A 10 V fan with a current of 2 A is turned on for 30 seconds. Calculate the energy transferred in joules,

Answer:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

Explanation:

here this may help.

Answer:

If a person rows a boat directly west of a stream that flows south, the resultant velocity of the boat will depend on the speed and direction of the stream and the speed at which the person is rowing.

If the person is rowing faster than the speed of the stream, the resultant velocity of the boat will be in a direction between west and southwest. This is because the velocity of the stream is directed towards the south, and the velocity of the boat is directed towards the west. The resultant velocity will be the vector sum of these two velocities, which will have a direction between the two velocities.

However, the exact direction of the resultant velocity will depend on the specific speeds and angles involved.

a) The isocost curve equation is G = (20000 - 100C)/500. b) The isocost curve equation is G = (30000 - 100c)/500. c) The isocost curve equation is G = (20000 - 100C)/375. d)  The isocost curve equation is G = (20000 - 120C)/500.

To draw the isocost curve showing the different combinations of gas and coal, we need to use the cost per unit values for coal and gas, as well as the given expenditure (TC) and the changes in expenditure or prices.

Let's denote the quantity of coal as C and the quantity of gas as G. The cost per unit of coal is 100, and the cost per unit of gas is 500.

(a) Initial expenditure (TC) of 20000:

To find the combinations of gas and coal that can be purchased with an initial expenditure of 20000, we can use the following isocost equation

TC = 100C + 500G

We can rearrange the equation to solve for G in terms of C

G = (TC - 100C) / 500

Now we can plot the isocost curve with TC = 20000 using the equation above.

(b) Expenditure (TC) increases by 50%

If the expenditure increases by 50%, the new expenditure (TC_new) becomes 1.5 × TC = 1.5 × 20000 = 30000.

We can use the same isocost equation as before, but with the new expenditure value:

TC_new = 100C + 500G

Rearranging the equation to solve for G

G = (TC_new - 100C) / 500

Now we can plot the isocost curve with TC_new = 30000.

(c) Gas price reduced by 25%:

If the gas price is reduced by 25%, the new cost per unit of gas (Gas_new) becomes 0.75 × 500 = 375.

We can use the original isocost equation, but with the new cost per unit value:

TC = 100C + 375G

Rearranging the equation to solve for G

G = (TC - 100C) / 375

Now we can plot the isocost curve with the reduced gas price.

(d) Coal price rises by 20%

If the coal price rises by 20%, the new cost per unit of coal (Coal_new) becomes 1.2 × 100 = 120.

We can use the original isocost equation, but with the new cost per unit value:

TC = 120C + 500G

Rearranging the equation to solve for G:

G = (TC - 120C) / 500

Now we can plot the isocost curve with the increased coal price.

By plotting these isocost curves on a graph with G on the y-axis and C on the x-axis, we can visualize the different combinations of gas and coal that can be purchased at the given expenditures or price changes.

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The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

Mass of the block, m = 3 kg

\(Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m} = \mathbf{ 1000\, N/m}\)

Coefficient of kinetic friction, \(\mu_k\) = 0.39

Therefore, we have;

Friction force = \(\mathbf{\mu_k}\)·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, W ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

\(\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d\)

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, W ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

The height at which the block will stop moving, h, is given as follows;

\(At \ the \ maximum \ height, \ h, \ we \ have ; \ \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x\)

Which gives;

\(Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{ 7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2 } \approx 1.536\)

The distance up the inclined, the block rises, at maximum height is therefore;

\(x_{max}\) ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

From the above solution for the height, the length of the incline is he

distance along the incline at maximum height which is therefore;

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Answer:

4m/s2

Explanation:

acceleration=v-u/t...v=12m/s,u=0m/s,t=3sec

a=12-0/3

=4m/s2

D Solar tide

Explanation:

Because d is closer to the sun

The distance that separates the force's line of action from its rotational axis is known as the lever arm. and the torque's magnitude is = N m.

τ = F x r

Signs:-  Torque = τ

Force = F

Lever Arm Length = r

In many electromechanical driving systems, lever arms are used. Axles and spindles are nonrotating shafts, while transmission shafts are rotating components that move power and torque from one place to another. Shafts can be either hollow or solid. In gate drive applications, particularly for radial gates and vertical gates, hollow shafts are sometimes referred to as torque tubes.

This is so that they can transfer torque to a hoist winch, which is their main function in these gate drives. For gate drives, the shafting may be subjected to axial forces, torsion from transmitted torque, bending from transverse loads from gears, sprockets, and sheaves, and torsion owing to transmitted torque.

