Answer:
weight, acceleration
Explanation:
weight = mass x gravity(meaning the direction of the mass)
acceleration = v-u/t
v-u is the change in velocity
explain why a meter bridge is unsuitable for comparison of low resistance
A meter bridge is not suitable for the comparison of low resistance.
A meter bridge is based on the principle of null deflection, the bridge is balanced by adjusting the known resistance until there is no current flowing through the galvanometer. But if they are low resistance values, then the current flowing through the bridge is relatively high. This situation leads to inaccurate measurements and potential damage to the components.
The meter bridge operation is based on the assumption that the resistance being measured is purely resistive. But in reality, low resistance values include parasitic inductance and capacitance. This will affect the accuracy of the measurement.
The sensitivity of a meter bridge decreases as the resistance values being compared decrease.
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A block of mass
m = 2.50 kg
is pushed
d = 2.30 m
along a frictionless horizontal table by a constant applied force of magnitude
F = 18.0 N
directed at an angle
= 25.0°
below the horizontal as shown in the figure below.
The work done on the block by the applied force is approximately -23.04 Joules.
The negative sign indicates that the work done is in the opposite direction of the displacement, as the force is directed below the horizontal.
In this scenario, the block of mass 2.50 kg is being pushed along a frictionless horizontal table by a constant applied force of magnitude 18.0 N, directed at an angle of 25.0° below the horizontal.
To determine the work done on the block by the applied force, we can use the formula:
Work = Force * Distance * cos(theta)
where the force is 18.0 N, the distance is 2.30 m, and theta is the angle between the force and the direction of displacement.
Since the force is directed at an angle below the horizontal, the angle theta is 180° - 25° = 155°, as the angle is measured with respect to the horizontal axis.
Now we can calculate the work done:
Work = 18.0 N * 2.30 m * cos(155°)
The cosine of 155° is negative, as it is in the third quadrant.
Work = 18.0 N * 2.30 m * (-0.57358)
Work ≈ -23.04 J
Therefore, the work done on the block by the applied force is approximately -23.04 Joules.
The negative sign indicates that the work done is in the opposite direction of the displacement, as the force is directed below the horizontal.
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What is the central idea of the passage? "Wild life on rockies"
The central idea of the passage "Wild life on rockies" is that Nature is a beautiful but dangerous force that should be treated with respect.
Option C is correct.
What is meant by wildlife?Wildlife refers to undomesticated animal species, but has come to include all organisms that grow or live wild in an area without being introduced by humans.
In the Wildlife, Animals that live independently of people, in natural conditions.
The Wildlife in the area includes deer, bears, and raccoons.
In conclusion, Nature is a beautiful but dangerous force but should be accorded respect.
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As the book fell through the air and eventually hit the ground,
A. potential energy was destroyed and kinetic energy was create
B. sound energy was created
C. both potential energy and kinetic energy were destroyed
D. the energies of the system were converted from one form to
another
As the book fell through the air and eventually hit the ground, the energies of the system were converted from one form to another.
Option D is correct.
What is energy?Energy is described as the quantitative property that is transferred to a body or to a physical system, recognizable in the performance of work and in the form of heat and light.
The forms of energy includes:
Chemical energy.
Electrical Energy.
Mechanical Energy.
Thermal energy.
Nuclear energy.
Gravitational Energy.
In conclusion, When the book fell through the air and hit the ground, the potential energy it possessed due to its position above the ground was converted into kinetic energy as it accelerated towards the ground.
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A circle loop of radius 2 m is positioned in a uniform magnetic field of magnitude 1.5 N/C so that the plane of the loop makes an angle of 65° with the magnetic field. Find the flux passing through the circle loop.
The flux through the circle loop of wire is determined as 7.96 N/C.m².
What is the flux passing through the circle loop?The flux through the circle loop of wire is calculated by applying the following formula.
