a graduated cylinder.measures 15.3 mL. Convert this measurement to DaL
Answers:

A. 0.0153
B. 0.00153
C. 0.000153
D. 0.153

Answers

Answer 1

Answer:

0.000153DaL  

Explanation:

We have been given:

         15.3mL to convert to DaL

DaL is a unit of volume which indicates a decaliter.

 This implies that;

             1 Da L  = 1 x 10²L

So:

               1 mL  = 1 x 10⁻³L

       So 15.3mL will give 15.3 x 10⁻³L

So;

           1 x 10²L   =  1 DaL  

      15.3 x 10⁻³L  will give [tex]\frac{15.3 x 10^{-3} }{1 x 10^{2} }[/tex]   = 15.3 x 10⁻⁵DaL

Therefore, this is 0.000153DaL  

           


Related Questions

A person pushes down on a lever with a force of 100 N. At the other end of the lever, a force of 200 N lifts a heavy object. What is the mechanical advantage of the lever?

A. 1/2, because the object will be lifted half the distance
B. -1, because the direction changes
C. 2, because the output force is twice the input force
D. 1, because the same amount of work is done​

Answers

Answer:

Explanation:

C 200÷100=2

Output ÷ Input= MA

The mass of 60 paper clips is 18.0 grams. What is the mass of one paper clip?

Answers

Answer:

3.333333333333333333333333333333333333333

Explanation:

3.3333333333333333333333333333333333

Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.

a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?

Answers

Answer:

a) ω₁ = ω₂ = 3.7 rad/sec

b) Δθ₁ = Δθ₂ = 18.5 rad

c) d₁ = 14.5 m  d₂ = 57.5 m

d) Fc1 = 273.9 N Fc2 = 1069.8 N

e) The boy near the outer edge.

Explanation:

a)

Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.However, linear speeds of points at different distances from  the center, are different.Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:

       [tex]v = \omega*r (1)[/tex]

Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:

       [tex]\omega = \frac{v_{out} }{r_{out} } = \frac{11.5m/s}{3.14m} = 3.7 rad/sec (2)[/tex]

As we have already said, ωout = ωin = 3.7 rad/sec

b)

Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:[tex]\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)[/tex]

        ⇒  Δθ₁ = Δθ₂ = 18.5 rad.

c)

The linear distance traveled by each child, will be related with the linear speed of them.Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:

       [tex]v_{inn} = \omega * r_{inn} = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)[/tex]

      vout is a given of the problem ⇒ vout = 11. 5 m/s

Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:

       [tex]d_{inn} = v_{inn} * t = 2.9m/s* 5.0 s = 14.5 m (5)[/tex]

      [tex]d_{out} = v_{out} * t = 11.5 m/s* 5.0 s = 57.5 m (6)[/tex]

d)

The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:

       [tex]F_{c} = m*\frac{v^{2}}{r} (7)[/tex]

Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:

      [tex]F_{cin} = m*\frac{v_{in}^{2}}{r_{in}} = 25.4 kg* \frac{(2.9m/s)^{2} }{0.78m} = 273.9 N (8)[/tex]

In the same way, we get Fcout (the force on the boy near the outer edge):

      [tex]F_{cout} = m*\frac{v_{out}^{2}}{r_{out}} = 25.4 kg* \frac{(11.5m/s)^{2} }{3.14m} = 1069.8 N (9)[/tex]

e)

The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.The maximum friction force is given by the product of the coefficient of static friction times the normal force.Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.As both boys have the same mass, the normal force is also equal.This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:[tex]F_{frs} = \mu_{s} * m* g (10)[/tex]If this force is greater than the centripetal force, the boy will be able to hold on.So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.

