5. A 950 kg car is driven up a hill at constant velocity of 7 m/s, where 1200 N of friction and drag oppose its motion. How do I do a?

Answer:4 m/s

Explanation:

Answer:

P = W / t    work/time

W(ork) = m g h = 80kg * 9.8 m.s^2 * 5 m = 3920 N-m

P = 3920 N-m / 3.5 s = 1120 Watts

Answer:

Mike can travel 80 Km in 4 hours

Answer:

c

Explanation:

Answer:

c

Explanation:

The simple harmonic motion and Hooke's law allows to find the displacement when hanging the mass of 2 kg is:

      x = 2.45 m

Simple harmonic motion is a motion where the restoring force is proportional to the displacement.

          x = A cos (wt + Ф)

Where x is the displacement, A is the amplitude, w is the angular velocity, t is the time

The angular velocity is

          w² = k / m

Where k is the spring constant and m is the mass.

From the graph we see that the oscillatory wave has half a period between the times t₁ = \(\frac{3}{4} \pi\)  and t₂ = \(\frac{1}{4} \pi\) , therefore the period is:

                T = π  s.

The angle in a complete oscillation is θ = 2π rad.

The angular kinematics defines the angular velocity with the angle in the unit of time.

        w = \(\frac{\Delta \theta}{\Delta t}\)  

        w = \(\frac{2\pi }{\pi }\)  

        w = 2 rad / s

Let's find the spring constant.

        k = w² m

Let's calculate.

        k = 2² 2

        k = 8 N / m

Hooke's law says that the restoring force of a spring is proportional to its displacement.

         F  = - k x

         mg = - k x

          x = \(\frac{mg}{k}\)  

          x = \(\frac{2 \ 9.8}{8}\)  

          x = 2.45 m

In conclusion using simple harmonic motion and Hooke's law we can find the displacement when hanging the 2kg mass is:

      x = 2.45 m

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Calculating the mass of the block requires a bit of work. The formula for the volume of a rectangular solid is V = l*w*h, where V is the volume, l is the length, w is the width, and h is the height. Using the dimensions given, we can calculate the volume of the block as 30*80*60 = 144000 cubic centimeters.

The density of lead is approximately 11.34 grams per cubic centimeter. To calculate the mass of the block, we can use the formula m = V*d, where m is the mass, V is the volume, and d is the density. Plugging in the values we get m = 144000*11.34 = 1,634,400 grams or approximately 1.63 metric tons.

So, the mass of the block of lead is approximately 1.63 metric tons.

Define the following terms

Metal :

Electric field:

When two objects held close together and then released, they move toward each other. The phenomenon can be explained by the Coulomb's Law. Coulomb's Law is given by: $F=k\frac{q_{1}q_{2}}{r^2}$ where F is the electrostatic force, q1 and q2 are the magnitudes of the charges, r is the distance between the centers of the charged objects, and k is Coulomb's constant.

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The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².

To calculate the average acceleration of the motorcycle, we can use the formula:

Average acceleration = (final velocity - initial velocity) / time

First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.

Converting the final velocity:

Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s

Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:

Average acceleration = (21.67 m/s - 0 m/s) / time

To find the time taken to reach this velocity, we need to use the formula for average speed:

Average speed = total distance/time

Rearranging the formula:

time = total distance / average speed

Plugging in the values:

time = 50 m / 21.67 m/s ≈ 2.31 seconds

Now we can calculate the average acceleration:

Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²

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The magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹ (Option E).

The electric flux through a surface is given by the product of the electric field strength and the area of the surface projected perpendicular to the electric field.

In this case, the electric field strength is 18 N C⁻¹, and the area of the sheet projected perpendicular to the electric field is 0.450 m²

(since the normal to the sheet makes an angle of 60° with the electric field). Multiplying these values gives the electric flux:

Electric flux = Electric field strength × Area

Electric flux = 18 N C⁻¹ × 0.450 m²

Electric flux = 8.1 N m² C⁻¹

In summary, the magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹. This value is obtained by multiplying the given electric field strength by the projected area of the sheet perpendicular to the electric field.

