The force on a current carrying wire in a magnetic field is proportional to the current, the length of wire in the field, and the field strength. Is this true or false?

Yes, if a body is floating in a liquid, the rise in the liquid level is equal to the body height. This phenomenon is known as Archimedes' principle.

Archimedes' principle says when a body is immersed in a fluid (liquid or gas), it experiences an upward buoyant force equal to the weight of the fluid displaced by the body. Buoyant forces act in the opposite direction to gravity.

When a body floats in a liquid, it displaces a volume of liquid equal to its volume. As a result, the liquid level rises by an amount equal to the height of the submerged part of the body.

This principle holds for objects that float or are partially immersed in a liquid, such as a buoyant boat or a floating object. However, if the body sinks completely into the liquid, the liquid level rise will no longer be equal to its height. Instead, it depends on the density and volume of the submerged object.

Answer:

t = 1.875 hours

Explanation:

Speed

The speed is the rate at which an object moves. If the speed is constant, then the formula to calculate it is:

\(\displaystyle v=\frac{d}{t}\)

Where d is the distance traveled in time t.

The jogger runs 6 Km plus 5 Km plus 4 Km north for a total of 15 Km.

It's given her average speed is 8 Km/h. We can calculate the time take to complete the run by solving the above formula for t:

\(\displaystyle t=\frac{d}{v}\)

\(\displaystyle t=\frac{15\ Km}{8\ Km/h}\)

t = 1.875 hours

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

the bullet is traveling at   150 m/s relative to the ground

Explanation:

The two velocities are pointing in the same direction (both pointing North) therefore, they should add.

Then, from the ground frame of reference the bullet is travelling at: 50 m/s + 100 m/s = 150 m/s

The finished compost can be used to enrich soil for planting and gardening. Mix the compost into the soil or use it as a top dressing around plants.

Substances Used to Fill It:

A mixture of "brown" and "green" organic materials such as leaves, grass clippings, fruit and vegetable scraps, coffee grounds, and shredded paper can be used to fill the bin.

Household Items That Can Go in It:

Most fruit and vegetable scraps, eggshells, tea bags, coffee grounds, yard waste such as grass clippings and leaves, and shredded paper can be added to the compost bin. Meat, dairy, and fatty foods should be avoided.

How Long It Takes Before It Is Ready:

The compost should be ready to use in approximately 6 to 12 months, depending on the environmental conditions and the frequency of turning and mixing the compost.

How You Use the Compost:

The finished compost can be used as a nutrient-rich soil amendment for planting and gardening.

Compost Bin Design:

To add household items to your compost bin, simply add any of the following organic materials to the bin: fruit and vegetable scraps, coffee grounds and filters, tea bags, eggshells, yard waste such as leaves and grass clippings, and shredded paper.

Compost is ready to use in this much time:

The compost should be ready to use in approximately 6 to 12 months, depending on the environmental conditions and the frequency of turning and mixing the compost.

Use your compost this way:

The finished compost can be used to enrich soil for planting and gardening. Mix the compost into the soil or use it as a top dressing around plants.

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The density of water is approximately 1 g/cm^3. Therefore, the volume of 2800 g of water would be 2800 cm^3 because density is mass/volume, and so volume is mass/density.

Since this volume is inside a cubic box, the length of each side of the cube (a, for instance) could be found by taking the cubic root of the volume. This is because the volume of a cube is calculated by a^3 (length of one side cubed). Hence, a = cube root of 2800 cm^3 ≈ 14.1 cm.

Converting centimeters to meters (as 1 meter is equal to 100 centimeters), we get approximately 0.141 meters.

So the filled cubic box has a side length of approximately 0.141 m.

The motorcycle had covered a distance of 16 meters in half the total time.

a) To calculate the acceleration, we can use the formula:

a = (v - u) / t

where a is the acceleration, v is the final velocity, u is the initial velocity (which is 0 since the motorcycle starts from rest), and t is the time.