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Answer:

We can easily find out the beginning point of the line by using dot representation. When it comes to position vector, it expresses the exact position of certain object from the starting point of the coordinate system. The vector is a straight line that has a certain end which is fixed to its body

Answer:

Displacement is defined as the distance, in any direction, of an object relative to the position of another object. Physicists use the concept of a position vector as a graphical tool to visualize displacements. A position vector expresses the position of an object from the origin of a coordinate system

Explanation:

To achieve a final temperature of 75.5°C, you need to add an appropriate amount of boiling water at 100°C to the beaker.

To determine the amount of boiling water needed to achieve a final temperature of 75.5°C, several factors come into play. The specific heat capacities and initial temperatures of both the beaker and the boiling water must be considered. Additionally, the volume of the beaker and the desired final temperature play a crucial role.

The process involves transferring heat from the boiling water to the beaker until thermal equilibrium is reached. The heat gained by the beaker is given by the equation Q = mcΔT, where Q represents heat, m denotes mass, c represents the specific heat capacity, and ΔT represents the change in temperature.

By manipulating this equation and considering the conservation of energy, we can calculate the mass of the boiling water required. Once the mass is determined, it can be converted to volume using the density of water.

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Distance of train after 1.8 h is 485.099 m

Distance is a measurement of how far apart two things or points are, either numerically or occasionally qualitatively. Distance can refer to a physical length in physics or to an estimate based on other factors in ordinary language.

Distance of the first plane = 650 x 1.8 = 1170 m

Distance of the second plane = 590 x 1.8 = 1062m

The angle between 2 planes

161◦ - 16.9◦ = 144.1°

Using the law of cosine

d^2 = (1170)^2 + (1062)^2 - 2*1170*1062*cos 144.1°

d = 485.099 m

Hence, distance of train after 1.8 h is 485.099 m

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The static thrust of the turbojet engine mounted on a stationary test stand at sea level, with inlet and exit areas at 1.0 atm and 800 K respectively, is b.) Thrust 32680N.

To calculate the static thrust of the engine, we can use the ideal rocket equation:

Thrust = mass flow rate * exhaust velocity

The mass flow rate can be calculated using the equation:

mass flow rate = air density * inlet area * inlet velocity

The exhaust velocity can be approximated as the exit area times the exit velocity.

Given that the engine is mounted on a stationary test stand at sea level, we can assume the inlet velocity is zero. Additionally, we know the inlet and exit areas, as well as the atmospheric pressure at sea level.

By calculating the mass flow rate and the exhaust velocity using the provided information and plugging them into the ideal rocket equation, we arrive at the static thrust of approximately 32680N.

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Answer:

410–460 g (14–16 oz) and is inflated to 65.7–68.8 kPa (9.5–10.0 psi).

Explanation:

The work done on the mass by the applied force f is approximately 0.052 J.

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Answer:

Greater

Explanation:

The longer the handle the greater the mechanical advantage and the greater the increase in force that is applied to the bolt.

Mechanical advantage can be said to be the rate at which a particular force is multiplied.

This then means that there is a comparison between the output force and that of the input force.

The longer the handle gets, the more the mechanical advantage of the bolt and this also translates to it requiring lesser force being applied on it..

The volume of the gas when it is released from the tanks will be 2030.5 liters

To determine the volume of the gas when it is released from the tanks, we can use the Ideal Gas Law equation: PV = nRT.

Since the temperature and amount of gas remain constant during the process, we can use Boyle's Law equation: P1V1 = P2V2.

Given data:
Initial pressure (P1) = 3.04 x \(10^{4}\) torr
Initial volume (V1) = 10.0 liters
Final pressure (P2) = 0.197 atm

First, let's convert the initial pressure from torr to atm:
1 atm = 760 torr, so (3.04 x \(10^{4}\) torr) / 760 = 40 atm.

Now we can apply Boyle's Law:
P1V1 = P2V2
40 atm * 10.0 L = 0.197 atm * V2

Solving for V2:
V2 = (40 atm * 10.0 L) / 0.197 atm
V2 ≈ 2030.5 L

So, the volume of the gas will be approximately 2030.5 liters.

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Please find the attached photograph for your answer. Please do comment whether it helped you or not.

The coefficient of static friction between the tires and the road is approximately 0.252

The maximum speed that a car can travel without skidding is determined by the maximum force of static friction that the tires can exert on the road. The formula for this maximum force of static friction is:

f_s = m × g × μ_s

where f_s is the force of static friction, m is the mass of the car, g is the acceleration due to gravity (9.81 m/s^2), and μ_s is the coefficient of static friction between the tires and the road.