Which of the following is a vector quantity?
speed
distance
acceleration

Answers

c) Acceleration
For example, if person A is the observer, then he/she can predict the distance or speed or time by simply looking at it. But at the same time, the observer can't predict the acceleration of a moving object as he/she can't determine the direction of the object. So, without a direction, the acceleration can't be predicted. For example, displacement, force, velocity, momentum, etc. are vector quantities.
◙ But, Distance, speed and time are only specified with their magnitude. For example, work, volume, density, mass, etc. don't need a direction for their representation.
So, (c)Acceleration is the answer.

is 0.8 kilograms bigger then 80 grams

Answers

Answer:

Yes

Explanation:

0.8 kilograms is equal to 800 grams

Answer:

Yes, 0.8 kilograms is greater than 80 grams

Explanation:

0.8 kilograms is equal to 800 grams and 80 grams is equal to 0.08 kilogrmas.

Sorry if I'm wrong, correct me.

take a picture of an object in your house, describe the
energy stores and transfers that happen with it. You can be as imaginative as you wish
with the object (choose something unusual), but the stores you identify and transfers
that happen must be real.


pls give me ideas of what to take a photo of for this I'm really stuck :(​

Answers

A charger or a battery

A vertical piston-cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 10 kg and a diameter for 14 cm. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will doublethe pressure of the gas inside the cylinder.

Answers

Answer:

the local atmospheric pressure is  93.63 kPa

the mass of the weights is 156.9 kg

Explanation:

Given that;

Initial pressure of gas = 100 kPa

mass of piston = 10 kg and diameter = 14 cm = 0.14 m

g = 9.81 m/s²

Now,

P_gas = P_atm + P_piston

100 = P_atm + P_piston --------- let this equation 1

P_piston = M_piston × g / A = (10 × 9.81) / π/4×d²

P_piston = 98.1 / (π/4×( 0.14 )²)

P_piston = 98.1 / 0.01539 = 6374,269 Pa = 6.37 kPa

now, from equation 1

100 = P_atm + P_piston

we substitute

100 = P_atm + 6.37

P_atm = 100 - 6.37

P_atm = 93.63 kPa

Therefore, the local atmospheric pressure is  93.63 kPa

Now for pressure of the gas in the cylinder ⇒ 2×initial pressure

Pgas_2 = 2 × 100 = 200 kPa

Pgas_2 = P_atm + P_piston + P_weight

Pgas_2 =  P_gas  + P_weight

we substitute

200 kPa =  100 kPa  + P_weight

P_weight =  200 kPa -  100 kPa

P_weight = 100 kPa =  100,000 Pa

Also;

P_weight = M×g / A

100,000 Pa = ( M × 9.81 ) / (π/4 × (0.14)²)

100,000 × 0.01539 = M × 9.81

1539 = M × 9.81

M = 1539 / 9.81

M = 156.9 kg

Therefore, the mass of the weights is 156.9 kg

please help asap!!

Explain the movement of a roller coaster in terms of potential and kinetic energy? When are these energies thegreatest? Smallest? Are they ever the same?​

Answers

Answer:

Potential energy: Greatest at the top of the hill

Kinetic energy: Greatest at the bottom of the hill

The two meet at some point on the way down!

Explanation:

Potential energy is energy that represents an object's potential for motion. Kinetic energy is that object's energy during motion. They're two sides of the same coin, and in fact, their sum gets a special name: mechanical energy. Potential energy builds up in reaction to working against certain forces - in the case of the roller coaster, that primary force is gravity. Gravity exerts a downward force on the roller coaster, and it takes work to pull it up the hill.

When it reaches the peak, the coasters potential energy is at its highest, and the moment it crests over the hill and begins its descent, that gravitational potential energy starts converting into kinetic energy: the coaster starts accelarating down the track, and the potential energy decreases at the same rate that the kinetic energy increases.

At the bottom of the hill, all of that potential energy has become kinetic energy, and the coaster zooms along the track, hopefully not giving too many riders nausea

why do players choose to follow the unconventional route of kicking down the middle

Answers

Answer:

My biggest reason is to make it a habit. Even if the ball goes into the endzone it is a live ball and the offensive players must down the ball. Don't leave any room for "I thought he downed it" or "I thought I heard the whistle" just run to the ball always.

