what is the equivalent resistance between point a and b in ohms? all the resistors are the same an each of them has a resistance

Answer:

Momentum is computed using below formula.

Momentum = Mass x Velocity,

The problem states that the final velocity is 6.0m/s, Substituting onto the equation we have

Momentum = 1000 kg x 6.0 m/s = 6000 kg · m/s

Explanation:

hope it help ;)

The final momentum of the car is the product of its mass and final velocity. The final momentum of the car of 1000 kg with the final velocity of 6 m/s is 6000 kg m/s.

Momentum of an object is its ability to bring the force to make the maximum displacement. In physics, momentum is the product of mass and velocity of the object.

The initial velocity of the car = 2 m/s

mass = 1000 kg

final velocity = 6 m/s.

The acceleration of the car when changing from a velocity of 2 to 6 m/s within 5 seconds is:

a = Δv/t = 4 m/s/5 s = 0.8 m/s²

The final momentum of the car = mass × final velocity

P = 1000 kg × 6 /s = 6000 kg m/s.

Therefore, the final momentum of the car is 6000 kg m/s.

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The new period of the pendulum on Venus would be approximately 0.947 times the old period on Earth.

The period of a pendulum is given by the formula:

T = 2π√(L/g)

where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

Let's assume that the length of the pendulum remains the same when it is transported from Earth to Venus. On Earth, the acceleration due to gravity is 9.81 m/s^2, and on Venus, it is 8.89 m/s^2.

The ratio of the new period to the old period can be found by taking the ratio of the square roots of the gravitational accelerations:

(new period / old period) = √(8.89 / 9.81)

= 0.947

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Answer:

Below

Explanation:

Momentum = m v

                    = 18 kg * 20 m/s =  360 kg m/s

Answer:

momentum = mass × velocity = 6× 20 =120 kg.ms-1

Explanation:

not sure if this is right

In Edgar Allan Poe's poem "The Raven," the speaker's description of the bird evolves throughout the poem, reflecting the changing emotions and perceptions of the speaker. By the end of the poem, the speaker's description of the bird is one of eerie and unsettling admiration.

Initially, the speaker describes the raven as a "stately" and "ebony" bird, highlighting its majestic and imposing presence. As the poem progresses, the speaker's description becomes more vivid and eerie. The bird is referred to as a "prophet" and a "devil," emphasizing its ominous and supernatural qualities. Its eyes are described as "burning" and "demon's," intensifying the sense of unease.

Towards the end of the poem, the speaker's attitude towards the bird shifts. The raven becomes a symbol of despair and sorrow, representing the speaker's torment. The speaker's repeated questioning of the raven about the possibility of being reunited with his lost love, Lenore, showcases his desperation and longing for answers.

In the final stanza of the poem, the speaker describes the bird as a "thing of evil" and a "demon." This description emphasizes the bird's malevolent nature and its power to haunt the speaker's thoughts and torment his soul. Despite its dark and foreboding presence, the speaker cannot help but be captivated and entranced by the raven's presence.

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Explanation:

Rate  X  Time  =   Distance

Distance  / Time  = Rate

225 km / 3.35 hr = 67.2 km/hr

a.) the ball was going up at 0.95 seconds and coming down at 4.58 seconds

b.) the distance of the wall from the plate is 34 m

Given that a ball player hits a home run, and the ball just clears the wall which is 22.0 m high. The ball is hit at an angle of 37.0" with a velocity of 45.0 m/s.

Since the ball is hit at an angle of 37.0 degrees, We need to find the vertical and horizontal component of the velocity.

\(U_{y}\) = 45 Sin 37 = 27.08 m/s

\(U_{x}\) = 45 cos 37 = 35. 94 m/s

Let us first calculate the maximum height reached.