The angle of 60° is taken into account to determine the effective area for calculating the flux.(Option E).

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Answer:

Tension = 227.6 N

Explanation:

Applying the equilibrium condition on the cylinder with the downward direction taken as positive:

\(Tension + Weight\ of\ Cylinder - Bouyant\ Force = 0\\Tension = \rho Vg - Weight\ of\ Cylinder\)

where,

ρ = density of water = 1000 kg/m³

g = acceleration due to gravity = 9.81 m/s²

V = Volume of water displaced = 70% of Volume of Cylinder

V = \(0.7(\frac{Mass\ of\ Cylinder}{Density\ of\ Cylinder} ) = 0.7(\frac{30\ kg}{395\ kg/m^3})\) = 0.0532 m³

Weight of Cylinder = (mass)(g) = (30 kg)(9.81 m/s²) = 294.3 N

Therefore,

\(Tension = (1000\ kg/m^3)(0.0532\ m^3)(9.81\ m/s^2) - 294.3\ N\\\)

Tension = 227.6 N

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

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Answer:

a. At point P, tension is to the right, gravity is pulling the mass down, and the centripetal force is directed toward the circle's center, so it is likewise to the right. Tension and centripetal force are both increasing at point Q, which is at the bottom of the circle. The force of gravity is pulling you down.

b. All forces are descending at point Z. Fc = mac, therefore. T + mg = m(Vmin)2 / R Vmin2 = R(T+mg) / m Vmin = radical R(T + mg) /m c T + mg = m(Vmin)2 / R Vmin2 = R(T+mg) / m Vmin = radical R(T + mg) /m c The only difference is that the maximum tension is at pt Q. mac = fc Tmax - mg = m(Vmax)2 / R Tmax - mg = m(Vmax)2 / R

c. The same as before, but with the maximum tension at pt Q Fc = mac Tmax - mg = m(Vmax)2 / R Tmax - mg = m(Vmax)2 / R R(Tmax - mg) / m d = radical R(Tmax - mg) / m d = radical R(Tmax - mg) / m It rises at the same rate as if it were in a circle. The only force acting on it is gravity. Is this accurate? In points P, Q, and Z, I'm not clear where the centripetal force is.

Explanation:

(:

The expression for the minimum speed the ball can have at point Z (top of the circle) without leaving the circular path is \(v_{min} = \sqrt{\frac{r(T + W)}{m} }\).

The minimum speed of an object moving in a circular path is calculated as follows;

Considering top of the horizontal circle;

T = Fc - mg

T = mv²/r - mg

mv²/r = T + mg

mv² = r(T + mg)

v² = r(T + mg)/m

\(v = \sqrt{\frac{r(T + mg)}{m} } \\\\v_{min} = \sqrt{\frac{r(T + W)}{m} }\)

where;

Thus, the expression for the minimum speed the ball can have at point Z (top of the circle) without leaving the circular path is \(v_{min} = \sqrt{\frac{r(T + W)}{m} }\).

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Answer:

A and B.

Explanation:

Answer C is wrong - there are no tall trees underwater

Answer D is wrong, there are mountain chains underwater due to plate tectonics.

The velocity of the coupled boxcars is 1.5 m/s after the collision.

What is Velocity?

Velocity is a measure of the rate of change of an object's position in a given direction. Velocity is a vector quantity, meaning it includes both size and direction. It is usually expressed in meters per second (m/s). Velocity is derived from the equation for acceleration, which is the rate of change of velocity. Velocity is a key concept in physics and is used to describe the motion of objects.

This problem involves an inelastic collision between two boxcars, where the cars stick together after colliding. In an inelastic collision, the total kinetic energy of the system is not conserved, as some energy is lost to deformation or other forms of energy transfer.

Using the conservation of momentum equation for this system, we have:

m1v1 + m2v2 = (m1 + m2)vf

where m1 is the mass of the first boxcar, v1 is its initial velocity, m2 is the mass of the second boxcar, v2 is its initial velocity (which is zero), and vf is the final velocity of the coupled boxcars.