Given:

u = 0 m/s (initial velocity)

v = ? (final velocity)

t = 4 s (time)

s = 64 m (distance)

Using the equation of motion:

s = ut + 1/2at^2

We can rearrange the equation to solve for acceleration:

a = 2s / t^2

a = 2(64) / (4)^2

a = 128 / 16

a = 8 m/s^2

Therefore, the acceleration of the motorcycle is 8 m/s^2.

b) To find the final velocity, we can use the formula:

v = u + at

v = 0 + (8)(4)

v = 32 m/s

Therefore, the final velocity of the motorcycle is 32 m/s.

c) To determine the time at which the motorcycle had covered half the total distance, we divide the total distance by 2 and use the formula:

s = ut + 1/2at^2

32 = 0 + 1/2(8)t^2

16 = 4t^2

t^2 = 4

t = 2 s

Therefore, the motorcycle had covered half the total distance at 2 seconds.

d) To calculate the distance covered in half the total time, we use the formula:

s = ut + 1/2at^2

s = 0 + 1/2(8)(2)^2

s = 0 + 1/2(8)(4)

s = 0 + 16

s = 16 m

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ASDA is found to be very proficient at detecting swirls in a variety of synthetic data with various levels of noise, implying our subsequent scientific results are astute.

Applying ASDA to photospheric observations with a pixel size of 39.2 km sampled by the Solar Optical Telescope on board Hinode suggests a total number of 1.62 × 105 swirls in the photosphere, with an average radius and rotating speed of ∼290 km and <1.0 km s−1, respectively.

Comparisons between swirls detected in Bifrost numerical MHD simulations and both ground-based and space-borne observations suggest that (1) the spatial resolution of data plays a vital role in the total number and radii of swirls detected, and (2) noise introduced by seeing effects could decrease the detection rate of swirls, but has no significant influences in determining their inferred properties.

All results have shown that there is no significant difference in the analyzed properties between counterclockwise or clockwise rotating swirls. About 70% of swirls are located in intergranular lanes.

Therefore, ASDA is found to be very proficient at detecting swirls in a variety of synthetic data with various levels of noise, implying our subsequent scientific results are astute.

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Answer:

When might an ASDS be most useful and why?

Autonomous Spaceport Drone Ships would be helpful when a spacecraft needs to land in the ocean, for example.  These prevent the rocket from being damaged by the water.  These are typically used when a spacecraft doesn’t have enough fuel to reach its planned landing spot.

How do Trappist-E, -F, and -G contrast with Earth?

Trappist-E, -F, and -G are the planets that orbit in the habitable zone.  They all have a high possibility of having water, similar to Earth.  These planets are about the same size as Earth but they have about 100 times more water.

Why is the greenhouse effect significant? Explain your reasoning.

Greenhouse gases is likely what has caused the increase of warmth on our planet (global warming).  People add to these greenhouse gases and they trap heat.  We don’t know exactly what will be the outcome of the rise in these gases, but so far they have largely contributed to global warming.

Do you agree that humans are contributing to greenhouse gas levels? Explain your answer.

Yes, I do agree.  People add to the heat-trapping greenhouse gases, like carbon dioxide, methane, nitrous oxide, and chlorofluorocarbons.  Methane, for example, is increased when landfills are made of coal is mined, both of which are human activities.

When might a rover be useful, and why?

A rover might be useful when we are exploring places that aren’t safe for people.  Another time rovers might be helpful is when we are exploring places that humans literally can’t go to or conduct research on because of what the planet is made out of.  Since a rover isn’t human it can’t experience feelings or get hurt (it can be damaged, but that wouldn’t lose a life) therefore, it’s safer to send a rover at least until we can find out if it’s safe or not.

Explanation:

I had to do this in class

The angle of the first bright interference maximum (m=1) can be calculated using the equation: θ = sin-1(mλ/d).

An equation is a mathematical expression that uses symbols to describe a relationship between two or more variables. Equations are typically used to express relationships between physical quantities, such as forces and masses, or distances and times. They are also used to describe chemical reactions and other phenomena. Equations can be used to solve for unknown values, or to describe patterns or trends in data. Equations are typically written using algebraic notation, which includes variables, constants, and operators.

Substituting the given values, we get,θ = sin-1(1(580 × 10-9m)/(0.000125m)),θ = 0.00463 radians or 0.2637° .

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The forces acting on the skier is 602.1 N in perpendicular direction and 421.6 N in parallel direction.

The forces acting on the skier at the inclination of the slope is calculated by applying Newton's second law of motion as shown below.

There are two forces acting on the skier;

The perpendicular force is calculated as;

Fn = mg cosθ

where;

Fn = 75 kg x 9.8 m/s² x cos (35)

Fn = 602.1 N

The parallel force on the skier;

F = mg sinθ

F = 75 kg x 9.8 m/s² x sin (35)

F = 421.6 N

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Answer:

I believe the answer is 5718.75. Respond if it is wrong please.

Explanation:

I used a calculator.