If the players slow down and the returner takes it out of the end zone it could be a big return. Players are on a full sprint for 40+ yards sometimes and instead of breaking down, they choose to contine through the goal line to slow down at a decreased rate (possibly limiting a muscle pull injury).

When the bowling ball has fallen halfway down the building (height = 20 m), it has a speed of 19.8 m/s.
How much potential energy does the bowling ball have?
How much kinetic energy does the bowling ball have?
How much total energy (potential + kinetic) does the bowling ball have?
Of the bowling ball’s total energy, is more in the form of potential or kinetic energy?

Answers

Answer:

I think the answer is 19.8 potential energy

Explanation:

NONE.

What Coulombs discovered almost 300
years ago

Answers

Answer:

ummm hehe this is my time to shine

Explanation:

  MERICIA!!!!!!!!!!!!!!!!!!!!!!!

Christopher Columbus discovered
America

what is the direction of the third force that would cause the box to remain stationary on the ramp ?

Answers

An arrow pointing from the bottom of the ramp to the top, I assume it would be friction.

The arrow on the bottom pointing down due to friction the bow would not be able to go down the ramp

A ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds in the opposite direction with a speed equal to one-fourth its original speed. what is the mass of the second ball?

Answers

See image for the answer.

When a ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds in the opposite direction with a speed equal to one-fourth its original speed, then mass of the second ball having v/3 is velocity after collision is 9m/4.

What is momentum ?

Momentum is defined as mass times velocity of body. it is denoted by p and its SI unit is Kg.m/s. It has both magnitude and direction. it is a vector quantity.  it tells about the moment of the body. it is denoted by p and expressed in kg.m/s. mathematically it is written as p = mv. A body having zero velocity or zero mass has zero momentum. its dimensions is [M¹ L¹ T⁻¹]. Momentum is conserved throughout the motion.

initial momentum = final momentum

Given,

mass of first body m₁ = m

initial velocity of first body = v₁' = v

final velocity of first body = v₁'' =v/4

mass of second body m₂ = ?

initial velocity of second body = v₂' = 0

final velocity of second body = v₂'' = v/3

According to conservation of momentum,

initial momentum = final momentum

m₁v₁' + m₂v₂' = m₁v₁'' + m₂v₂''

putting al above values

m₁v + 0 = m₁v/4 + m₂v/3

m₁v - m₁v/4 = m₂v/3

m (1 - 1/4)v = m₂v/3

3m/4 = m₂/3

m₂ = 9m/4

Hence mass of the second body is 9m/4.

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When researchers replicate a study, they are seeking to __________.
A.
prove that the hypothesis upon which the study was founded is untestable
B.
develop a new hypothesis
C.
change the study to provide new results
D.
support or reject the hypothesis upon which the study was founded



Please select the best answer from the choices provided


A
B
C
D

Answers

The best answer I think is D) it’s the best one

Answer:

D

Explanation:

right edge 2022

A stone is dropped from the top of a high cliff with zero initial velocity. In which system is the net momentum zero as the stone falls freely

Answers

Answer:

A system that includes the stone and the earth.

Explanation:

If the system of being dropped from the height of the cliff consists of just the stone alone, then it means that its momentum will certainly undergo changes as it falls freely. However, If the system is now expanded to include not only the stone but also the Earth, then it implies that the momentum of the stone which is in the downward direction will be equal and opposite to the momentum of the Earth in the upwards direction towards the stone. Therefore, the momentum will cancel out and net momentum will be zero.

A system of stone and earth can result to a net zero momentum.

Conservation of linear momentum

The principle of conservation of linear momentum states that the sum of the initial momentum is equal to the sum of final momentum.

[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2 v_2[/tex]

A system that consists a linear system of stone and earth can result to a net zero momentum.

Thus, a system of stone and earth can result to a net zero momentum.

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Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0N, the second child exerts a force of 90.0 N, friction is 12.0 N, and the most of the third child plus wagon is 23.0 kga)what is the system of interest if the acceleration of the child in the wagon is to be calculated

Answers

Answer:

Explanation:

75 N and 90 N are acting in opposite direction so net force = 90 - 75 = 15 N .

Friction force will act in the direction opposite to the direction of net force .

