1. Which of the following does not affect the resistance of a wire?
a) Length
b) Temperature
c) Usage time
d) Cross-sectional area
2. If a 12V battery is passing current through a resistor with a current of 2A, what is the value of the resistor?
a 24resistance
b) 14resistance
c) 10resistance
d) 6resistance
3. Describe the differences between series and parallel circuits.
4. A circuit contains resistors of 8resistance and 4resistance,what is combined resistance if the resistors are combined:
a) In series
b) In parallel
5. A 0.5A current is passing across three resistors of 8resistance, 4resistance and 12resistance that are linked in series.
What is the potential difference of the circuit?
6. Wire A has a resistance of 24resistance. If wire B is double the length and has a diameter four times as large as wire A, what is the resistance of wire B?

Answers

Answer 1
The answer right there is b

Related Questions

a group of campers decide to take a canoe trip down the river. the campers traveled 25 miles in five hours. at what average speed did the group of campers travel down the river?

Answers

The average speed of the campers down the river was 5 miles per hour.

Total distance traveled by the campers = 25 miles

Total time taken by the campers = 5 hours

Speed is defined as the rate of change in distance or altitude reached. It is a time-based quantity. In the given question, a camper's average speed can be calculated by dividing the total distance travelled by the time taken.

Calculating the average speed of the campers -

Average speed = Total distance / Time taken

Substituting the values -

= 25 miles / 5 hours

= 5

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A plate of iron at 20 °C has shown in the figure below. If the temperature
raised to 100 °C and the coefficient of linear expansion of iron is 1.1 x 10-7 o
1, then what is the final area of the plate?
(5
2 m
2 m

Answers

The final area of the plate is 4.0000352 [tex]m^2[/tex] if the temperature raised to 100 °C and the coefficient of linear expansion of iron is 1.1 x 10-7.

Expecting that the plate of iron is rectangular, we can involve the recipe for warm extension of solids to compute the last region of the plate. The equation for direct warm development is given by ΔL = αLΔT, where ΔL is the adjustment of length, α is the coefficient of straight extension, L is the first length, and ΔT is the adjustment of temperature.

Since the region of the plate is given by A = L*W, where L is the length and W is the width, we can involve the equation for straight warm extension to compute the adjustment of length of the plate and afterward use it to compute the last region.

ΔL = αLΔT = [tex](1.1 x 10^-7 m/oC)(2 m)(80 oC) = 1.76 x 10^-5 m[/tex]

The last length of the plate is L + ΔL = 2 m + 1.76 x [tex]10^-5[/tex] m = 2.0000176 m (approx.)

The last width of the plate is thought to be unaltered as it isn't impacted by the adjustment of temperature.

Thusly, the last region of the plate is A = L*W = (2.0000176 m)(2 m) = 4.0000352 [tex]m^2[/tex] (approx.)

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What is the physics of a domino po pls in tagalog po asap langg​

Answers

the physics of a domino chain shows how momentum and kinetic energy can combine to produce an amazing display of dominoes falling.

First of all, momentum is a vector that denotes an object's mass and speed. When a domino chain is pushed over, the initial domino that was knocked over transfers momentum to the next domino, which then transfers momentum to the next, and so on. Each domino does not have to travel at the same pace as the others as long as they are in time with one another since the group of dominoes offers a coordinated transfer of momentum.

The second definition of kinetic energy is the energy that an item has as a result of motion. Kinetic energy increases each domino's speed and the power necessary to topple the subsequent domino in a chain of dominoes as momentum is transferred from one domino to the next. The largest domino, which frequently topples onto a designated catch or platform that intensifies the effect for spectators, is where the chain's greatest amount of kinetic energy is stored when it reaches its conclusion.

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You will need to know that Force (N) is equal to mass (kg) multiplied by acceleration (m/s2) for this problem. A fearless space explorer has discovered a new planet with a frictionless surface! He pushes a large crate with a mass of 220kg a distance of 5.3 km, as he does so, it accelerates at a rate of 2m/s2. How much work has our intrepid hero done?

Answers

Our intrepid hero has done 2332 kJ of work pushing the crate on the frictionless surface of the newly discovered planet.

The work done by the space traveler can be determined utilizing the recipe W = F x d, where W is work, F is power, and d is distance. To find the power, we can utilize the recipe F = m x a, where m is mass and an is speed increase. Connecting the given qualities, we get F = 220 kg x 2 m/s^2 = 440 N.