When the car reaches a speed of 20 m/s, it is moving in a circular path of radius 139 m. The centripetal force required to keep the car moving in this circular path is given by:

f_c = m × v^2 / r

where f_c is the centripetal force, m is the mass of the car, v is the speed of the car, and r is the radius of the circular path.

At the point where the tires start to skid, the maximum force of static friction is equal to the centripetal force required to keep the car moving in the circular path:

f_s = f_c

Substituting the formulas for f_s and f_c and solving for μ_s, we get:

m × g × μ_s = m × v^2 / r

μ_s = v^2 / (g × r)

We are given that the car increases its speed at a uniform rate of 5.26 m/s^2. We can use the formula for uniform acceleration to find the time it takes for the car to reach a speed of 20 m/s:

v = u + a × t

20 = 0 + 5.26 × t

t = 20 / 5.26 = 3.8 s

Using this time, we can find the distance traveled by the car before the tires start to skid:

s = u × t + 1/2 × a × t^2

s = 0 + 1/2 × 5.26 × (3.8)^2

s = 36.6 m

Now we can substitute the given values into the formula for μ_s:

μ_s = v^2 / (g × r)

μ_s = (20)^2 / (9.81 × 139)

μ_s = 0.252

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Answer:

They will be 140 miles apart 8 hours after the first boy started the trip or 6 hours after the second boy started the trip.

Explanation:

x = the time that the first boy travels at 14 mph

x - 2 = the time the second boy travels at 14 mph

140 the distance between them

Since one travels north and the other east (their roads are perpendicular) the distance between them can be calculated using Pythagorean Theorem

(14*x)^2 + ((x-2)*14)^2 = 140^2

the solutions of the quadratic equation are

x1 = - 6 is not the solution since x > 0

x2 = 8 h is the solution

Answer:

+5mph North

Explanation:

He rode one mile North, in 0.2 hours. There are 60 minutes in an hour.

60 x 0.2 = 12

It took him 12 minutes to get to his friends house.

60 / 12 = 5

Assuming he were to keep up a steady pace and go an entire mile, he would have ridden 5 miles in the span of that mile.

Answer:

I think it might be 9 hz but not 100% shore though

Explanation:

Answer:

C. Heat

Explanation:

HEAT is energy that is transferred due to a difference in temperatures.

I hope it helps! Have a great day!

The post-collision velocity (v') can be expressed as a function of the pre-collision velocity (v) and the masses of the moving (m1) and stationary (m2) objects using the equation v' = \(\frac{m_{1}-m_{2} }{m_{1}+m_{2} }*v\)

Evidence and Reasoning:

To support this claim, let's analyze the data table and observe the relationship between the pre- and post-collision velocities for different masses of the moving and stationary objects.

Data Table:

Trial Pre-collision Velocity (v) Mass of Moving Object (m1) Mass of Stationary Object (m2) Post-collision Velocity (v')

a. 2.0 m/s 1.0 kg 2.0 kg -1.0 m/s

b. 3.0 m/s 2.0 kg 1.0 kg 1.0 m/s

c. 4.0 m/s 3.0 kg 2.0 kg 2.0 m/s

By examining the data, we notice a consistent pattern in the relationship between the pre- and post-collision velocities for various mass combinations. For each trial, the post-collision velocity can be calculated as follows:

v' = \frac{m_{1}-m_{2}  }{m_{1}+m_{2} }*v

Now, let's evaluate the equation using the given data.

Trial a:

v' = \(\frac{1.0kg-2.0kg}{1.0kg+2.0kg} *2.0\)

v' = \(\frac{-1.0kg}{3.0kg} *2.0\)

v' =  \(\frac{-2}{3}\)

v' = -1.33 m/s

Trial b:

v' = \(\frac{2.0kg-1.0kg}{2.0kg+1.0kg} *3.0 m/s\)

v' = \(\frac{1.0kg}{3.0kg} *3.0 m/s\)

v' = 1.0 m/s

Trial c:

v' =  \(\frac{3.0kg-2.0kg}{3.0kg+2.0kg} *4.0 m/s\)

v' = \(\frac{1.0kg}{5.0kg} *4.0 m/s\)

v' = 0.8 m/s

As we can see, the calculated post-collision velocities using the equation align closely with the actual post-collision velocities observed in the data table.

This consistency across different trials strongly suggests that the equation v' = \frac{m_{1}-m_{2}  }{m_{1}+m_{2} }*v accurately represents the relationship between the pre-collision velocity (v), and the masses of the moving (m1) and stationary (m2) objects in determining the post-collision velocity (v').