So friction force will act in the direction in which 75 N is acting .

Total force acting in the direction of 75 =  75 + 12 = 87 N

Net force acing on the third child = 90 - 87 = 3 N  

Its direction will be that in the direction of 90 N .

A 500 kg wrecking ball is knocking down a wall. When it is pulled back to its highest point, it is at a height of 6.2 m. When it hits the wall, it is moving at 3.1 m/s. How high is the wrecking ball when it hits the wall? (Show your work and follow all of the steps of the GUESS method. Check your answer after you submit the form - it's in the feedback for this question.) |​

Answers

The first is that you have the time to write a letter ✉️ and a lot more of the same, and the like are the same time as a result of the most popular connection and a half ago I was in a way ↕️ and a few other people are paying for new cars at the time of his death own or manage Hotel in a way ↕️ and the second half of the season ❄️ and a half ago I had a lot of people the first time I have to admit I have to say I am a little more time with my own personal information on how the hell out of the box house and a few other people and the second one of the most popular and a half ago I had to do it again in the first.

A student is driving through a mountainous region where the road is at some times flat, at some times inclined upward, and at some time inclined downward. The student maintains a speed of 20 m/s on the roadway, but is required to make an emergency stop on the three sepearte occasions. On levels roadway, it takes 25 m to stop. On a downward-sloping roadway, it takes 40 m to stop. On an upward-sloping roadway, it takes 18 m to stop. Explain why the stopping distances are different. (Focus answer using work and energy, other concepts may be used as well but be sure work and energy are included.)

Answers

Answer:

Explanation:

It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .

At levelled road , for stoppage

Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .

At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .

At upward inclined  road , for stoppage

Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force  = (friction force + gravitational force ) x displacement .

Hence displacement is less .

At downward slopping road ,  friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle  also .

At downward inclined  road , for stoppage

Kinetic energy of vehicle + work done by gravitational force  = Work done by frictional force = friction force  x displacement .

Hence displacement is more .

Hence displacement is more in the downward slopping.

What is Displacement?

Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.

It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .

At levelled road , for stoppage

Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .

At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .

At upward inclined  road , for stoppage

Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force  = (friction force + gravitational force ) x displacement .

Hence displacement is less .

At downward slopping road ,  friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle  also .

At downward inclined  road , for stoppage

Kinetic energy of vehicle + work done by gravitational force  = Work done by frictional force = friction force  x displacement .

Hence displacement is more in the downward slopping.

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Two metal bricks are held off the edge of a balcony from the same height above the ground. The bricks are the same size but one is made of Titanium (density of 4.5 g/cm%) and one is made of Lead (density of 11.3 g/cm3) so the Lead is about twice as heavy as the Titanium. The time it takes the bricks to reach the ground will be:________.
a. less but not necessarily half as long for the heavier brick
b. about half as long for the lighter brick
c. less but not necessarily half as long for the lighter brick
d. about half as long for the heavier brick
e. about the same time for both bricks

Answers

Answer:

e.

Explanation:

Assuming that the air resistance is neglectable, both bricks are only accelerated by gravity, which produces a constant acceleration on both bricks, which is the same, according  Newton's 2nd Law, as we can see below:[tex]F_{g} = m*g = m*a (1)[/tex]⇒a = g = 9.8m/s² (pointing downward)Since acceleration is constant, if both fall from the same height, we can apply the following kinematic equation:

       [tex]\Delta y = v_{o} * t - \frac{1}{2} *g*t^{2} (2)[/tex]

Since both bricks are held off the edge, the initial speed is zero, so (2) reduces to the following equation:

        [tex]h =\frac{1}{2} *g*t^{2} (3)[/tex]

Since h (the height of the balcony) is the same, we conclude that both bricks hit ground at exactly the same time.If the air resistance is not negligible, due both bricks have zero initial speed, and have the same shape, they will be affected by the drag force in similar way, so they will reach the ground at approximately the same time.

