Present at least one example that illustrates acceleration. Is this a scalar or vector quantity? Explain why. ...?

Correct answer choice is :

C) It is a negative ion that has one more valence electron than a neutral bromine atom.

Explanation:

A bromide is a synthetic composite including a bromide ion or ligand. Potassium bromide (KBr) is a salt, usually selected as an anticonvulsant and a drug in the late 19th and early 20th centuries, with over the stand value increasing to 1975 in the US. Potassium bromide is applied as a veterinary drug, as an antiepileptic medicine for dogs.

C. It is a negative ion that has one more valence electron than a neutral bromine atom.

Explanation of how a lever works and how to calculate its mechanical advantage when lifting a load with a given input force.

A lever is a simple machine that consists of a rigid bar pivoted at a fixed point called the fulcrum. It helps in exerting a smaller force over a longer distance to lift a heavier load over a shorter distance.

The mechanical advantage of a lever is calculated by dividing the length of the effort arm by the length of the load arm (MA = Le/Lr). It tells how many times a lever multiplies the input force to lift a load.

In the described scenario, with an input force of 500 N and a load of 2000 N, the mechanical advantage of the lever can be calculated using the formula MA = Load / Effort, which will be 4.

Answer:

If you break a bar magnet in half, each half becomes a magnet.

Explanation:

A substance which attracts or repels another similar substance is known as magnet. When domains align in single direction in a substance, it acts as a magnet. A magnet has two poles- North pole and South pole. Like poles repel each other and unlike poles attract each other. Mono-poles do not exist. So, when a magnet is broken into two halves, each half forms another magnet with two poles.

Answer:

If you break a bar magnet in half, each half will have a new south pole, a north pole and a neutral zone.  In other words, two new magnets will be generated.

Explanation:

A magnet is a body of any material capable of producing a magnetic field and attracting itself or being attracted to another magnet or to any other body of iron, cobalt or other ferromagnetic metals (ferromagnetism is a physical phenomenon in which magnetic ordering occurs of all the magnetic moments of a sample, in the same direction and direction. In other words, the ferromagnetic interaction is the magnetic interaction that makes the magnetic moments tend to be arranged in the same direction and direction. This property makes the ferromagnetic materials they are intensely magnetized when they are placed in a magnetic field, and retain part of their magnetization when said field disappears.). Then the magnet is a material with natural or artificial ferromagnetic properties, which generate a continuous magnetic field.

Magnets are magnetically charged bodies, which generate a magnetic field around them oriented according to two poles: negative pole (also called South pole) and positive pole (also called North pole). Opposites attract each other (positive-negative) and equal poles repel each other (positive-positive or negative-negative). The line that joins both poles is called the magnetic axis. This line located in the central zone located between both poles has no attraction or repulsion capacity.

In the event that a magnet is broken into pieces, each new fragment will have a new south pole, a north pole and a neutral zone. That is, equal and opposite poles appear on each side of the breaking point. This happens even if the fragments are of different size.

Answer: B. heat

Explanation:

The nuclear reaction involves the simultaneous fission and fusion of the heavy metal nuclei such as uranium in control conditions of the nuclear reactor so as to produce energy in the form of heat. The heat or thermal energy is used to boil the water and convert it the form of steam. The steam is used to run the turbine of the generator which produces electricity.  

Casey can organize physics formulas by categorizing them based on topics, arranging them sequentially, using color-coded tabs, prioritizing key formulas, and creating a concise index for quick reference during the open book test.

Casey can adopt a systematic approach to organize his physics formulas for the upcoming open book test, promoting efficiency and accessibility during the examination. Firstly, he can categorize the formulas based on the topics or chapters they belong to. This helps create a structured framework that aligns with the course syllabus, making it easier for Casey to locate specific formulas relevant to different sections of the test.

Within each category, Casey may further arrange the formulas in a logical sequence, following the order of topics covered in the class. This sequential arrangement aids in maintaining a smooth flow while navigating through the formulas during the test, reducing the time spent searching for the right information. Additionally, he could use color-coded tabs or highlighters to distinguish between different formula categories, providing a visual aid for quick identification.

Considering the importance of some formulas over others, Casey might prioritize key formulas or those frequently used in class. By placing these prominently at the beginning of each category, he ensures immediate access to crucial information. Moreover, creating a concise index or summary at the front of his formula compilation can serve as a quick reference guide, aiding Casey in quickly identifying the page numbers corresponding to specific formulas.

By implementing these strategies, Casey can enhance his organization of physics formulas, facilitating a more efficient and stress-free experience during the open book test.

The question probable maybe: What organizational method could Casey employ to effectively categorize and arrange the various physics formulas for his open book test, ensuring easy access and streamlined use during the examination?