\(V^{2}\) = \(U_{y} ^{2}\) - 2gH

At maximum height, V = 0

0 = \(27.08^{2}\) - 2 x 9.8H

19.6H = 733.3

H = 733.33/19.6

H = 37.4 m

If the ball is hit from a height of 0.750 m above the ground,

(a) To calculate the time the ball stays in the air until it  clears the wall while on its way down, we will use the formula below.

h =  \(U_{y}\)t - 1/2g\(t^{2}\)

Substitute for all the parameters

22 - 0.75 = 27.08t - 0.5 x 9.8 x \(t^{2}\)

21.25 = 27.08t - 4.9\(t^{2}\)

4.9\(t^{2}\) - 27.08t + 21.25

We will use quadratic formula

a = 4.9

b = - 27.08

c = 21.25

t =  \(\frac{-b+/- \sqrt{b^{2} - 4ac } }{2a}\)

t = \(\frac{27.08 +/- \sqrt{27.08^{2} - 4 * 4.9 * x^{2} 21.25 } }{2 * 4.9}\)

t = \(\frac{27.08 +/-\sqrt{733.3 - 416.5} }{9.8}\)

t = \(\frac{27.08 +/- 17.8}{9.8}\)

t = 44.88/9.8 or 9.28 / 9.8

t = 4.58 s  or 0.95 s

This means that the ball was going up at 0.95 seconds and coming down at 4.58 seconds

(b) The distance of the wall from home plate will be the range which is

R = \(U_{x}\)t

R = 35.94 x 0.95

R = 34.143 m

Therefore, the distance of the wall from the plate is 34 m approximately

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The atoms of these noble gases have stable arrangements of electrons, the elements are all inert.

The term inert gases has to do with those gases that are not reactive. The atoms of these elements have a complete octet of electrons hence they are unreactive.

As such,  the atoms of these noble gases have stable arrangements of electrons, the elements are all inert.

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Answer:

d. Ice

Explanation:

Answer:

The drum struck with a lot of energy would be louder than the drum struck with little energy because there is a harder impact on it

Explanation:

Answer:

Curvature

Explanation:

Answer:

It would be C.scale, because

A. is for liquid

B. is for measuring 2 things

and D. is for i don't know(lol)

but i think the scale would be the easiest way to measer an iPad

Answer:

Please see below as the answer is self-explanatory.

Explanation:

The width of the outer core is 1,350 miles.

Earth's outer core is a fluid layer about 2,200 km thick, composed of mostly iron and nickel that lies above Earth's solid inner core and below its mantle.

The outer core, about 2,200 kilometers (1,367 miles) thick, is mostly composed of liquid iron and nickel.

The width of the outer core is calculated as follows;

Δx = (3991 mi ) - (32 mi + 1821 mi + 788 mi)

Δx = 1,350 mi

Thus, from the data given, the width of the outer core is 1,350 miles.

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(i) The work done on the point charge by the electric force when moved from A to B is 2.1 x 10⁻⁶ J.

(ii) The work done on the point charge by the electric force when moved from A to C is -5.5 x 10⁻⁶ J.

The work done by an electric force is equal to the negative of the change in potential energy, which is given by the product of the charge and the change in potential. The change in potential between two points is equal to the potential difference between those points.

For (i), the potential difference between A and B is 6 V, so the work done is (3.9 x 10⁻⁷ C) x (-6 V) = -2.1 x 10⁻⁶ J (negative because the charge moves from higher to lower potential).

For (ii), the potential difference between A and C is -15 V, so the work done is (3.9 x 10⁻⁷ C) x (-(-15 V)) = -5.5 x 10⁻⁶ J (negative because the charge moves from lower to higher potential).

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Answer:

The options that increase greenhouse gases are:

Coal burning power plant

Exhaust from a car

Wind power and solar power do not increase greenhouse gases, as they are considered renewable energy sources that do not produce direct emissions of greenhouse gases during their operation.

Explanation:

Answer:

Explanation:

Coal burning power plants and exhaust from a car increase greenhouse gases

Wind power and solar power do not produce greenhouse gases

I hope this helps!

The magnitude of the torque that slows the turntable is equal to 2.17 × 10⁻³ N.m.

Torque can be described as the measure of the force that can cause the rotation of an object about an axis. The force causes an object to accelerate similarly, torque causes an angular acceleration. Therefore, torque is the rotational equivalent of linear force.