Presently we can compute the work done by increasing the power by the distance: W = 440 N x 5.3 km = 2332 kJ. Accordingly, our fearless legend has done 2332 kJ of work pushing the container on the frictionless surface of the newfound planet.

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during SHM an object has_____________at mean position
(a)P.E=0,K.E=0
(b)P.E=K.E
(c)P.E=MAX,K.E=0
(d)P.E=0,K.E=MAX

Answers

During Simple Harmonic Motion (SHM), an object at the mean position has potential energy (P.E)=0 and kinetic energy (K.E)=MAX. So, the correct option is (d).

In Simple Harmonic Motion (SHM), an object oscillates about a mean position, with the motion characterized by a restoring force proportional to its displacement from the mean position.

When the object is at the mean position, it has maximum kinetic energy (K.E) because it is at its maximum velocity, and it has zero potential energy (P.E) since it is not displaced from the equilibrium position.

As the object moves further from the mean position, its P.E increases, and K.E decreases. The correct answer to the question is option (d), where P.E=0 and K.E=MAX at the mean position.

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3.
A steel container (the coefficient of linear expansion = 10-5 °C-1) with a volume
of 6 liters filled with acetone (the coefficient of volume expansion = 1.5 x 10-
3 °C-1). If the container and acetone are heated from 0 °C to 40 °C, what is the
volume of spilled acetone?
(6)

Answers

The amount of acetone that was spilt is around 0.36 liters, or 360 milliliters.

How to determine volume?

To solve this problem, use the formula for volumetric thermal expansion:

ΔV = V₀βΔT

Where:

ΔV = change in volume

V₀ = initial volume

β = coefficient of volumetric expansion

ΔT = change in temperature

Also use the formula for linear thermal expansion to find the change in length of the container:

ΔL = L₀αΔT

Where:

ΔL = change in length

L₀ = initial length

α = coefficient of linear expansion

ΔT = change in temperature

Initial length can be calculated as follows:

V = L³ ⇒ L = ∛V = ∛6 L ≈ 1.82 meters

Now calculate the change in length of the container and the change in volume of acetone:

ΔL = L₀αΔT = (1.82 m)(10⁻⁵ °C⁻¹)(40 °C) ≈ 0.00073 meters

ΔV = V₀βΔT = (6 liters)(1.5 x 10⁻³ °C⁻¹)(40 °C) ≈ 0.36 liters

Since the acetone spills out of the container, its final volume is equal to the initial volume minus the change in volume:

Vf = Vi - ΔV = 6 L - 0.36 L = 5.64 L

Therefore, the volume of spilled acetone is approximately 0.36 liters or 360 milliliters.

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What is the minimum amount ( in kg) of liquid water at 26 degrees that would be required to completely melt 41 grams of ice? The specific heat capacity of liquid water is 4180 J/kg/°C and the specific heat of fusion of ice is 3.33×105 J/kg.

Answers

Approximately 0.123 kg of liquid water at 26 degrees Celsius would be needed to melt 41 grams of ice.


To calculate the minimum amount of liquid water required to melt 41 grams of ice at 0°C, we need to consider the energy required for the phase change from solid to liquid, which is known as the specific heat of fusion of ice.

The energy required to melt 1 kg of ice is 3.33×105 J/kg.

Therefore, the energy required to melt 41 grams of ice is (3.33×105 J/kg) × (41/1000) kg = 13653 J.

To calculate the amount of liquid water required, we use the specific heat capacity of water, which is 4180 J/kg/°C.

Assuming the initial temperature of water is 26°C, the amount of water needed can be calculated as (13653 J) ÷ (4180 J/kg/°C) ÷ (26°C) = 0.123 kg or approximately 123 ml of water.

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4 Starting from rest on a level road a girl can reach a speed of 5 m/s in 10s on her bicycle. a Find the acceleration b Find the average speed during the 10s. c Find the distance she travels in 10 s. [2] Eventually, even though she still pedals as fast as she can, she stops accelerating and her speed reaches a maximum value. Explain in terms of the forces acting why this happens. ​

Answers

a. To find the acceleration, we can use the equation:

acceleration = change in velocity / time taken

Here, the initial velocity is 0 m/s (since she starts from rest), the final velocity is 5 m/s, and the time taken is 10 s. Therefore:

acceleration = (5 m/s - 0 m/s) / 10 s = 0.5 m/s^2

So the acceleration of the girl on her bicycle is 0.5 m/s^2.