Final answer:

The rate at which the area of the triangle increases when the length of one of the equal sides is 10mm is 0 mm²/s.

Explanation:

To find the rate at which the area of the triangle is increasing, we can use the formula for the area of a right triangle, which is (1/2) * base * height. Since the triangle is isosceles, the base and height are equal. Let's call the length of the equal side x. The hypotenuse is also related to x by the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

We can set up the equations:

x² + x² = c²

2x² = c²

Taking the derivative of both sides concerning time, we get:

4x * (dx/dt) = 2c * (dc/dt)

Plugging in the given values, where dx/dt = 0 (since x is constant) and dc/dt = 2mm/s, we can solve for the rate at which the area is increasing:

4(10) * (0) = 2(10.3) * (dA/dt)

0 = 20.6 * (dA/dt)

(dA/dt) = 0 mm²/s

By definition we have to:

A system or frame of reference are those conventions used by an observer (usually standing at a point on the ground) to be able to measure the position and other physical magnitudes such as speed and acceleration of one or several objects.

The numerical value of some magnitudes can also be relative to the reference system when we refer to the relative movement. There are always mathematical relationships between the observer and the relative magnitudes.

Answer:

C. a position from which something is observed

The spring constant can be calculated using Hooke's Law. By determining the force exerted by the added mass on the spring and dividing that by the distance the spring is stretched, the spring constant is found to be 6.37 N/m.

This question is regarding the concept of Hooke's law in physics, which states that the force needed to extend or compress a spring by some distance is proportional to that distance. Given that the spring stretches an additional 10 cm when a 65g mass is added, we calculate the force exerted by the mass on the spring as F = m*g = 0.065 kg * 9.79 m/s² = 0.637 N.

Then, using the equation from Hooke's Law, F = kx, where F is the force, k is the spring constant, and x is the distance the spring is stretched, we can calculate the spring constant as k = F / x = 0.637 N / 0.1 m = 6.37 N/m.

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Answer:

15.49m/s

Explanation:

Using one of the equations of motion;

[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2gs

Where;

v = the final velocity of the rock.

=> This is the speed at which the rock strikes the water.

u = initial velocity of the rock.

=> The rock is just dropped from that height. That means the initial velocity is zero (0).

=> u = 0

g = the acceleration due to gravity.

=> Since the rock moves downwards, the acceleration due to gravity is positive and let it have a value of 10m/[tex]s^{2}[/tex]

=> g = 10m/[tex]s^{2}[/tex]

s = the distance covered by the rock

=> s = 12m

Substituting these values into the equation above gives

[tex]v^{2}[/tex] = [tex]0^{2}[/tex] + 2(10 x 12)

[tex]v^{2}[/tex] = 240

v = [tex]\sqrt{240}[/tex]

v = 15.49m/s

Therefore the velocity(speed) with the rock strikes the water is 15.49m/s

Using the work-energy principle, the average force exerted by water is found to be approximately 1639.52 N. This force is non-conservative because it involves energy dissipation in the water.

(a) Determine the magnitude of the average force that the water exerts on the diver:

The average force acting underwater is bringing the diver to rest. Thus, we can say that the force (F) will be negative, or it acts opposite to the direction of displacement (d). In this case, the work done underwater should also be negative, as W = Fdcosθ, where θ = angle between force and displacement = 180° (for this case).

Displacement under water = d = 1.14 m

According to the work-energy principle, the change in kinetic energy of the diver is equal to the work done by the force (F) exerted by the water.

First, we need to find the velocity of the person just before hitting the water.

We can use the kinematic equation:

v² = u² + 2gh

where u is the initial velocity (which is 0, since the person jumps from rest), g is the acceleration due to gravity (9.8 m/s² downward), and h is the height (2.98 m downward).

Therefore:

v² = 0 + 2 × (-9.8 m/s²) × (-2.98 m)
v² = 58.408 m²/s²
v = √58.408 m/s
v ≈ 7.64 m/s

Applying the work-energy principle, we can write:

Work = Fdcosθ = -Fd = 1/2 × m × v²

Rearranging this equation to solve for F:

F = (1/2 × m × v²) / d

Substituting the values:

F = (0.5 × 64.0 kg × (7.64 m/s) ²) ÷ 1.14 m
F = (0.5 × 64.0 × 58.424) ÷ 1.14
F ≈ 1639.52 N

(b) Is this force non-conservative?

A non-conservative force is one where the work done depends on the path taken. In this case, the force exerted by the water is non-conservative because it involves dissipative forces like drag and other resistances which convert mechanical energy into heat and other forms of energy.

The strength of the magnetic field is approximately 8.22 T.

