Two electrical circuits have the same 15-volt batteries. Circuit A has 2 ohms of resistance, and circuit B has 6 ohms of resistance. Which of the following statements is correct?
 (1 point)
• Circuit B has the same amount of current as circuit A.
• Circuit B has more current than circuit A, as it has less resistance.
• Circuit B has less current than circuit A, as it has more resistance.
• Circuit B has less current than circuit A, as it has less resistance.

The relative velocity of swimmer to the bank is [tex]\boxed{10.21\text{ m/s}}[/tex] and [tex]\boxed{21.54^\circ}[/tex] downstream.

Further Explanation:

The swimmer swims across the river from one end to the other. The river is flowing downstream.

The swimmer will cross the river with some relative velocity.

Given:

The velocity of swimmer is [tex]9.50\text{ m/s}[/tex].

The velocity of river is [tex]3.75\text{ m/s}[/tex] downstream.

Concept:

Consider the direction of velocity of swimmer in positive x-direction.

The velocity of swimmer in vector form:

[tex]V_s=9.50\hat i[/tex]

Consider the direction of downstream current of river in negative y-direction.

The velocity of downstream current in vector form:

[tex]V_r=- 3.75\hat j[/tex]  

The relative velocity of swimmer to the bank is the resultant of two vectors in x and y direction.

Magnitude of relative velocity:

[tex]{V_{sr}}=\sqrt{V_r^2+V_s^2}[/tex]  

Substitute [tex]9.50\text{ m/s}[/tex] for [tex]V_s[/tex] and [tex]3.75\text{ m/s}[/tex] for [tex]V_r[/tex] in above equation.

[tex]\begin{aligned}{V_{sr}}&=\sqrt{3.75^2+9.50^2}\text{ m/s}\\&=10.21\text{ m/s}\end{aligned}[/tex]

The magnitude of relative velocity of swimmer is [tex]10.21\text{ m/s}[/tex] .  

The direction of relative velocity:

[tex]\theta={\tan^{-1}}\left({\dfrac{{{V_r}}}{{{V_s}}}}\right)[/tex]  

Substitute [tex]9.50\text{ m/s}[/tex] for [tex]V_s[/tex] and [tex]-3.75\text{ m/s}[/tex] for [tex]V_r[/tex] in above equation.

[tex]\begin{aligned}\theta&={\tan^{-1}}\left({\dfrac{{{-3.75}}}{{{9.50}}}}\right)\\&=21.54^\circ\end{aligned}[/tex]

The direction of relative velocity is [tex]21.54^\circ[/tex] downstream.

Thus, the relative velocity of swimmer to the bank is [tex]\boxed{10.21\text{ m/s}}[/tex] and [tex]\boxed{21.54^\circ}[/tex] downstream.

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Answer Details:

Grade: Middle School

Subject: Physics

Chapter: Scalars and vectors

Keywords:

Swimmer, still, water, speed, 9.50 m/s, intends, directly, river, downstream, current, 3.75 m/s, velocity, relative, bank, vector, direction, 21.5degree and 10.21 m/s.

Answer:

2 m

Explanation:

Because there will be the greatest amount of force when they are closest together.

The distance at which the greatest gravitational force is exerted by one object on another if Two objects of the same mass are in a room should have the shortest distance, so option A is correct.

The universal force of attraction operating on all matter is known as gravity, usually spelled gravitation, in mechanics. The weakest known force in nature, it is by far and does not influence how everyday matter's internal properties are determined.

Measurements of gravity around the Earth, other planets, and the Moon, as well as experiments on the nature of gravitation, have greatly improved as a result of the launch of spacecraft and the research that has resulted from it.

The formula for the gravitational force is inversely proportional to the square of the distance between two objects,

Thus, to get the maximum gravitational force, the mass is to be at the lowest separation.

Therefore, the distance at which the greatest gravitational force be exerted by one object on another if Two objects of the same mass are in a room should have the shortest distance.

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Answer:

80 ft/s

Explanation:

Use III equation of motion

V^2 = U^2 + 2g h

Here, U = 0, g = 32 ft/s^2, h = 100 ft

V^2 = 0 + 2 × 32 ×100

V^2 = 6400

V = 80 ft/s

Answer: Cricket chirping increases with an increase in temperature.

Explanation:

The independent variable in the experiment is the variable which can be altered, manipulated and changed so as to determine it's influence over the dependent variable, which is the outcome of the experiment.

Here in the given experiment, the number of chirps is the dependent variable which is increasing due to the increase in the temperature which is the independent variable.

The object fell a distance of 396.9 meters before hitting the ground.