b. The average speed during the 10 s can be found by dividing the total distance traveled by the time taken. We don't know the distance traveled yet, so we can use another equation:

distance = (initial velocity x time taken) + (0.5 x acceleration x time taken^2)

Here, the initial velocity is 0 m/s, and the time taken is 10 s. We already calculated the acceleration in part (a) as 0.5 m/s^2. Plugging these values in, we get:

distance = (0 m/s x 10 s) + (0.5 x 0.5 m/s^2 x (10 s)^2) = 25 meters

So the girl traveled 25 meters in 10 seconds.

c. As the girl pedals, she applies force to the pedals, which in turn transfers the force to the rear wheel. This force drives the bicycle forward, causing it to gain speed. However, when the bicycle picks up speed, air resistance (also called drag) comes into play, which increases. Eventually, the force of air resistance becomes equal and opposite to the force that propels the bicycle forward. This means that the net force on the bicycle becomes zero, stopping its acceleration and reaching its maximum speed, known as the terminal velocity. Even if the girl pedals faster, she won't be able to increase her speed because the drag force counteracts the propelling force.

Four cuboids are shown in the diagram below. The position of their center of mass is also shown. Which of the cuboids is the most stable.

Answers

The position of their center of mass is also shown. the square cuboids is the most stable. Hence option B is correct.

A cuboid is a six-sided solid known as a hexahedron in geometry. Quadrilaterals make up its faces. Cuboid is short for "like a cube". A cuboid is similar to a cube in that a cuboid may become a cube by varying the lengths of the edges or the angles between the faces.

The square cuboid has its center of mass on the center of square, the masses are uniformly distributed about it.

Hence option B is correct.

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Car A is moving at a speed of 45km/h towards car B which is moving at a speed of 55km/h. if the two car were initially separated at a distance of 150km, determine how long it will take the two cars to meet?​

Answers

The time taken by the cars to meet is 5.4 x 10³ s.

Speed of car A, v₁ = 45 km/h = 12.5 m/s

Speed of car B, v₂ = 55 km/h = 15.27 m/s

Distance between the cars, d = 150 km = 15 x 10⁴m

The expression for the time taken by the cars to meet can be given as,

Time = Distance/Average speed

t = d/(v₁ + v₂)

Applying the values of d, v₁ and v₂.

t = 15 x 10⁴/(12.5 + 15.27)

t = 15 x 10⁴/27.77

t = 5.4 x 10³ s

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A flat-bottomed barge loaded with coal has a mass of 4.80 × 105 kg. The barge is 20.0 m long and 10.0 m wide. It floats in fresh water. What is the depth of the barge below the waterline?

Answers

The depth of the barge below the waterline is 2.40 m.

To calculate the depth of the barge below the waterline, we need to consider the buoyancy force acting on the barge. The buoyancy force is equal to the weight of the water displaced by the barge.

First, we need to calculate the volume of water displaced by the barge.

Since the barge is flat-bottomed, we can assume that the shape of the displaced water is rectangular with a length of 20.0 m, a width of 10.0 m, and a depth of d (which is what we're trying to find).

Therefore, the volume of water displaced is V = 20.0 m x 10.0 m x d = 200.0 m³.

The weight of the displaced water can be calculated using its density and volume. In fresh water, the density of water is approximately 1000 kg/m³.

Therefore, the weight of the displaced water is W = 1000 kg/m³ x 200.0 m³ = 2.00 × 10⁵ kg.

Since the buoyancy force is equal to the weight of the displaced water, we have [tex]F_b[/tex] = W = 2.00 × 10⁵ kg.

The weight of the barge is [tex]W_b[/tex] = 4.80 × 10⁵ kg. According to Archimedes' principle, the buoyancy force acting on an object in a fluid is equal to the weight of the fluid displaced by the object, so we can write:

[tex]F_b[/tex] = [tex]W_b[/tex] - [tex]W_d[/tex]

where [tex]W_d[/tex] is the weight of the water displaced by the submerged part of the barge. Solving for [tex]W_d[/tex], we get:

[tex]W_d[/tex] = [tex]W_b[/tex] - [tex]F_b[/tex] = 4.80 × 10⁵ kg - 2.00 × 10⁵ kg = 2.80 × 10⁵ kg.