Answer:

C. \(p_{A} = -p_{B}\) and \(K_{A} = K_{B}\).

Explanation:

The two hockey pucks travels in opposite sides with velocities of same magnitude, by definitions of linear momentum and translational kinetic energy:

Linear momentum

Hockey puck A

\(p_{A} = m\cdot v\) (1)

Hockey puck B

\(p_{B} = -m\cdot v\) (2)

\(p_{A} = -p_{B}\)

Translational kinetic energy

Hockey puck A

\(K_{A} = \frac{1}{2}\cdot m\cdot (v)^{2}\)

\(K_{A} = \frac{1}{2}\cdot m \cdot v^{2}\) (3)

Hockey puck B

\(K_{B} = \frac{1}{2}\cdot m\cdot (-v)^{2}\)

\(K_{B} = \frac{1}{2}\cdot m \cdot v^{2}\) (4)

\(K_{A} = K_{B}\)

Hence, the correct answer is C.

A VfL (Vertical Field LED) backlighting system is commonly used in LCD (Liquid Crystal Display) televisions.

LCD TVs rely on a backlight to illuminate the liquid crystal layer, which controls the passage of light to create the visual image. The VfL technology is a specific type of LED backlighting arrangement used in certain LCD TV models. In a VfL backlighting system, the LEDs (Light-Emitting Diodes) are positioned vertically along the edges of the LCD panel.

The light emitted by these LEDs is directed across the panel using light guides or optical films, illuminating the liquid crystal layer uniformly. One advantage of VfL backlighting is its ability to provide consistent illumination across the LCD panel, reducing any potential inconsistencies in brightness or color uniformity. The vertical orientation of the LEDs allows for more precise control over light distribution, improving overall image quality.

Additionally, VfL backlighting offers potential advantages in terms of power efficiency. By selectively dimming or turning off specific zones of LEDs, local dimming techniques can be employed to enhance contrast and black levels, resulting in improved picture quality while conserving energy. It's important to note that VfL backlighting is just one of several backlighting technologies available for LCD TVs.

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Answer:

C) Carpet

Explanation:

If you were trying to build a soundproof room, the material that would absorb the most sound would be carpet. Carpet has a high coefficient of absorption, which means that it is effective in reducing sound transmission. Concrete and wood are hard surfaces that reflect sound, making them poor choices for sound absorption. Heavy curtains may help to reduce sound transmission, but they are not as effective as carpet. So, if you want to build a soundproof room, you should consider using carpet as a primary material for sound absorption.

Answer:

We can use the formula:

q = mcΔT

where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The heat transferred from the gold bar to the water is equal to the heat transferred from the water to the gold bar, since they reach thermal equilibrium. Therefore:

q_gold = q_water

We can solve for the temperature change of the gold bar:

q_gold = mcΔT_gold

q_water = mcΔT_water

Since the heat transferred is equal:

mcΔT_gold = mcΔT_water

Rearranging and solving for ΔT_gold:

ΔT_gold = ΔT_water(m_water/m_gold)

ΔT_water is the temperature change of the water, which is 17°C. m_water is 0.220 kg, and m_gold is 3 kg. c_gold is given as 129 J/kg K.

ΔT_gold = 17°C(0.220 kg/3 kg)(1/129 J/kg K) = 0.025°C

Therefore, the temperature change of the gold bar is 0.025°C when it is placed into 0.220 kg of water and thermal equilibrium is reached.

Answer:

Skyscraper

Explanation:

A human has less inertia than a skyscraper. The ability of matter to resist changes in motion is known as inertia, and it is inversely proportional to mass. A building has more inertia than a person since it has a larger mass.

According to Newton's third law of motion, if you apply a force of 100 N to the side of a skyscraper, the structure will respond by applying an equal and opposing force of 100 N to you. This is due to the fact that the force you exert on the tower is transmitted through your body, to your feet, and then into the ground. The skyscraper experiences an equal and opposite force from the earth, which is reflected back up your body through your feet.