Plugging in the given values, we have:

(9000 kg)(3 m/s) + (9000 kg)(0 m/s) = (9000 kg + 9000 kg)vf

Simplifying, we get:

27000 kg m/s = 18000 kg vf

Dividing both sides by 18000 kg, we get:

vf = 1.5 m/s

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Answer:

headlight of vehicles use less energy and gives light to very limited area hence electricity can be conserved by doing this

Answer:

771 .4  kg

Explanation:

Remember    F = ma      so     F/a = m    

m = 27000 N / 35 m/s^2 = 771 .4 kg

Answer:

The mutual inductance is  \(M = 0.000406 \ H\)

Explanation:

From the question we  are told that

    The  number of turns per unit length  is  \(N = 1800\)

    The radius is  \(r = 0.0165 \ m\)

     The  number of turns of the solenoid is  \(N_s = 210 \ turns\)

Generally the mutual inductance of the  system is mathematically represented as

       \(M = \mu_o * N * N_s * A\)

Where A is the cross-sectional area of the system which is mathematically represented as

       \(A = \pi * r^2\)

substituting values

      \(A = 3.142 * (0.0165)^2\)

       \(A = 0.0008554 \ m^2\)

also   \(\mu_o\) is the permeability of free space with the value  \(\mu_o = 4\pi * 10^{-7} N/A^2\)

So  

      \(M = 4\pi * 10^{-7} *1800 * 210 * 0.0008554\)

      \(M = 0.000406 \ H\)

Answer:

a)

f  = 1.59*10⁸ Hz.

λ = 0.42 m

vp = 0.67*10⁸ m/s = 0.22 c

b)

γ = 0.31 1/m

Explanation:

a)

       \(\lambda = \frac{2*\pi }{k} =\frac{2*\pi }{(15)1/m} = 0.42 m (4)\)

       \(f = \frac{\omega }{2*\pi } =\frac{10e9rad/sec }{(2*\pi rad} = 1.59*e8 (6)\)

       \(v_{phase} =\lambda * f = 0.42m * 1.59e8 1/sec = 0.67e8 m/s (7)\)

b)

        \(V = 10 V *e^{-\gamma*x} (8)\)

       \(10 V *e^{-\gamma*3m} = 4 V (9)\)

         \(-\gamma*3m* ln e = -\gamma*3m = ln (\frac{4V}{10V}) = ln 0.4\\\gamma = -\frac{ln 0.4}{3m} = 0.31 (1/m) (10)\)

       γ = 0.31 1/m

Inquiry-based pedagogy empowers students to explore, question, and construct knowledge through active engagement. It encourages curiosity, critical thinking, and independent investigation, fostering a deeper understanding of concepts and skills that extend beyond the classroom.

Inquiry-based pedagogy is an approach to teaching and learning that emphasizes the active engagement of students in the exploration of meaningful questions, problems, or phenomena. It encourages students to ask questions, investigate, and construct their own knowledge through critical thinking, problem-solving, and hands-on experiences. Here's an example to illustrate the concept of inquiry-based pedagogy:

Example: Exploring Ecosystems

In a biology class, the teacher introduces the topic of ecosystems using an inquiry-based approach. The teacher poses a driving question to the students: "How do living organisms interact with their environment to form ecosystems?"

1. Questioning and Investigation: Students begin by generating their own questions related to ecosystems. They might wonder about the roles of different organisms, energy flow, or the impact of human activities. Guided by their questions, they conduct research, gather information from various sources, and share their findings.

2. Hands-on Exploration: The teacher organizes hands-on activities to allow students to observe and explore ecosystems firsthand. For example, they could set up mini-ecosystems in terrariums or conduct field trips to local habitats. Through these experiences, students can make observations, collect data, and analyze patterns.

3. Collaborative Learning: Students work in groups or pairs to analyze the data they have collected and draw conclusions. They engage in discussions, share their ideas, and challenge each other's thinking. This collaborative learning environment promotes critical thinking, communication, and teamwork.

4. Reflection and Presentation: Students reflect on their findings and insights gained from their investigations. They are encouraged to synthesize their learning into presentations, reports, or visual representations. These presentations provide opportunities for students to articulate their understanding and demonstrate their learning outcomes.