The volume of water displaced by the submerged part of the barge is equal to the volume of the rectangular prism with a length of 20.0 m, a width of 10.0 m, and a depth of d. Therefore, we can write:

[tex]V_d[/tex] = 20.0 m x 10.0 m x d = 200.0 m³ x (d/10.0)

The weight of the displaced water is also equal to its density times its volume, so we have:

[tex]W_d[/tex] = 1000 kg/m³ x [tex]V_d[/tex]

Substituting [tex]V_d[/tex] in terms of d and solving for d, we get:

d = ([tex]W_d[/tex] / (1000 kg/m³ x 200.0 m²)) x 10.0 m = (2.80 × 10⁵ kg / (1000 kg/m³ x 200.0 m²)) x 10.0 m = 2.40 m

Therefore, the depth of the barge below the waterline is 2.40 m.

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Rotation axis. Big disk on the bottom M2 and small disk on the top M1 or mass1. Angular frequency of M₁ is zero.  M₂ initial angular frequency is Wi = 6π rad/s. Two objets both can spine freely in the axis someone is holding M1  and M2 is fixed on the axis. So M1 can slide down the axis and M₂ is fixed its rotating. Surface of M₂ is rough like sand paper but bottom of M1 surface is soft and it can rub on M₂ rough surface. then M1 disk is dropped on M2. then both disks are rotating and M1 cant slipe because of M2 rough surface. find wf final anular frequency of the system​

Answers

The final angular frequency of the system is 3π/2 rad/s. Moment of inertia increased, frictional torque decreased angular momentum, conservation of energy equation used.

At the point when M1 is dropped onto M2, the two circles will be coupled together and pivot as a solitary framework. The complete precise force of the framework should be saved. At first, just M2 was turning with a precise recurrence of Wi = 6π rad/s, while M1 was very still, so the underlying rakish force was L = I2 * Wi, where I2 is the snapshot of dormancy of M2.

At the point when M1 is dropped onto M2, the snapshot of latency of the framework expands, so to preserve rakish energy, the precise recurrence of the framework should diminish. The last rakish recurrence can be tracked down utilizing the preservation of precise force condition: L = I1 * wf + I2 * wf, where I1 is the snapshot of dormancy of M1 and wf is the last rakish recurrence of the framework.

Since M1 is sliding on the unpleasant surface of M2, there is frictional power that goes against its movement. This frictional power creates a force on the framework, which makes the precise energy decline. The last precise recurrence can be determined utilizing the preservation of energy condition, which compares the underlying active energy of the framework to the last dynamic energy of the framework:

0.5 * I1 * 0^2 + 0.5 * I2 * (6π)^2 = 0.5 * (I1 + I2) * wf^2

Improving on the situation and tackling for wf, we get:

wf = 3π/2 rad/s

Thusly, the last rakish recurrence of the framework is 3π/2 rad/s.

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Work is done on a wooden crate by pushing it across a floor. The work done is equal to the force applied parallel to the floor multiplied by the

A. force of friction
B. distance moved
C. mass of the crate
D. crate's direction

Answers

The work done is equal to the force applied parallel to the floor multiplied by the distance moves. Hence, option B is correct.

When a wooden crate is pushed across a floor, the work done on it is given by the product of the force applied parallel to the floor and the distance moved by the crate.

This is because work is defined as the product of force and displacement in the direction of the force, which in this case is the distance moved by the crate. The force of friction and the mass of the crate are not relevant to the calculation of work done in this scenario

It is important to understand the concept of work and the factors that influence it, as it is a fundamental concept in physics and is used in many real-world applications.

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Which example best describes Newton's third law of motion?

F
When a glass slid across a table, it spilled water when it stopped suddenly.

G
An engine used less work to move a lighter car than when it moved a heavier car.

H When a passenger stepped from a boat to the shore, the boat moved away from the shore .

Answers

The best example of Newton's third law of motion is, When a passenger stepped from a boat to the shore, the boat moved away from the shore. Thus, option C is correct.

Sir Issac Newton gives three laws of motion. The first law states that an object remains at rest or in continuous motion unless an external force acted on it. The second law stated that the force is directly proportional to the acceleration of the object. Newton's third law states that, for every action, there is an equal and opposite reaction.

From the given, Newton's third law is applicable, When a passenger stepped from a boat to the shore, the boat moved away from the shore. This shows the action and reaction of the boat and shore.

Thus, the ideal solution is option C.