Answer:

11,000 kg

(a) 11.2 m/s

(b) 1.6 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(2200 kg) (60.0 km/h) + m (0 km/h) = (2200 kg) (10 km/h) + m (10 km/h)

132,000 kg km/h = 22,000 kg km/h + m (10 km/h)

110,000 kg km/h = m (10 km/h)

m = 11,000 kg

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(m) (-v) + (2m) (5v) = (m) (v₁) + (2m) (v₂)

-mv + 10mv = m v₁ + 2m v₂

9mv = m (v₁ + 2 v₂)

9v = v₁ + 2 v₂

Since the collision is elastic, kinetic energy is also conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(m) (-v)² + (2m) (5v)² = m v₁² + (2m) v₂²

mv² + 50mv² = m v₁² + 2m v₂²

51mv² = m (v₁² + 2 v₂²)

51v² = v₁² + 2 v₂²

We know v = 1.60 m/s.  So the two equations are:

14.4 = v₁ + 2 v₂

130.56 = v₁² + 2 v₂²

Solve the system of equations using substitution.

130.56 = (14.4 − 2 v₂)² + 2 v₂²

130.56 = 207.36 − 57.6 v₂ + 4 v₂² + 2 v₂²

0 = 6 v₂² − 57.6 v₂ + 76.8

0 = v₂² − 9.6 v₂ + 12.8

v₂ = [ 9.6 ± √(9.6² − 4(1)(12.8)) ] / 2(1)

v₂ = 1.6 or 8

If v₂ = 1.6 m/s, then v₁ = 14.4 − 2 v₂ = 11.2 m/s.

If v₂ = 8 m/s, then v₁ = 14.4 − 2 v₂ = -1.6 m/s.

We know v₁ can't be -1.6 m/s, since that would mean puck A didn't change speeds after the collision.  Therefore, v₁ = 11.2 m/s and v₂ = 1.6 m/s.

Solution :

Given :

Diameter, D = 30 m

∴ Radius, R = 15 m

Angular acceleration, α = 0.044 \($rad/s^2$\)

a). Velocity of the rider just as he completes a quarter of a turn is :

\($\omega_i = 0 \ rad/s$\)

\($\theta = \frac{\pi}{2}\ rad$\)

V = R ω

∴\($\omega^2_f=\omega^2_i + 2 \alpha \theta$\)

 \($\omega^2_f=0+ 2 \times 0.04 \times \frac{\pi}{2}$\)

\($\omega^2_f=0.1256$\)

\($\omega = \sqrt{0.1256}$\)

   \($=0.354 \ rad/s$\)

∴ \($V=R \omega_f$\)

     \($= 15 \times 0.354$\)

    = 5.32 m/s up

b). Tangential acceleration

   \($a_T= \alpha R$\)

        = 0.04 x 15

       \($= 0.600 \ m/s^2$\)  up

Radial acceleration,

\($a_r=\frac{V^2}{R}$\)

   \($=\frac{(5.32)^2}{15}$\)

   \($= 1.88 \ m/s^2$\) towards center.

c). Final angular velocity

   given : \($V_0 = 6 \ m/s$\)

              \($\omega_i = 0.354 \ rad/s$\)

              \($\alpha = 0.0400 \ rad/s^2$\)

\($\omega_f = \frac{6}{15} \ rad/s$\)

\($\omega^2_f=\omega^2_i + 2 \alpha \theta$\)

\($\left(\frac{6}{15}\right)^2=(0.354)^2 + 2 \times 0.04 \times \theta$\)

\($\theta = \frac{0.16-0.1256}{2 \times 0.04}$\)

  = 0.43 rad

or \($\theta = 0.43 \times \frac{180}{\pi}$\)

      \($24.6^\circ$\)

In a DC generator, the generated electromotive force (emf) is directly proportional to the rotational speed of the generator's armature and the strength of the magnetic field within the generator.

This relationship is described by the equation for the generated emf in a DC generator:

Emf = Φ * N * A * Z / 60

Where:

Emf is the generated electromotive force (in volts),

Φ is the magnetic flux density (in Weber/meter^2\(meter^2\) or Tesla),

N is the number of turns in the armature winding,

A is the effective area of the armature coil (in square meters),

Z is the total number of armature conductors, and

60 is a constant representing the conversion from seconds to minutes.

From this equation, we can see that the generated emf is directly proportional to the magnetic flux density (Φ) and the product of the number of turns (N), effective area (A), and the total number of armature conductors (Z). This means that increasing any of these factors will result in a higher generated emf.

The magnetic flux density (Φ) can be increased by using stronger permanent magnets or increasing the strength of the field windings in the generator.

The number of turns (N) and the effective area (A) are design parameters and can be optimized for a specific generator. Increasing the number of turns or the effective area will result in a higher generated emf.

Similarly, the total number of armature conductors (Z) can be increased to enhance the generated emf.