Given, the rotational speed of the turntable, ω = 78 rpm = 8.16 rad/s

The time turntable takes to stop, t = 45 s

The rotational inertia of the turntable, I = 1.2 × 10⁻² Kg.m²

From the rotational kinetics equation, the angular acceleration:

\(\omega_f = \omega +\alpha t\)

The final speed of the turntable = 0

\(\alpha =\frac{\omega_f-\omega}{t}\)

\(\alpha =\frac{0-8.16}{45}\)

\(\alpha =-0.1814 \;rad/s^2\)

Calculation of torque by using inertia and angular acceleration:

\(\tau = I\alpha\)

\(\tau =1.2 \times 10^{-2} \times (-0.1814)\)

\(\tau = 2.17\times 10^{-3} \; N.m\)

Therefore, the magnitude of the torque that slows the turntable is 2.17 × 10⁻³ N.m.

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Answer:

(a). The kinetic friction force is 1950 N.

(b). The magnitude of force will be equal of friction force

Explanation:

Given that,

Constant speed = 6.38 m/s

Force \(F=7.50\times10^{3}\ N\)

Kinetic friction = 0.26

(a). We need to calculate the friction force

Using formula of friction force

\(f_{k}=\mu F_{N}\)

Put the value into the formula

\(f_{k}=0.26\times7.50\times10^{3}\)

\(f_{k}=1950\ N\)

(b). If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh,

We need to calculate the magnitude of this force

According to given data,

The same force will be applied to keep constant velocity.

Hence, (a). The kinetic friction force is 1950 N.

(b). The magnitude of force will be equal of friction force.

(a). The kinetic friction force is 1950 N.

(b). The magnitude of force will be equal of friction force

a. The magnitude of the kinetic friction force experienced by the sleigh is

\(= 0.76 \times 7.50 \times 10^3\)

= 1950 N

b. It should be equivalent to the friction force.

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The spring constant of the spring is 178.4 N/m.

In this problem, we are given a mass of 0.25 kg that is attached to a spring and set into vibration with a period of 0.22 s.

We can use the equation for the period of a mass-spring system, which relates the period of oscillation to the mass and spring constant, to calculate the spring constant.

We can use the equation for the period of a mass-spring system:

T = 2π√(m/k)

where T is the period, m is the mass, and k is the spring constant.

Rearranging this equation to solve for k, we get:

k = \((4\pi^2m) / T^2\)

Substituting the given values, we get:

k = \((4\pi^2 \times 0.25 \:kg) / (0.22 s)^2\)

k = 178.4 N/m

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Answer:

The drop in temperature overnight causes a decrease in the average kinetic energy of the air molecules inside the tires. According to the ideal gas law, this leads to a decrease in tire pressure. The low tire pressure light in vehicles is triggered when the pressure falls below a certain threshold, alerting the driver to check and adjust the tire pressure.

Explanation:

The ideal gas law, represented by the equation PV = nRT, relates the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas. In this case, we can analyze how the drop in temperature affects the tire pressure.

When the temperature drops, according to the ideal gas law, the pressure of a gas will decrease if the volume and the number of moles remain constant. This is because the decrease in temperature causes a decrease in the average kinetic energy of the gas particles, leading to less frequent and less forceful collisions with the tire walls, resulting in a decrease in pressure.

In the context of the tire pressure, the air inside the tires behaves as an ideal gas. When the temperature drops overnight, the air inside the tires also cools down, causing a decrease in its temperature. As a result, the average kinetic energy of the air molecules decreases, leading to a decrease in pressure inside the tires.

The low tire pressure light comes on as a result of this drop in pressure. The tire pressure monitoring system in modern vehicles is designed to detect significant deviations from the recommended tire pressure. When the pressure drops below a certain threshold, typically due to temperature changes or a puncture, the light is triggered to alert the driver to check and adjust the tire pressure.

Therefore, the drop in temperature overnight causes a decrease in the average kinetic energy of the air molecules inside the tires, resulting in a decrease in tire pressure, which triggers the low tire pressure warning light.

Hope this helps!

Glass is the material that would provide the most electrical resistance.  

Elements which are good conductors of electricity are referred to as

Metals. They usually have low resistance.

Non metals are are low conductors of electricity and have high resistance.

Electrolytes such as salt water are also good conductors of electricity.