Two steamrollers begin 105 mm apart and head toward each other, each at a constant speed of 1.20 m/s. At the same instant, a fly that travels at a constant speed of 2.50 m/s starts from the front roller of the southbound steamroller and flies to the front roller of the northbound one, then turns around and flies to the front roller of the southbound once again, and continues in this way until it is crushed between the steamrollers in a collision.

Required:
What distance does the fly travel?

Answers

Answer: 109.4 mm

Explanation: Distance is a scalar quantity and it is the measure of how much path there are between two locations. It can be calculated as the product of velocity and time:  d = vt

The separation between the two steamrollers is 105 mm or 0.105 m. They collide to each other at the middle of the separation:

location of collision = [tex]\frac{0.105}{2}[/tex] = 0.0525 m

To reach that point, both steamrollers will have spent

[tex]v=\frac{\Delta x}{t}[/tex]

[tex]t=\frac{\Delta x}{v}[/tex]

[tex]t=\frac{0.0525}{1.2}[/tex]

t = 0.04375 s

The fly is travelling with speed of 2.5 m/s. So, at t = 0.04375 s:

d = 2.5*0.04375

d = 0.109375 m

Until it is crushed, the fly will have traveled 109.4 mm.

When a moving object collides with an object that isn't moving, what happens to the kinetic energy of each object?

Answers

The kinetic energy of the object being hit will increase and the the potential energy will decrease. The kinetic energy of the object that collided with the other said object will increase and there will be not potential energy. This is my personal knowledge on the matter.

Although the internet states otherwise:

In the extreme case, multiple objects collide, stick together, and remain motionless after the collision. Since the objects are all motionless after the collision, the final kinetic energy is also zero; the loss of kinetic energy is a maximum.

All the objects are motionless, so kinetic energy of each object is zero after the collision.

What is Kinetic Energy?

The kinetic energy of an object is defined as the energy which is  possesses due to its motion. It is the work required to accelerate a body of a given mass from rest to its stated velocity. This energy is gained during its acceleration, the body maintains the kinetic energy as long as its momentum does not change.

Kinetic Energy can be expressed as

[tex]K.E.=[/tex] [tex]1/2 mv^2[/tex]

Where, m is the mass of the object

v is the velocity.

It is expressed in joules (J).

After the collision all the objects are at rest, therefore, the final kinetic energy is also zero which shows maximum loss of kinetic energy. Such collisions are called perfectly inelastic.

Thus, all the objects are motionless, so kinetic energy of each object is zero after the collision.

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As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock. As the waves recede, they carry the sediment away. In this scenario, which process represents weathering, and which process represents erosion?

Answers

Answer:

WEATHERING is represented by the scenario (As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock).

Erosion is represented by the scenario (As the waves recede, they carry the sediment away).

Explanation:

A wave is a disturbance which travels through a medium and transfers energy from one point to another. When wind blows over a water body like the ocean, ocean waves are formed. As the generated energy from the wind is transported through the water by the waves, the can hit against rocks on the shores leading to its break down with time. WEATHERING occurs when tiny bit of sediments from rocks are loosened due to the impact of ocean waves.

Erosion can be described as the wearing away of the earth's surface due to the impact of wind, rainfall ( water) or waves. There are different types of erosion which is classified according it's cause of formation.

Wave erosion occurs when sediments such as sand, shell and other materials are carried to the shoreline by ocean waves. This erodes the shore over time as the sediments act like sandpapers.

A car is traveling on a straight road at a constant 35 m/sm/s, which is faster than the speed limit. Just as the car passes a police motorcycle that is stopped at the side of the road, the motorcycle accelerates forward in pursuit. The motorcycle passes the car 13.5 ss after starting from rest. What is the acceleration of the motorcycle (assumed to be constant)

Answers

Answer:

2.59m/s

Explanation:

Using the equation of motion

v = u+at

v is the final velocity = 35ms

u is the initially velocity = 9m/s

t is the time = 13.5s

a is the acceleration

Substitute into the formula

35 = 0+13.5a

a = 35/13.5

a = 2.59m/s²

Hence the acceleration of the motorcycle is 2.59m/s

How much kinetic energy does a 0.104 kg hamster have if it is moving at 24.0 m/s?

