/svf = 60 m/st = 4 sy1 = 300 my2 = 600 mFrmulas:(1) vf = v0 + gt(2) y = v0t + gt(3) vf - v0 = 2ghEnunciado del ejercicio n 3Un observador situado a 40 m de altura ve pasar un cuerpo hacia arriba con una cierta velocidad y al cabo de 10 s lo ve pasar hacia abajo, con una velocidad igual en mdulo pero de distinto sentido.a) Cul fue la velocidad inicial del mvil?b) Cul fue la altura mxima alcanzada?Usar g = 10 m/sDesarrolloDatos:t = 10 sy = 40 mFrmulas:(1) vf = v0 + gt(2) y = y0 + v0t + gt(3) vf - v0 = 2ghEnunciado del ejercicio n 4Desde un 5 piso de un edificio se arroja una piedra verticalmente hacia arriba con una velocidad de 90 km/h, cunto tardar en llegar a la altura mxima?Usar g = 10 m/sDesarrolloDatos:v0 = 90 km/hv0 = 25 m/sFrmulas:(1) vf = v0 + gt(2) y = v0t + gt(3) vf - v0 = 2ghEnunciado del ejercicio n 5Un auto choca a 60 km/h contra una pared slida, desde qu altura habra que dejarlo caer para producir el mismo efecto?Usar g = 10 m/sDesarrolloDatos:vf = 60 km/hvf = 16,67 m/sv0 = 0 m/sFrmulas:(1) vf = v0 + gt(2) y = v0t + gt(3) vf - v0 = 2ghEnunciado del ejercicio n 6Se lanza una pelota hacia arriba y se recoge a los 2 s, calcular:a) Con qu velocidad fue lanzada?b) Qu altura alcanz?Usar g = 10 m/sDesarrolloDatos:t = 2 sFrmulas:(1) vf = v0 + gt(2) y = v0t + gt(3) vf - v0 = 2ghEnunciado del ejercicio n 7Se lanza una pelota de tenis hacia abajo desde una torre con una velocidad de 5 m/s.a) Qu velocidad tendr la pelota al cabo de 7 s?b) Qu espacio habr recorrido en ese tiempo?Usar g = 10 m/sDesarrolloDatos:v0 = 5 m/st = 7 sFrmulas:(1) vf = v0 + gt(2) y = v0t + gt(3) vf - v0 = 2ghEnunciado del ejercicio n 8Se lanza un cuerpo verticalmente hacia arriba con una velocidad de 60 km/h, se desea saber la altura mxima alcanzada, la velocidad que posee al cabo de 4 s y 30 s, la altura alcanzada a los 8 s, el tiempo total que se encuentra en el aire.DesarrolloDatos:v0 = 60 km/h = (60 km/h)(1.000 m/km)(1 h/3.600 s) = 16,67 m/st1 = 4 st2 = 30 st3 = 8 sUsar g = 10 m/sFrmulas:(1) vf = v0 + gt(2) y = v0t + gt(3) vf - v0 = 2ghEnunciado del ejercicio n 9Se dispara verticalmente hacia arriba un objeto desde una altura de 60 m y se observa que emplea 10 s en llegar al suelo. 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