The intrepid hero has done 2.332 x[tex]10^6[/tex] Joules of work in pushing the crate.
To ascertain the work done by the traveler, we first need to find the power he applied on the case. As per Newton's subsequent regulation, force is equivalent to mass times speed increase, so the power applied by the traveler on the container is:
Force = mass x speed increase
Force = 220 kg x 2 [tex]m/s^2[/tex] = 440 N
Then, we really want to work out the distance the case was moved. The pilgrim pushed the box a distance of 5.3 km, or 5,300 m.
At long last, we can compute the work done by the pioneer utilizing the equation:
Work = force x distance
Work = 440 N x 5,300 m
Work = [tex]2.332 x 10^6[/tex] Joules
Thusly, the valiant legend has done 2.332 x [tex]10^6[/tex] Joules of work in pushing the case.
The space pilgrim takes care of business on the case by applying a power that makes it speed up. The work done is equivalent to the power duplicated by the distance over which the power is applied. Involving the recipe for force, F=ma, and the given qualities for mass and speed increase, we can ascertain the power applied. Then, at that point, involving the recipe for work, W=Fd, and the given distance, we can ascertain the work done. The work done by the adventurer is 2.332 x [tex]10^6[/tex] J.
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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)
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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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
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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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?
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)
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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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?
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?"
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
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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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?
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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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.
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).For more questions on thermometer
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20N
10N
40N
30N
Balanced or Unbalanced?
Answer:
Unbalanced
Explanation:
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?
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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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.
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.In which circuit would ammeter A shows the greatest current?
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? --
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?
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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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
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.
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
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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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.
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.
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 .
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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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.
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.
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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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?
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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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.
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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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
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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