Answers

Answer:

30J

Explanation:

Given parameters:

Mass of hamster  = 0.104kg

Velocity  = 24m/s

Unknown:

Kinetic energy  = ?

Solution:

Kinetic energy is the energy due to the motion of a body. It is mathematically derived by;

  Kinetic energy  = [tex]\frac{1}{2}[/tex] m v²  

m is the mass

v is the velocity

  Kinetic energy  = [tex]\frac{1}{2}[/tex] x 0.104 x 24²   = 30J

A neutral metal bob is hanging on the bottom of a pendulum that is 15 cm long. A charged balloon is held near the metal bob and the pendulum is pulled up to a vertical angle of 20-deg. If the mass of the metal bob is 0.025kg, what is the charge on the balloon.

Answers

Answer:

Explanation:

See the figure attached

F is electrostatic force .

T cos20 = mg

T sin20 = F

Tan20 = F / mg

F = mg tan 20 = .025 x 9.8 tan20

= .09 N

Distance between bob and balloon

= 15 sin20 = 5.1 cm = .051 m

If q be the charge on balloon

F = 9 x 10⁹ x q² / .051²

= 3460 x 10⁹ q² = .09

q² =  26 x 10⁻⁶ x 10⁻⁹

q = 16.12 x 10⁻⁸ C .

One disadvantage to experimental research is that experimental conditions do not always reflect reality.


Please select the best answer from the choices provided

T
F

Answers

Answer:

It's true I took the test on Edge.

Explanation:

Answer:

True

Explanation:

Got it right on edg

If there is "waste" energy, does the Law of Conservation of Energy still apply? ​

Answers

Explanation:

Yes, the law of conservation of energy still applies even if there is waste energy.

The waste energy are the transformation products of energy from one form to another.

According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".

But of then times, energy is lost as heat or sound within a system.

If we take into account these waste energy, we can see that energy is indeed conserved. The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.

This table shows the mass and volume of four different objects.

A two-column table with 4 rows. The first column titled objects has entries W, X, Y, Z. The second column titled Measurements has entries Mass: 16 grams Volume: 84 centimeters cubed in the first cell, Mass: 12 grams Volume: 5 centimeters cubed in the second cell, Mass: 4 grams Volume: 6 centimeters cubed in the third cell, Mass: 408 grams Volume: 216 centimeters cubed in the fourth cell.

Which ranks the objects from most to least dense?

Answers

Answer:

Here its right but its also better than Barney's response

Explanation:

W, Y, Z, X or C

Answer:

W, Y, Z, X

Explanation:

In which number are the zeros not significant?
100.0
O 0.0003
O 4.00005
O 1.0004

Answers

Answer:

0.0003

Explanation:

In the rules of Sig Figs, all zeros before with decimals are not sigificant. I.E. 0.00000000000000009. Despite how many 0's there are, only the 9 is significant. Zeros before a number is not significant. In 100, only the one is signficant in 100. with a dot at the end, the one and the two zeros are significant. hope this helps.

Answers:

the second option

Explanation:

Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the battery connected).

a. Capacitance
b. Charge on the plates
c. Voltage across the plates
d. Net electric field between the plates
e. Energy stored in the capacitor

Answers

Answer:

a. Capacitance

b. Charge on the plates  

e. Energy stored in the capacitor

Explanation:

Let A be the area of the capacitor plate

The capacitance of a capacitor is given as;

[tex]C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\[/tex]

where;

V is the potential difference between the plates

The charge on the plates is given as;

[tex]Q = \frac{V\epsilon _0 A}{d}[/tex]

The energy stored in the capacitor is given as;

[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2[/tex]

Thus, the physical variables listed that will change include;

a. Capacitance

b. Charge on the plates  

e. Energy stored in the capacitor

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