To calculate the distance an object falls in a given time ignoring air resistance, we use the formula for the distance covered under constant acceleration: d =  rac{1}{2} g t^2, where d is the distance, g is the acceleration due to gravity (

9.8 m/s2), and t is the time in seconds. Substituting the given time of 9.0 seconds into the equation, the distance is calculated as d =  rac{1}{2}  imes 9.8  imes 9.0^2, which equals 396.9 meters. This equation is derived from basic kinematic principles that govern the motion of objects in free fall, assuming gravity is the only force acting on the object.

Potential energy is stored energy due to an object's position or state, while kinetic energy is the energy of an object in motion. They differ in their state, with potential energy being at rest and kinetic energy being in action. However, both can be transformed into each other.

Potential energy and kinetic energy are both forms of mechanical energy, but they are different in many ways. Potential energy is stored energy that an object has due to its position or state, like a rock perched at the top of a hill has gravitational potential energy. It has the potential to convert into kinetic energy but for now, it remains at rest.

On the other hand, kinetic energy is the energy of motion. An object in motion has kinetic energy— for example, a moving car or a flowing river has kinetic energy. Unlike potential energy, it is already in action. So the key difference is that potential energy is stored, while kinetic energy is in motion.

However, they are similar in that they both can be transformed into each other. For instance, when that rock on the hill begins to roll downhill, its potential energy is converting into kinetic energy.

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You could tell the difference between two envelopes containing 4 and 5 sheets of paper by comparing their weights. An envelope with 5 sheets should feel heavier than the one with 4 sheets, assuming that all materials involved are identical.

One way to tell the difference between two envelopes containing 4 and 5 sheets of paper without opening them would be to use weight as a distinguishing factor. Given that the weight of paper is constant, one sheet of paper would add a perceivable amount of weight to an envelope. Therefore, the envelope with 5 sheets should be heavier than the envelope containing 4 sheets. The assumption here is that both the envelopes and the sheets of paper are identical in every other aspect including size, material, weight, etc.

In theory, if you were to hold each envelope in either hand, the difference in weight should be detectable. It's a common practice in mailrooms where workers are adept at figuring out the number of papers inside envelopes just by weight comparison. This method, however, requires a certain level of sensitivity and practice. The real application of this method can vary based on multiple factors such as individual's sensitivity to weight change, uniformity in the weight of the paper etc.

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Answer:

The Atmosphere

Explanation:

The speed of the satellite is approximately [tex]\(3311 \, \text{m/s}\)[/tex] when the force acting on it is 87.4 N.

The gravitational force [tex](\(F_{\text{gravity}}\))[/tex] acting on the satellite is given by [tex]\(F_{\text{gravity}} = \frac{GMm}{r^2}\)[/tex], where [tex]\(G\)[/tex] is the gravitational constant, [tex]\(M\)[/tex] is the mass of the Earth, [tex]\(m\)[/tex] is the mass of the satellite, and [tex]\(r\)[/tex] is the distance from the center of the Earth to the satellite.

The centripetal force [tex](\(F_{\text{centripetal}}\))[/tex] required for circular motion is [tex]\(F_{\text{centripetal}} = \frac{m v^2}{r}\)[/tex], where [tex]\(v\)[/tex] is the speed of the satellite.

In a circular orbit, the gravitational force provides the centripetal force, so [tex]\(F_{\text{gravity}} = F_{\text{centripetal}}\)[/tex]:

[tex]\[\frac{GMm}{r^2} = \frac{m v^2}{r}\][/tex]

Solve for \(v\):

[tex]\[v = \sqrt{\frac{GM}{r}}\][/tex]

Now, substitute the given values:

[tex]\[v = \sqrt{\frac{(6.67 \times 10^{-11} \, \text{m}^3/\text{kg}\cdot\text{s}^2) \times (5.97 \times 10^{24} \, \text{kg})}{(3.00 \times 10^7 \, \text{m} + 6.37 \times 10^6 \, \text{m})}}\][/tex]

Calculating the expression to find the speed of the satellite. I'll do the math for you.

[tex]\[v = \sqrt{\frac{(6.67 \times 10^{-11} \, \text{m}^3/\text{kg}\cdot\text{s}^2) \times (5.97 \times 10^{24} \, \text{kg})}{(3.00 \times 10^7 \, \text{m} + 6.37 \times 10^6 \, \text{m})}}\][/tex]

[tex]\[v = \sqrt{\frac{(6.67 \times 10^{-11} \times 5.97 \times 10^{24})}{(3.00 \times 10^7 + 6.37 \times 10^6)}}\][/tex]

[tex]\[v = \sqrt{\frac{3.9879 \times 10^{14}}{3.637 \times 10^7}}\][/tex]

[tex]\[v \approx \sqrt{1.097 \times 10^7}\][/tex]

[tex]\[v \approx 3311 \, \text{m/s}\][/tex]