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A falling object accelerates from
-10.0 m/s to -30.0 m/s. How much
time does that take?
(Unit = s)
Remember: Falling = -v
a = -9.80 m/s²
Lost height = -Ay
Be careful with minus signs!!!/for acellus

Answers

It takes 2.04 seconds for the object to accelerate from -10.0 m/s to -30.0 m/s.

Velocity is a measure of an object's speed in a specific direction. It is calculated by dividing the distance traveled by the time taken and specifying the direction of the motion.

We can use the following formula to find the time taken for an object to change velocity under constant acceleration:

Δv = a × Δt

where Δv is the change in velocity, a is the acceleration, and Δt is the time taken.

In this case, the initial velocity is -10.0 m/s, and the final velocity is -30.0 m/s. So, the change in velocity is:

Δv = (-30.0 m/s) - (-10.0 m/s) = -20.0 m/s

The acceleration of the falling object is -9.80 m/s² (negative because the object is falling downward).

Now, we can rearrange the formula to solve for Δt:

Δt = Δv / a

Substituting the values we have:

Δt = (-20.0 m/s) / (-9.80 m/s²) = 2.04 seconds

Therefore, The object accelerates from -10.0 m/s to -30.0 m/s in 2.04 seconds.

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Waves portfolios I will give brainliest!

Now, leave the frequency constant, but play with the amplitude. What is the affect of changing the
amplitude on the tone generated?

Answers

Now, leave the frequency of wave constant, but play with the amplitude. then the effect of changing the amplitude on the tone generated is its loudness. more the amplitude more it will be loud.

Wave is is a disturbance in a medium that carries energy as well as momentum . wave is characterized by amplitude, wavelength and phase. Amplitude is the greatest distance that the particles are vibrating. especially a sound or radio wave, moves up and down. Amplitude is a measure of loudness of a sound wave. More amplitude means more loud is the sound wave.

Wavelength is the distance between two points on the wave which are in same phase. Phase is the position of a wave at a point at time t on a waveform. There are two types of the wave longitudinal wave and transverse wave.

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How many kilocalories of heat would be needed to melt 0.32 kg of ice at 0°C and increase the temperature to 25°C? The specific heat of water is 1 cal/g.°C, specific heat of ice is 0.5 cal/g.°C, the latent heat of ice to water is 80 cal/g. Keep one digit after the decimal.

Answers

To calculate the heat needed to melt the ice and raise the temperature, we need to break it down into two steps:

Step 1: Heat needed to melt the ice
The mass of ice is 0.32 kg, and the latent heat of ice to water is 80 cal/g. Therefore, the heat needed to melt the ice is:

0.32 kg x 80 cal/g = 25.6 kcal

Step 2: Heat needed to raise the temperature from 0°C to 25°C
The specific heat of water is 1 cal/g.°C, and the mass of water produced by melting the ice is also 0.32 kg. Therefore, the heat needed to raise the temperature from 0°C to 25°C is:

0.32 kg x 25°C x 1 cal/g.°C = 8 kcal

Total heat needed = Heat needed to melt the ice + Heat needed to raise the temperature
Total heat needed = 25.6 kcal + 8 kcal = 33.6 kcal

Therefore, 33.6 kilocalories of heat would be needed to melt 0.32 kg of ice at 0°C and increase the temperature to 25°C.

The side of a cube of metal is measured to be (1.00±0.06) cm and its mass is measured to be (41.0±0.4) g. Determine the uncertainty in the density of the solid in kilograms per cubic meter.

Answers

The density of the solid is (4.10 ± 0.78) × 10^3 kg/m^3.

To calculate the density of the cube below formula can be used:

ρ = m/V

where ρ is density, m is mass, and V is volume. For a cube, the volume is given by:

V = (side)^3

Therefore, the uncertainty in density can be calculated using the formula:

δρ/ρ = sqrt[(δm/m)^2 + 3(δs/s)^2]

where δρ is the uncertainty in density, δm is the uncertainty in mass, δs is the uncertainty in side, and s is the value of the side.