Ф = EA cosθ
where;
B is the magnitude of the electric fieldA is the area of the circular loopθ is the direction of the loopThe area of the wire is calculated as follows;
A = πr²
A = π (2 m)²
A = 12.57 m²
The flux through the circle loop of wire is calculated as;
Ф = EA cosθ
Ф = 1.5 x 12.57 x cos (65)
Ф = 7.96 N/C.m²
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Determine the energy released per kilogram of fuel used.
Given MeV per reaction, calculate energy in joules per kilogram of reactants.
Consider 1 mile of tritium plus 1 mole of deuterium to be a mole of "reactions" ( total molar mass = 5 grams)
The energy released per kilogram of fuel used is 3.39 * 1014 J/Kg
Why is the energy released in a reaction?Energy is released in a reaction because of the breaking of bonds are well as formation of bonds.
The quantity of energy released in reactions differs according to the reaction type involved.
The energy released in nuclear reactions are far larger than that released in chemical reactions due to the release of nuclear energy from the nucleus.
The energy, E released in nuclear reactions is given by the formula below:
Energy per kilogram of reactants (in joules) = Energy per mole of reactants (in joules) / Total molar mass of reactants (in kg)
Energy per kilogram of reactants (in joules) = [tex]1.60218 x 10^-^1^3 joules[/tex] / 5 grams
Energy per kilogram of reactants (in joules) = 3.39 * 1014 J/Kg
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The energy released per kilogram of fuel used is 5.632 × 10^-14 J/kg.
5 grammes, 0.005 kilogrammes, of reactants are provided.
Since 1 mile of tritium and 1 mile of deuterium are equal to 1 mole of "reactions," the sum of the "reactions" in 0.005 kilogrammes of reactants may be computed as follows:
Total moles of "reactions" in 0.005 kg of reactants = (0.005 kg / 5 g/mol)
= 0.001 mole.
The MeV per reaction must now be multiplied by the total number of "reactions" in order to get the total energy released by the "reactions."
Next, the energy must be converted from MeV to Joules.
MeV to Joules conversion factor is 1.6 10-13 J/MeV.
Total energy released = (MeV per reaction) x (number of reactions) x (conversion factor)
Total energy released = (17.6 MeV/reaction) x (0.001 mole) x (1.6 × 10^-13 J/MeV)
Total energy released = 2.816 × 10^-16 J
The total energy released by the "reactions" is 2.816 × 10^-16 J.
To determine the energy released per kilogram of fuel used, we need to divide the total energy by the mass of fuel used.
Total energy released per kilogram of fuel used = (total energy released) / (mass of fuel used)
Total energy released per kilogram of fuel used = (2.816 × 10^-16 J) / (0.005 kg)
Total energy released per kilogram of fuel used = 5.632 × 10^-14 J/kg
Therefore, the energy released per kilogram of fuel used is 5.632 × 10^-14 J/kg.
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Two objects gravitationally attract with force of 1,926N. If mass 1is multiplied by 0.38,mass2 is multiplied by 3.21, and the distance is multiplied by 2.25,the new force of attraction is:
The new force of attraction is found to be 813.25 N.
How do we calculate?We that F_ = G * (m1 * m2) / r²
F_ = force of gravitational attraction,
G = gravitational constant = [tex]6.67430 * 10^-^1^1[/tex]N(m/kg)²
m1 and m2 = masses of the two objects, and
r = distance between the centers of the two objects.
new mass = 0.38 * m1
new mass = 3.21 * m2_o
new distance = 2.25 * r
new force = F * (m1 * m2) / (m1 * m2) * (r² / r²)
new force= 1926 N * (0.38 * m1_original * 3.21 * m2_original) / (m1_original * m2_original) * (r_original²/ (2.25 * r_original)²)
new force = 1926 N * 0.38 * 3.21 / (2.25²)
new force = 813.25 N
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Unger homework!!!!! please guys can you help me? I don't have the textbook.
Textbook: the developing person through the Lifespan 11th edition.
Characters 20, 21, 22
you are to pick one area (biosocial, cognitive, or psychosocial) of development and apply it to your specific development. Once you have read the chapters, some aspects of one of these three chapters or areas should resonate with you such that you see yourself in your textbook readings. That's what I want you to center your discussion around.