To solve for the magnetic field strength B, we use the equation that relates the force due to gravity to the magnetic force.

When the rod is floating, the gravitational force is balanced by the magnetic force. The gravitational force is given by [tex]\( F_g = m \cdot g \)[/tex], and the magnetic force is given by [tex]\( F_m = I \cdot L \cdot B \), where \( B \)[/tex] is the magnetic field strength.

Setting the gravitational force equal to the magnetic force, we have:

[tex]\[ m \cdot g = I \cdot L \cdot B \][/tex]

Now we can solve for B:

[tex]\[ B = \frac{m \cdot g}{I \cdot L} \][/tex]

Given the values:

[tex]- \( m = 40.6 \) \\g \( = 40.6 \times 10^{-3} \) kg (since 1 g \( = 10^{-3} \) kg),\\ - \( g = 9.81 \) m/s^2,\\ - \( I = 0.229 \) A\\ - \( L = 2.09 \) m,[/tex]

we can plug these into the equation:

[tex]\[ B = \frac{40.6 \times 10^{-3} \text{ kg} \cdot 9.81 \text{ m/s}^2}{0.229 \text{ A} \cdot 2.09 \text{ m}} \] \[ B = \frac{40.6 \times 10^{-3} \cdot 9.81}{0.229 \cdot 2.09} \] \[ B = \frac{0.4 \cdot 9.81}{0.47711} \] \[ B = \frac{3.924}{0.47711} \] \[ B \approx 8.22 \text{ T} \][/tex]

Setting the gravitational force equal to the magnetic force, we have:

[tex]\[ m \cdot g = I \cdot L \cdot B \][/tex]

Now we can solve for B:

[tex]\[ B = \frac{m \cdot g}{I \cdot L} \][/tex]

Given the values:

[tex]- \( m = 40.6 \) g \( = 40.6 \times 10^{-3} \) kg (since 1 g \( = 10^{-3} \) kg), - \( g = 9.81 \) m/s², - \( I = 0.229 \) A, - \( L = 2.09 \) m,[/tex]

we can plug these into the equation:

[tex]\[ B = \frac{40.6 \times 10^{-3} \text{ kg} \cdot 9.81 \text{ m/s}^2}{0.229 \text{ A} \cdot 2.09 \text{ m}} \] \[ B = \frac{40.6 \times 10^{-3} \cdot 9.81}{0.229 \cdot 2.09} \] \[ B = \frac{0.4 \cdot 9.81}{0.47711} \] \[ B = \frac{3.924}{0.47711} \] \[ B \approx 8.22 \text{ T} \][/tex]

Therefore, the strength of the magnetic field is approximately 8.22 T.

To find the ball's speed in uniform circular motion, we can use the concept of tension, gravitational force, and trigonometry. By finding the tension in the string and using the formula for speed, we can calculate the ball's speed.

To find the ball's speed, we can use the concept of circular motion. The tension in the string provides the centripetal force that keeps the ball moving in a circle. We can use the vertical component of the tension to find the gravitational force acting on the ball. The relationship between the tension, gravitational force, and the angle can be used to solve for the speed of the ball.

Using trigonometry, we can determine that the tension is equal to the gravitational force divided by the cosine of the angle. So, T = m * g / cos(26°).

Once we have the tension, we can use it to find the speed of the ball using the formula v = √(T / m), where v is the speed, T is the tension, and m is the mass of the ball. Plugging in the values, we can calculate the speed of the ball.

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By comparing the Waxing Crescent and Waxing Gibbous lunar phases, it can be seen that the Waxing Crescent is more illuminated while the Waxing Gibbous is less illuminated. Hence option A is correct.

The second stage of the cycle of phases is the Waxing Crescent. Once a month, this Moon phase lasts for 7.38 days before transitioning into the First Quarter phase. It rises at 9 AM and sets at 9 PM. The reason this phase is known as the Waxing Crescent is because the region of the Moon's surface that is lit resembles the shape of a crescent, and waxing refers to growth. The Earth, Moon, and Sun are practically perpendicular at this phase because it is one cycle away from the First Quarter phase. This indicates that the gravitational attraction of the tides from the Sun and Moon cancels out, resulting in a smaller tidal pull. At this time, the Earth's tides are practically at neap tide.

Hence option A is correct.

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The correct answer to compare the lunar phase of the Waxing Crescent to the Waxing Gibbous is D: The Waxing Crescent is less than half illuminated and the Waxing Gibbous is more than half illuminated.

Both phases are increasing in illumination as the visible portion of the moon grows. During the Waxing Crescent, there is a growing small portion, about 1/4, on the right side of the moon that is lit. As the moon moves towards the Waxing Gibbous phase, the illuminated portion increases to more than half, approximately 3/4, on the right side of the moon.