Now, putting in the given values:

s = (1.00 ± 0.06) cm = 0.01 ± 0.0006 m

m = (41.0 ± 0.4) g = 0.0410 ± 0.0004 kg

Volume, V = (0.01 m)^3

                  = 1.0 × 10^-6 m^3

Therefore, the density is:

ρ = m/V

  = 0.0410 kg/1.0 × 10^-6 m^3

  = 4.10 × 10^4 kg/m^3

Now substituting the values and calculating the uncertainty in density:

δρ/ρ = sqrt[(δm/m)^2 + 3(δs/s)^2]

δρ/ρ = sqrt[(0.0004/0.0410)^2 + 3(0.0006/0.01)^2]

δρ/ρ = 0.019

Therefore, the uncertainty in density is:

δρ = (0.019)(4.10 × 10^4 kg/m^3)

     = 779 kg/m^3

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The diagram shows two sets of vectors that result in a
single vector.
R
100 m
200 m
250 m
50 m
What are the first two steps for finding the magnitude of
the resultant vector?
find the square of the first horizontal vector and the
square root of the first vertical vector
O find the square root of the first horizontal vector and
the square root of the second horizontal vector
O find the sum of the two horizontal vectors and the
sum of the two vertical vectors
O find the difference between the two horizontal vectors
and the difference between the two vertical vectors

Answers

The magnitude of the resultant vector is obtained by finding the sum of two horizontal vectors and the sum of two vertical vectors. Thus, option C is correct.

The resultant vector is the single vector that has the same effect in the number of vectors collectively produced. The resultant vector in the horizontal and vertical direction is obtained by drawing a diagonal and hence by using the Pythagoras theorem.

To find resultant vector is obtained by finding the sum of two horizontal vectors and vertical vectors and then using the Pythagoras theorem.

Thus, the ideal solution is option C.

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Move numbers to the boxes to show the factor pairs of 14. Response area with 4 blank spaces Blank space 1 empty times Blank space 2 empty equals 14 Blank space 3 empty times Blank space 4 empty equals 14 Answer options with 14 options. Answer Options 1 2 3 4 5 6 7 8 9 10 11 12 13 1

Answers

Move numbers to the boxes to show the factor pairs of 14:

Blank space 1: 1

Blank space 2: 1

Blank space 3: 2

Blank space 4: 7

A factor pair of a number is a pair of whole numbers that can be multiplied together to give the original number. For the number 14, the factor pairs are (1,14) and (2,7). So, we can put 1 in the first blank, 14 in the second blank, 2 in the third blank, and 7 in the fourth blank to show the factor pairs of 14.

To show the factor pairs of 14 in the given response area with 4 blank spaces, we need to find the two numbers that can be multiplied together to give 14. These two numbers are called factor pairs of 14.

To begin, we can start listing the factors of 14. The factors of 14 are 1, 2, 7, and 14. We can then use these factors to form factor pairs by multiplying them together. The factor pairs of 14 are (1, 14) and (2, 7).

To show these factor pairs in the given response area, we can put the first factor of each pair in the first and third blank spaces, and the second factor of each pair in the second and fourth blank spaces. Therefore, we can put 1 in the first blank, 14 in the second blank, 2 in the third blank, and 7 in the fourth blank to show the factor pairs of 14.

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A skier slides horizontally along the snow for a distance of 22 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is 0.15. Initially, how fast was the skier going, in meters per second?

Answers

The skier was initially going at 8.046 m/s velocity.

Assume that the skier's initial velocity is v.

The skier travelled 22 metres before coming to a stop.

The kinetic friction coefficient between the skier and the snow is 0.15.

The force of friction acting on the skier is given by:

friction = coefficient of friction multiplied by normal force

The normal force is equal to the weight of the skier, which can be calculated as:

weight = mass x gravity

Assuming the mass of the skier to be m and the acceleration due to gravity to be g, we get:

weight = m x g

The force of friction can then be calculated as:

friction = 0.15 x m x g

The work done by the force of friction is equal to the initial kinetic energy of the skier, which can be expressed as:

work done by friction = 0.5 x m x [tex]v^2[/tex]

Equating the work done by friction to the initial kinetic energy, we get:

0.5 x m x [tex]v^2[/tex] = 0.15 x m x g x 22

Simplifying the equation, we get:

[tex]v^2[/tex] = 2 x 0.15 x 9.8 x 22

When we take the square root of both sides, we get:

v = 8.046 m/s

As a consequence, the skier began moving at 8.046 m/s.

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The reflection in a clear window of a store
is a(n)

Answers

The reflection in a clear window of a store is a(n) image.

Why are images seen as reflection?

Images are seen as reflections because of the behavior of light. When light strikes a smooth, reflective surface such as a mirror or still water, it bounces off the surface at the same angle at which it hit it. This process is called reflection. The reflected light rays then travel to our eyes, creating an image.

The angle of incidence (the angle at which the light strikes the surface) is equal to the angle of reflection (the angle at which the light bounces off the surface). This causes the reflected image to be a mirror image of the original object.