What did you learn about yourself while reading one or all three of these chapters?
Are there any areas of the reading that made you think, "That's me! or "That's where I am in my life right now!" or "Wow, I hadn't considered that!"
Take a deep dive into the chapter content, and really apply and integrate what you've read to your personal life and developmental experiences.
Biosocial development focuses on the physical and biological changes that occur throughout the lifespan. It encompasses aspects such as growth, motor skills, and changes in the body's systems.
Cognitive development refers to the growth and changes in thinking, memory, problem-solving, and intelligence. Psychosocial development involves the social, emotional, and psychological aspects of development, including identity formation, relationships, and moral reasoning.
When reading these chapters, you might come across various concepts and theories that shed light on different stages of development and the challenges individuals may face during those stages.
By reflecting on the content, you can gain insights into your own developmental journey, identifying areas that resonate with your experiences, thoughts, and behaviors.
It can provide a deeper understanding of yourself and potentially open up new perspectives on your personal growth.
Remember, it's important to engage with the material and consider how it relates to your own life experiences rather than relying solely on the textbook.
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inquiry based pedagogy is being embraced in principle across the globe . in the last decade , it has been supported by an increasing body of research on its effectiveness . illustrate your understanding of the concept of inquiry based pedagogy with a relevant example.( 10 marks)
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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How does uplift change the surface of Earth?
Answer options with 4 options
A.
Snow melts on the surface of Earth, causing rivers to form.
B.
Magma rises from the surface of Earth, causing lava beds to form.
C.
Wind blows across the surface of Earth, causing sand dunes to form.
D.
Pressure builds under the surface of Earth, causing mountains to form.
Pressure builds under the surface of Earth, causing mountains to form.The correct answer is option D.
Uplift refers to the geological process that elevates the Earth's surface, resulting in the formation of mountains. This process is primarily driven by tectonic forces, including the movement and collision of Earth's lithospheric plates.
When two plates converge, immense pressure builds up beneath the surface, causing the crust to buckle and fold. This deformation leads to the formation of mountains, as rocks are pushed upward and displaced vertically.
As the uplift process continues over millions of years, mountains gradually take shape. Erosion and weathering play significant roles in shaping their features, but it is the initial uplift that initiates the formation of mountains.
As the Earth's surface is elevated, a wide range of landforms can emerge, including rugged peaks, deep valleys, and steep slopes.
Uplift has a profound impact on the Earth's surface and ecosystems. Mountains alter local climates, influencing precipitation patterns and creating variations in temperature and wind patterns.
Therefore, uplift plays a crucial role in shaping the Earth's surface and influencing various geological, biological, and climatic processes.
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1000 cm³ of air at 10°c is heated to 80°c. what will be the new volume if the pressure remained atmospheric?
The new volume of the air is 1247.35 cm³.
What is volume?Volume is the space occupied by a matter. The S.I unit of volume is cubic meter (m³).
To calculate the volume of the air, we use the formula below.
Formula:
V/T = V'/T'............................ Equation 1Where:
V = Initial volume of airV' = Final volume of the airT = Initial temperature of airT' = Final temperature of airFrom the question,
Given:
V = 1000 cm³T = 10 °C = 283 KT' = 80 °C = 353KSubstitute these values into equation 1 and solve for V'
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What is the difference between classical mechanics and quantum mechanics?
Classical mechanics describes the motion of objects on a macroscopic scale, while quantum mechanics deals with the behavior of particles on a microscopic scale. Classical mechanics is deterministic, meaning that it predicts precise outcomes based on initial conditions, while quantum mechanics is probabilistic, providing probabilities of different outcomes. Classical mechanics follows the principle of causality, where every effect has a specific cause, whereas quantum mechanics introduces inherent uncertainty and wave-particle duality. Classical mechanics is well-suited for describing everyday objects, while quantum mechanics is necessary to explain the behavior of particles at the atomic and subatomic levels.
~~~Harsha~~~