By controlling and optimizing these factors, the generated emf in a DC generator can be increased, resulting in higher electrical output. However, it is important to note that there are practical limits to these factors based on the design and construction of the generator.

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The ball exerts a force on the wall, and the wall exerts an equal and opposite force on the ball. So, the correct answer is A.

Newton's 3rd law of motion states that for every action, there is an equal and opposite reaction. The action is the force that the ball exerts on the wall, and the reaction is the force that the wall exerts back on the ball. When the ball hits the wall, it exerts a force on the wall. This force is equal in magnitude but opposite in direction to the force that the wall exerts back on the ball. This is why the ball reverses course. Therefore, option A is correct.

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--The complete Question is, Which of the following statements is true according to Newton's 3rd law when a ball hits a wall and reverses course?

A) The ball exerts a force on the wall, and the wall exerts an equal and opposite force on the ball.

B) The ball exerts a force on the wall, but the wall does not exert a force on the ball.

C) The wall exerts a force on the ball, but the ball does not exert a force on the wall.

D) The ball and the wall do not exert any forces on each other. --

Answer:

Explanation:

Looking at the left string node

sum vertical forces to zero

T₁cos35 - 40 = 0

T₁ = 40/cos35 = 48.83 N

Sum horizontal forces to zero

T₂ - T₁sin35 = 0

T₂ = 48.83sin35 = 28.01 N

Now consider the right string node

Sum horizontal forces to zero

T₃sinθ - T₂ = 0

Τ₃ = T₂/sinθ

sum vertical forces to zero

 T₃cosθ - 50 = 0

(T₂/sinθ)cosθ = 50

          T₂cotθ = 50

              cotθ = 50/28.01

                   θ = 29.256067...°

                   θ = 29°

Τ₃ = T₂/sinθ

Τ₃ = 28.01/sin29.256067

Τ₃ = 57.3102...

Τ₃ = 57 N

The value of  the missing resistance, R₃ = 10.35 ohms.

The value of the missing voltages, V₁ = 6 V, V ₃ = 24 V.

The value of the missing currents, I₁ = 3 A, I₃ = 2.32 A.

The values of the missing component of the circuit is calculated by applying the following formula.

The total resistance of the circuit;

For R₂, R₃, 1/R = 1/R₂ + 1/R₃

1/R = 1/12 +  1/R₃

1/R = (R₃ + 1)/(12R₃)

R = 12R₃ / (R₃ + 1)

For, R₁, R₂ and R₃, total resistance;

R = 12R₃ / (R₃ + 1) + R₁

R = [12R₃ / (R₃ + 1)] + 2

R = (12R₃ + 2(R₃ + 1) ) / (R₃ + 1)

R = (12R₃ + 2R₃ + 2 ) / (R₃ + 1)

R = (14R₃ + 2 ) / (R₃ + 1)

The total current in circuit is calculated as;

I = V/R

I = 30 / R

I = ( 30 ) / (14R₃ + 2 ) / (R₃ + 1)

I = (30R₃ + 30) / (14R₃ + 2) ------- (1)

The voltage in parallel circuit is the same

V₂ = V₃ = 24 V

V₃ = IR₃

24 = IR₃

I = 24/R₃  --------- (2)

Solve (1) and (2) together as follows;

24/R₃ = (30R₃ + 30) / (14R₃ + 2)

30R₃² - 306R - 48 = 0

Solve the quadratic equation, using formula method.

R₃ = 10.35 ohms

I₃ = V₃/R₃

I₃ = 24 V / 10.35

I₃ = 2.32 A

If the voltage drop at R₂ and R₃ = 24 V, the voltage drop at R₁ = 30V - 24 V = 6 V

The current in R₁ = V₁/R₁ = 6 V / 2 V = 3 A

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The displacement of the train after 2.23 seconds is 25.4 m.

The resultant velocity of the train is calculated as follows;

R² = vi² + vf² - 2vivf cos(θ)

where;

R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)

R² = 129.75

R = √129.75

R = 11.39 m/s

The displacement is calculated as follows;

Δx = vt

Δx = 11.39 m/s x 2.23 s

Δx = 25.4 m

Thus, the displacement of the train after 2.23 seconds is 25.4 m.

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The coefficient of static friction places a limit of approximately 3.42 seconds on the minimum time required for the car to accelerate from rest to 60.0 mph (26.8 m/s).

To find the limit that the coefficient of static friction places on the minimum time required for a car to accelerate from rest to 60.0 mph (26.8 m/s), we need to consider the maximum acceleration the car can achieve due to the friction between the rear tires and the pavement.