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Pls solve 50 points!!!!!!!!!​

Answers

Refer to the attachment for solution

Which statement describes the law of conservation of energy?
A. Energy cannot change forms.
B. Energy cannot be created or destroyed.
C. Air resistance has no effect on the energy of a system.
D. The total energy in a system can only increase.

Answers

D i think I took the exam I think abt 2 years ago?

Answer:

The correct answer is B. Energy cannot be created or destroyed.

The law of conservation of energy states that energy cannot be created or destroyed, but it can be transformed from one form to another. This means that in any physical process, the total amount of energy in a system remains constant. Energy can be converted from one form to another, such as from kinetic energy to potential energy or from electrical energy to light energy, but the total amount of energy in the system remains the same.

Option A is incorrect because energy can change forms, as mentioned above.Option C is incorrect because air resistance is a type of force that opposes the motion of an object, and therefore, it can cause a decrease in the kinetic energy of a system.Option D is incorrect because the law of conservation of energy states that the total energy in a system remains constant, it cannot increase or decrease without an external force acting on it.

A particle of mass 2unit moves along space curve defined by ~r(t) = (4t 2 − t 3 )ˆi − 5tˆj + (t 4 − 2)ˆk. Find the force acting on it at any time to work out​

Answers

The force acting on the particle at t = 1 s is 24.4 N.

Mass of the particle, m = 2 units

Distance of the curve, r(t) = (4t²- t³)i - 5tj + (t⁴- 2)k

The velocity of the particle,

v = d[r(t)]/dt = (8t - 3t²)i - 5j + 4t³k

The acceleration of the particle,

a = d²[r(t)]/dt² = (8 - 6t)i + 12t²k

Let the time for which the force is acting be 1 second.

Therefore, acceleration at t = 1 is,

a₁ = 2i + 12k

Hence, the magnitude of acceleration,

|a₁| = √(2²+ 12²)

|a₁| = √148

|a₁| = 12.2 unit/s²

Therefore, the force acting on the particle at t = 1 s is,

F₁ = m x |a₁|

F₁ = 2 x 12.2

F₁ = 24.4 N

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The resistance RT of a platinum varies with temperature T(°C), as measured on the constant-volume gas thermometer according to the equation RT = Ro(1+AT+BT^2). Where A = 3.8×10^-3°C^-1 and B = -5.6×10^-7°C^-2. Calculate the temperature that would be on indicated on a platinum thermometer, when the gas scale reads 200°C.​

Answers

The resistance indicated by the platinum thermometer at 200°C is 1.648 times the reference resistance Ro at 0°C.

The given equation is RT = Ro(1+AT+BT²), where A = 3.8×10⁻³°C⁻¹ and B = -5.6×10⁻⁷°C⁻². To determine the temperature that would be indicated on a platinum thermometer when the gas scale reads 200°C, we will have to use the given formula. RT = Ro(1+AT+BT²) .....(i)We know that the gas scale reads 200°C. Therefore, we can substitute T = 200°C in equation (i).RT = Ro (1 + A × 200 + B × 200²) = Ro (1 + 0.76 - 0.112) = Ro (1.648)Thus, the resistance that the platinum thermometer would indicate is 1.648 times the reference resistance Ro at 0°C. This is the solution to the problem.In summary, The given equation is RT = Ro(1+AT+BT²), where A = 3.8×10⁻³°C⁻¹ and B = -5.6×10⁻⁷°C⁻². To determine the temperature that would be indicated on a platinum thermometer when the gas scale reads 200°C, we substituted T = 200°C in equation (i) to get RT = Ro (1 + A × 200 + B × 200²) = Ro (1 + 0.76 - 0.112) = Ro (1.648).

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Two objects, X and Y, experience external net torques that vary over a period of 5 seconds. Object X has a moment of inertia I0, and Object Y has a moment of inertia 2I0. The average value of the magnitude of the external net torque exerted on Object X from time t=0 to t=5s is torquex. Similarly, the average value for ObjectY is torquey. The magnitudes of the angular momenta L of Objects X and Y versus t are shown in the graph. Which of the following expressions correctly relates torquey to torquex?