By engaging in inquiry-based learning, students develop essential skills such as critical thinking, problem-solving, communication, and self-directed learning. They become active participants in their own education, taking ownership of their learning process and developing a deeper understanding of the subject matter.

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Hello and Good Morning/Afternoon:

What makes a unit a fundamental SI unit:

Thus the seven fundamental S.I. units are:

Hope that helps!

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This is actually an advantage of a parallel circuit. In a parallel circuit, the resistance in each branch is the same as if there was just one light bulb in the whole circuit.

In a parallel circuit, the current has multiple pathways to flow through. Each component, such as a light bulb, is connected across its own branch. The total resistance in a parallel circuit is determined by the sum of the reciprocals of the individual resistances in each branch.

Having the same resistance in each branch ensures that the current is distributed equally among the branches. This means that each component receives the same amount of current as if it were the only component in the circuit. As a result, the brightness of the light bulbs in a parallel circuit will be the same.

This characteristic of parallel circuits is advantageous because it allows for independent operation of each component. If one light bulb were to burn out or be removed, the other bulbs would continue to operate unaffected. Additionally, parallel circuits offer lower overall resistance compared to a series circuit, enabling the flow of more current.

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A bob is the mass on the end of a pendulum found most commonly, but not exclusively, in pendulum clocks.

A basic pendulum consists of a light, flexible, inextensible thread with a heavy but tiny item, known as a "bob," at one end. Graph: Pendulum clocks employ it. It's used to calculate the acceleration caused by gravity.

Clock pendulums are typically composed of a weight or bob attached to the bottom end of a rod, with the top linked to a pivot so it can swing, despite the fact that a pendulum's shape or any rigid item hanging on a pivot is theoretically possible.

The benefit of this design is that it places the centre of mass farthest from the pivot, nearer to the actual end of the pendulum. This reduces the length of the pendulum needed for a specific period and increases moment of inertia. 

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Answer:

7 m/s²

Explanation:

From the question given above, the following data were:

Force applied (Fₐ) = 15 N

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Acceleration (a) of block =?

Next, we shall determine the frictional force. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Normal reaction (R) = mg = 1.5 × 10 = 15 N

Frictional force (Fբ) =?

Fբ = μR

Fբ = 0.3 × 15

Fբ = 4.5 N

Next, we shall determine the net force acting on the block. This can be obtained as follow:

Force applied (Fₐ) = 15 N

Frictional force (Fբ) = 4.5 N

Net force (Fₙ) =.?

Fₙ = Fₐ – Fբ

Fₙ = 15 – 4.5

Fₙ = 10.5 N

Finally, we shall determine the acceleration produced in the block. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Net force (Fₙ) = 10.5 N

Acceleration (a) of block =?

Fₙ = ma

10.5 = 1.5 × a

Divide both side by a

a = 10.5 / 1.5

a = 7 m/s²

Therefore, the acceleration produced in the block is 7 m/s²

Answer:6

Explanation:

Answer:

6

Explanation:

Answer:

b

Explanation:

The size of the angle θ of a point object moving from point A to point B along a circular path is 2πR / L.

To understand this, consider a simple example. Suppose that a point object that moves from point A to point B along a circular path with a radius of 1 meter. The distance between points A and B is also 1 meter. Therefore, the size of the angle θ is equal to 2π × 1 / 1 = 2π radians.

In general, the size of the angle θ = ratio of the circumference of the circle to the distance between points A and B. The circumference of the circle is equal to 2πR, where R = radius of the circle. Therefore, the size of the angle θ is equal to 2πR / L.

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Complete question:

A point object moves from point A to point B along a circular path with a radius R. What is the size of the angle θ?

The eastern component of the plane's displacement is 592.35 m.

The given parameters;

The eastern component of the plane's displacement is calculated by applying Pythagoras theorem as follows;

X²  + Y² = 599²

X² + 89² = 599²

X² = 599²  - 89²

X² = 350880

X = √350880

X = 592.35 m

Thus, the eastern component of the plane's displacement is 592.35 m

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