The maximum acceleration can be determined using the formula:

a_max = μs * g

where μs is the coefficient of static friction and g is the acceleration due to gravity (approximately 9.8 m/s²).

In this case, since the car's engine only turns the two rear wheels, the maximum acceleration is limited by the friction force between the rear tires and the pavement.

Now, to calculate the minimum time required to accelerate to 60.0 mph (26.8 m/s), we can use the following kinematic equation:

v = u + a * t

where v is the final velocity (26.8 m/s), u is the initial velocity (0 m/s), a is the acceleration, and t is the time.

Rearranging the equation, we have:

t = (v - u) / a

Plugging in the values, we get:

t = (26.8 m/s - 0 m/s) / a_max

t = 26.8 m/s / (μs * g)

Substituting the given value for the coefficient of static friction (μs ≈ 0.800) and the acceleration due to gravity (g ≈ 9.8 m/s²), we can solve for the minimum time required:

t = 26.8 m/s / (0.800 * 9.8 m/s²)

t ≈ 3.42 seconds

Therefore, the coefficient of static friction places a limit of approximately 3.42 seconds on the minimum time required for the car to accelerate from rest to 60.0 mph (26.8 m/s).

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The probability of getting a prime number ( 2, 3, 5, 7) between 1 to 8 is 1/2. The total outcomes possible for spinning the spinner and tossing the coin is 16.

Branch of mathematics concerning numerical descriptions of how likely an event is to occur is called probability.

Total number of outcomes = 8.

Let, the probability of getting a prime number ( 2, 3, 5, 7) between 1 to 8

P(E) = (no. of outcomes) /( total no. of outcomes)

=4/8

=1/2

When a coin is tossed,  there are two possible outcomes that are H(head), and T(tail).

If you spin the spinner , there are 8 possibilities.

Total outcomes = 8 * 2

=16

There are 16 outcomes, when spinning a spinner numbered from 1 to 8 and tossing a coin.

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Subsequently, the value of the kinetic energy of the stone when it reached the target is 2.25 joules.

The kinetic energy (KE) of an question is given by the equation:

KE= 1/2 * mass * velocity²

In this case, the mass of the stone is 20 grams, which is rise to 0.02kg and the speed of the stone is 15m/s

Plugging these values into the equation, we are able calculate the active vitality of the stone:

KE = 1/2 * 0.02kg * (15.0mls)²

KE = 1/2 * 0.02kg * 225 m²/ s²

KE = 2.25J

Subsequently, the value of the kinetic energy of the stone when it reached the target is 2.25 joules.

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Answer:

It takes 4 seconds for the camera to fall 250 feet

Explanation:

4 seconds by using the quadratic formula

The initial volume of the liquid is 31.4 cm³. The volume coefficient of thermal expansion of the liquid is 0.002 per degree Celsius. A temperature increase of 109.5°C is needed for the liquid to rise to a height of 49cm.

The initial volume of the liquid can be found using the formula for the volume of a cylinder:

V = πr²h

where r is the radius (half the diameter), h is the height, and π is approximately 3.14. Plugging in the given values, we get:

V = π(0.5 cm)²(40 cm)

V = 31.4 cm³

The volume coefficient of thermal expansion (β) is defined as the fractional change in volume per degree Celsius change in temperature. It can be calculated using the formula:

β = ΔV/(VΔT)

where ΔV is the change in volume, V is the initial volume, and ΔT is the change in temperature. We can rearrange this formula to solve for ΔV:

ΔV = βVΔT

We know that ΔT = -10°C (a decrease of 10°C) and that the height decreased from 40cm to 37cm, or by 3cm. The change in volume can be found using the formula for the volume of a cylinder again, with the new height of 37cm:

ΔV = π(0.5 cm)²(40 cm - 37 cm)

ΔV = 0.59 cm³

Plugging in all the values, we get:

0.59 cm³ = β(31.4 cm³)(-10°C)

β = 0.002

To find the change in temperature needed for the liquid to rise to a height of 49cm, we can use the same formula as before, but solve for ΔT:

ΔT = ΔV/(βV)

We know that ΔV is the difference between the initial volume and the volume at the new height, which is:

ΔV = π(0.5 cm)²(49 cm - 40 cm)

ΔV = 6.86 cm³

Plugging in all the values, we get:

ΔT = 6.86 cm³/(0.002)(31.4 cm³)

ΔT = 109.5°C

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