Answers

Answer: τy = 2τx

Given: graph of L vs t
Ly = 20 at t = 5
Lx = 10 at t = 5
All L = 0 at t = 0
So change in angular momentum of y is 20 and x is 10

Explanation: ΔL = τ(average) * Δt
Change in angular momentum = average torque * change in time

solve for average torque for each objects
τ(average) = ΔL / Δt

Object y average torque
τy = ΔLy / Δt = 20 / 5 = 4
τy = 4

Object x average torque
τx = ΔLx / Δt = 10 / 5 = 2
τx = 2

Relates τy and τx
2τx = τy


Two objects, X and Y, experience external net torques that vary over a period of 5 seconds. Object X has a moment of inertia I0, and Object Y has a moment of inertia 2I0. The average value of the magnitude of the external net torque exerted on Object X from time t=0 to t=5s is torquex. Similarly, the average value for ObjectY is torquey.

The magnitudes of the angular momenta L of Objects X and Y versus t are shown in the graph. The precise relation between torquey and torquex is torquey = 2 * torquex.

To relate torquey to torquex, we are able to use the concept of angular momentum and torque. Angular momentum is described because the manufactured from the moment of inertia and angular velocity:

L = I * ω

Differentiating this equation with an appreciation of time, we get:

dL/dt = d(I * ω)/dt

Using the product rule of differentiation, we've got:

dL/dt = I * dω/dt + ω * dI/dt

Now, we realize that torque (τ) is described because of the charge of the exchange of angular momentum:

τ = dL/dt

Substituting the expression for dL/dt in terms of angular velocity and second of inertia:

τ = I * dω/dt + ω * dI/dt

Let's denote the common price of torque for item X as torquex. Since object X has a moment of inertia I0, we can write:

torquex = I0 * dω/dt + ω * dI0/dt

Now, let's consider item Y. It has a moment of inertia 2I0. Using the identical expression, we will write:

torquey = (2I0) * dω/dt + ω * d(2I0)/dt

torquey = 2I0 * dω/dt + ω * (2 * dI0/dt)

torquey = 2I0 * dω/dt + 2ω * dI0/dt

Comparing the expressions for torquex and torquey, we will see that:

torquey = 2 * torquex

Therefore, the precise relation between torquey and torquex is;

torquey = 2 * torquex.

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The correct question is;

"Two objects, X and Y, experience external net torques that vary over a period of 5 seconds. Object X has a moment of inertia I0, and Object Y has a moment of inertia 2I0. The average value of the magnitude of the external net torque exerted on Object X from time t=0 to t=5s is torquex. Similarly, the average value for ObjectY is torquey. The magnitudes of the angular momenta L of Objects X and Y versus t are shown in the graph. Which of the following expressions correctly relates torquey to torquex?"

The diagram below shows the velocity vectors for two cars that are moving
relative to each other.
45 m/s west
25 m/s east
Car 1
Car 2
From the frame of reference of car 1, what is the velocity of car 2?
OA. 20 m/s west
OB. 70 m/s east
C. 70 m/s west
D. 20 m/s east

Answers

The relative velocity of the car 2 with respect to car 1 is 70 m/s west. So, the correct option is C.

Velocity of the car 1 with respect to the ground, v₁ = 45 m/s west

Velocity of the car 2 with respect to the ground, v₂ = 25 m/s east

Let the direction towards the east be positive and the opposite direction towards the west be negative.

The expression for the relative velocity of the car 2 with respect to car 1 is given by,

v₂₁ = v₂ - v₁

v₂₁ = 25 - (-45)

Therefore, the relative velocity,

v₂₁ = 70 m/s west

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Attaching the image file here.

In which circuit would ammeter A shows the greatest current?

Answers

In the circuit with two 10Ω resistors in parallel, ammeter A would show the greatest current. This is because, in a parallel circuit, the total resistance is lower than in a series circuit, which means that the current can flow more easily.

In this case, the two 10Ω resistors in parallel create a total resistance of 5Ω (1/Rtotal = 1/10 + 1/10 = 2/10, Rtotal = 10/2 = 5), while in the series circuit,https://brainly.com/question/11409042?referrer=searchResults the total resistance would be 20Ω (10 + 10). Ohm's law states that the current is directly proportional to the voltage and inversely proportional to the resistance, so the circuit with lower resistance will allow for greater current flow.

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--The complete Question is, In which circuit would ammeter A show the greatest current: a circuit with one 6V battery and two 10Ω resistors in parallel or a circuit with one 6V battery and two 10Ω resistors in series? --

20N
10N
40N
30N
Balanced or Unbalanced?

Answers

Answer:

Unbalanced

Explanation:

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