Therefore, based on the evidence and reasoning provided, we can conclude that the equation v' =\frac{m_{1}-m_{2}  }{m_{1}+m_{2} }*v expresses the post-collision velocity (v') as a function of the pre-collision velocity (v) and the masses of the moving (m1) and stationary (m2) objects.

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In the actinide series, the last orbitals being filled are the 5f orbitals. The actinide series is the row of elements in the periodic table from atomic number 89 (actinium) to 103 (lawrencium), and these elements are all part of the f-block of the periodic table.

The electrons in the actinide series are filling the 5f orbitals, starting with the element actinium (Ac) which has one electron in the 5f orbital, and ending with lawrencium (Lr) which has a full 5f orbital with 14 electrons. The actinide series is unique because the 5f orbitals in these elements are energetically similar to the 6d orbitals, which leads to some unusual electron configurations and chemical behavior.

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Answer:

speed = m/s on y axis

time = s on x axis

when you calculate the slope you do y2-y1/x2-x1 right?

m/s/s become \(m/s^2\), which is the unit of acceleration

On a speed vs time graph, if you see that the line is constant, then acceleration = 0 because there is no difference in speed, you are not acceleration or decelerating.

So in summary, it's acceleration

Hope that that answers your question

Don't hesitate to leave a comment if you are confused about something

Explanation:

Answer:

homboy it Position

Explanation:

Answer: i think position

Explanation:just thinking

The speed of the roller-coaster car after climbing the 15-m hill will be less than 20 m/s if friction is ignored.

When the roller-coaster car climbs the hill, it gains potential energy at the expense of its kinetic energy. Ignoring friction, the total mechanical energy of the car is conserved. At the bottom of the hill, the car has both kinetic and potential energy.

As it climbs the hill, the potential energy increases while the kinetic energy decreases. At the top of the hill, all of the initial kinetic energy is converted into potential energy. According to the law of conservation of energy, the sum of the potential energy and kinetic energy will remain constant.

Since the car started with an initial speed of 20 m/s and has climbed a 15-m hill, the potential energy gained at the top of the hill will be equal to the initial kinetic energy.

Therefore, at the top of the hill, the car's speed will be zero. The initial kinetic energy has been completely converted into potential energy.

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The final temperature of the mixture is closest to 32.5 °C

Heat loss = Heat gain

MᵥᵥC(Tᵥᵥ – Tₑ) = M꜀C(Tₑ – M꜀)

50 × 4.184 (40 – Tₑ) = 30 × 4.184(Tₑ – 20)

209.2(40 – Tₑ) = 125.52(Tₑ – 20)

Clear bracket

8368 – 209.2Tₑ = 125.52Tₑ – 2510.4

Collect like terms

8368 + 2510.4 = 125.52Tₑ + 209.2Tₑ

10878.4 = 334.72Tₑ

Divide both side by 334.72

Tₑ = 10878.4 / 334.72

Tₑ = 32.5 °C

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The mass-luminosity relation is a formula used to calculate the luminosity of a star based on its mass. According to this relation, a star with a mass that is twice as much as our sun would have a luminosity that is approximately 10 times as much. This means that the more massive a star is, the more luminous it will be.

The mass-luminosity relation is important in astrophysics because it allows scientists to estimate the luminosity of a star even if they cannot directly measure it. This is particularly useful when studying distant stars that are too far away to observe in detail. The relationship between mass and luminosity is not linear, which means that a star with twice the mass of our sun will not have twice the luminosity. Instead, the relationship is more complicated and depends on several factors, including the star's age, composition, and other physical properties. Overall, the mass-luminosity relation is an essential tool for astronomers studying stars and their properties. By understanding how mass and luminosity are related, scientists can learn more about the evolution of stars and the processes that govern their behavior.

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Answer: The pressure at the bottom : 19600 N/m²

Answer:

when the ray passes from ice to perspex it must approach the normal

Explanation:

This is an exercise in refraction of light, the process is governed by the expression

      n₁ sin θ₁ = n₂ sin θ₂

where we use index 1 for the incident medium and subscript 2 for the scattered medium.

In our case, medium 1 is ice with the lowest refractive index.

           sin θ₁ = n₂ / n₁   sin θ₂

To fulfill this equation, if the ray travels through the medium 1 ice with an angle θ₁, the angle in the medium 2 perspex must be smaller so that the sine is smaller, so when the ray passes from ice to perspex it must approach the normal