Therefore, the speed of the satellite is approximately [tex]\(3311 \, \text{m/s}\)[/tex] when the force acting on it is 87.4 N.

the force is gravity holding them

Answer:

B.  

more than 150 meters

Explanation:

As we know that the range of projectile motion on the surface of earth is given by the formula

[tex]R = \frac{v^2sin(2\theta)}{g}[/tex]

so here on the surface of earth the gravity is given as g = 9.8 m/s/s

now if the golf ball is hit on the surface of earth then it range is given as R = 150 m

while if we hit the ball at surface of mercury then on the surface of mercury the gravity will be g = 3.7 m/s/s

so due to decrease in the magnitude of gravity the range on the surface of mercury will be more than 150 m

Answer:

B. More that 150 meters.

Explanation:

:)

The multiplication of (6.2 × 10^4) by (3.3 × 10^2) results in 2.046 × 10^7, when expressed in scientific notation.

The result of the multiplication (6.2 × 104) × (3.3 × 102), expressed in scientific notation, involves multiplying the coefficients (6.2 and 3.3) and then adding the exponents of 10 (4 and 2). It's important to recall that we must perform both steps to arrive at the correct answer in scientific notation.

The product of the coefficients is 20.46 and when we add the exponents the result is 106. So the multiplication gives us 20.46 × 106, but in correct scientific notation, we want the coefficient to be between 1 and 10. To achieve this we adjust the coefficient to 2.046 and increase the exponent by 1, giving us the result: 2.046 × 107.

The relationship between height and gravitational potential energy is directly proportional, following the equation G.P.E. = mgh. GPE increases as the height of an object increases, and it is measured in Joules, reflecting the energy due to the object's position above a reference level.

The relationship between height (h) and gravitational potential energy (GPE) is directly proportional. This means, as an object is raised to a higher elevation above the ground, its GPE increases. The formula for calculating GPE is G.P.E. = mgh, where m represents the mass of the object in kilograms, g is the acceleration due to gravity (approximately 10 m/s² on Earth's surface), and h is the height in meters. The unit of GPE is Joules, which quantifies the energy associated with the object's position in the gravitational field.

In physics, only the change in GPE (ΔGPE) is of significance, which is measured as the difference in energy when an object's height changes relative to a reference level. For instance, if a 0.500-kg mass from a cuckoo clock is lifted by 1.00 m, its GPE changes by mgh. The concept of GPE is vital in various applications, including understanding how energy is conserved within systems influenced by gravitational forces.

Answer:

B. C + O2 ------> CO2

Explanation:

i took the test

The right option is; C) the abyssal zone of the ocean

The anglerfish are likely to be found in the abyssal zone of the ocean.  

The abyssal zone is the bottomless layer of the pelagic zone of the ocean that has the bathyal zone located on it.  The abyssal zone is at depths of 4,000 to 6,000 metres, and it remains permanently in darkness. This zone is a food limited environment because it lacks nutrients. The temperature of the zone is around 2 to 3 °C and has continuous cold.

Two rocks hit against each other in a fast-flowing stream and break apart. Mosses grow on a rock and produce acids that wear away the rock over time. These statements describe physical weathering. Therefore, option A and D are correct.

Physical weathering, often known as mechanical weathering, is the process that fractures rocks without altering their chemical makeup. The following instances show physical weathering: water moving quickly. For brief periods of time, swiftly rushing water has the ability to raise rocks out of the streambed.

Rocks break down into tiny fragments due to abrasive weathering processes like wind and water. Exfoliation weathering rocks shatter and expand as pressure from unloading is relieved. As water freezes and expands, frost-wedging rocks crack.

When a rock is subjected to physical processes like temperature changes or exposure to the effects of wind, rain, and waves, physical weathering takes place.

Thus, option A and D are correct.

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Final answer:

Statements A and C describe physical weathering, which involves the purely physical breakdown of rocks, such as through collision in a stream or animal activity, without altering their chemical composition.

Explanation:

The question asks which statements describe physical weathering. Physical weathering, or mechanical weathering, involves the breakdown of rocks into smaller pieces without changing their chemical composition. Examples of physical weathering include the expansion and contraction of rocks due to temperature changes, the abrasive action of water or wind, and biological factors like plant roots breaking rocks apart.

A. Two rocks hit against each other in a fast-flowing stream and break apart.

C. Small rocks are pushed together when a mole digs an underground den.

Statement A is an example of physical weathering because it describes the physical action of rocks breaking apart due to collision in a stream. Statement C also illustrates physical weathering, showing how animal activity can physically move and potentially break rocks. In contrast, statements B and D describe chemical weathering, which involves the chemical alteration of rocks, such as mineral changes through oxidation or the action of acids.