In both waxing phases, the angle formed by pointing one arm at the Moon and one arm at the Sun demonstrates an increase in the lit portion of the moon we see. In the Waxing Crescent phase, this angle is acute, and it becomes obtuse during the Waxing Gibbous phase, indicating the moon's journey towards a full moon, when the angle is 180°, and the entire near side of the moon, as viewed from Earth, is illuminated.

Answer:

x is the displacement of spring

Explanation:

The Hooke's law gives the force acting on the spring when it is compressed or stretched. The mathematical expression for force is given by :

[tex]F=-kx[/tex]

Where

k is the spring constant of and it shows the stiffness of spring.

x shows the displacement of spring when it is stretched or compressed from equilibrium position and it acts in opposite direction.

Answer:

At 15cm to the left of the pivot

Explanation:

The weight 0.15kg(1.5N) and 3.2N are balanced on a meter stick. This weights are parallel to each other on the stick. Questions on parallel forces acting on a body are solved using the *principle of moment."

Moment is the turning effect of force about a point.

Principle of moment states that the sum of clockwise moment is equal to the sum of anticlockwise moments.

According to the question, the meter (100cm) stick is balanced with a pivot @ 18cm mark {this is the shorter end}

If the 3.2N weight is hung at xcm from the shorter end, this weight will turn about the pivot in the anticlockwise direction.

Since moment = Force × perpendicular distance from the force

Moment of the 3.2N weight = 3.2N × x = 3.2x

The weight of the stick (1.5N) will be positioned at the center of the stick i.e @ the 50cm mark which is at 32cm to the right of the pivot.

The weight of the stick will turn in the clock wise direction

The moment of the weight in the clockwise direction = 1.5×(50-18) = 1.5×32 = 48Ncm

Using the principle of moment, we will equate both moments to have;

3.2x = 48

x = 48/3.2

x = 15cm

This means the 3.2N weight will be positioned at the 15cm mark to the left if the pivot to balance the meter stick.

Answer:

For this we use general equation for gases. Our variables represent:

p- pressure

v-volume

t- temperature

P1V1/T1 = P2V2/T2

in this equation we know:

P1,V1 and T1, T2 and V2.  

We have one equation and 1 unknown variable.

P2 = T2P1V1/T1V2 = 1.1atm

Explanation:

the guy above me is VERY correct

Answer:

36.88 light years apart

Explanation:

use law of cosines to plug in A and B and use x as the C value you need to find with cosC = cos76.04

Final answer:

To determine the vertical distance climbed by the mountain hiker, we can use the relationship between air pressure and altitude. By applying the barometric formula and substituting the given values, we can solve for the change in altitude. The vertical distance climbed is 12.1 meters.

Explanation:

To determine the vertical distance climbed by the mountain hiker, we can use the relationship between air pressure and altitude. The change in air pressure, ΔP = P₁ - P₂, can be related to the change in altitude, Δh. Assuming the air density to be constant, we can use the barometric formula to find the change in altitude. The formula is given by ΔP = ρgh, where ρ is the air density, g is the acceleration due to gravity, and h is the change in altitude.

By substituting the given values into the formula, we can solve for Δh. We have ΔP = 930 mbars - 780 mbars = 150 mbars and ρ = 1.20 kg/m³. We know that g is approximately 9.8 m/s². Solving for Δh, we get Δh = ΔP / (ρg) = 150 mbars / (1.20 kg/m³ * 9.8 m/s²) = 12.1 m.

Therefore, the vertical distance climbed by the mountain hiker is 12.1 meters.

Final answer:

The vertical distance climbed by the hiker is approximately 1269 meters, calculated using the barometric pressure difference and average air density.

Explanation:

To determine the vertical distance climbed by the hiker, we can use the barometric pressure reading along with the average air density. The pressure difference (ΔP) can be used to calculate the height difference (h) using the barometric formula ΔP = ρgh, where ρ is the density of the air, g is the acceleration due to gravity (9.81 m/[tex]s^2[/tex]), and h is the height difference.

First, we convert the pressure difference from millibars to pascals (since 1 mbar = 100 pascals):
ΔP = (930 mbar - 780 mbar) × 100 Pa/mbar = 15000 Pa

Now, we can solve for h:
15000 Pa = (1.20 kg/[tex]m^3[/tex]) × (9.81 m/[tex]s^2[/tex]) × h
h = 15000 Pa / ((1.20 kg/[tex]m^3[/tex]) × (9.81 m/[tex]s^2[/tex]))
h ≈ 1269 meters

Therefore, the vertical distance climbed by the hiker is approximately 1269 meters.