Glass are generally regarded as Insulators and are poor conductors of

electricity which makes it the material with the most electrical resistance.

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To better understand crash dynamics we have to look at "the laws of motion."

The laws of motion

The laws of motion were introduced by Sir Isaac Newton in 1687 in his book Philosophiæ Naturalis Principia Mathematica ("Mathematical Principles of Natural Philosophy"), which defined the laws of motion, or three fundamental laws that govern the movement of bodies. The laws of motion, according to Newton, govern the motion of an object or a system of objects that interact.

It defines the concepts of force and mass, and the fundamental dynamics of motion.The following are the laws of motion:Every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. The velocity of an object changes proportional to the force applied to it, and the acceleration of an object is proportional to both its force and its mass. For every action, there is an equal and opposite reaction.

Therefore, these laws are necessary to fully grasp crash dynamics because they explain how objects respond to outside forces that cause them to accelerate or decelerate.

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The energy stored in the circuit at any time t is given by \(U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).\)The units are in seconds.

The total energy stored in the circuit can be calculated using the formula: U = (1/2)L*I² + (1/2)Q²/C, where L is the inductance, I is the current, Q is the charge on the capacitor, and C is the capacitance.

Initially, the capacitor is uncharged, so the second term is zero.

Therefore, the initial energy stored in the circuit is U₀ = (1/2)L*I₀², where I₀ is the initial current, which is zero.

When the magnetic field is switched on, a current begins to flow in the solenoid.

This current increases until it reaches its maximum value, given by I = V/R, where V is the voltage across the solenoid and R is its resistance.

Since the solenoid is connected in series with the capacitor, the voltage across the solenoid is equal to the voltage across the capacitor, which is given by V = Q/C, where Q is the charge on the capacitor.

The charge on the capacitor is given by Q = C*V, where V is the voltage across the capacitor at any time t.

Therefore, we have I = V/R = Q/(R*C) = dQ/dt*(1/R*C), where dQ/dt is the rate of change of charge on the capacitor.

This is a first-order linear differential equation, which can be solved to give \(Q(t) = Q_{0} *(1 - e^{(-t/(R*C)}))\), where Q₀ is the maximum charge on the capacitor, given by Q₀ = C*V₀, where V₀ is the voltage across the capacitor at t=0.

The current in the solenoid is given by I(t) = \(dQ/dt*(1/R*C) = (V_{0} /R)*e^{(-t/(R*C)}).\)

The energy stored in the circuit at any time t is given by\(U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).\)

The time t at which the energy stored in the circuit drops to 10% of its initial value can be found by solving the equation U(t) = U₀/10, or equivalently, \((1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C)}) + (1/2)C*V_{0} /R)^{2}*(1 - e^{(-2t/(R*C)})) = (1/20)L*I_{0} /R)^{2}.\)

This equation can be solved numerically using a computer program, or graphically by plotting U(t) and U₀/10 versus t on the same axes and finding their intersection point.

The solution is t = 1.74 ms.

The units are in seconds.

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Answer:

9 hr

Explanation:

Speed = distance/time

Let the time taken by the train be t.

=> 70 = 630/t

=> t = 630/70

=> t = 9

it take 9 hr to complete its journey

Answer:

9 hours

Explanation:

Using the formula to solve motion problems: d=r*t (Distance = Rate * Time)

Given:

Distance = 630 km

Rate (Speed) = 70 km/h

Time = Distance / Time

Time = 630/70

               = 9

Answer: A

Explanation: We know that f=p*n

f=50*300=15000 Hz = 15kHz.

Have a great day! <3

If the number of revolutions is 300 and the paired poles are 50 , then the frequency would be 15 kHz, therefore the correct answer is option A.

It can be defined as the number of cycles completed per second. It is represented in hertz and inversely proportional to the wavelength.

The frequency of a pendulum is the reciprocal of the time period can be given by the following relation,

F = 1 / T

As given in the problem, we have to calculate frequency if the number of revolutions is 300 and the paired poles are 50.

F = 300 × 50

  = 1500 kHz

Thus, If the number of revolutions is 300 and the paired poles are 50, then the frequency would be 15 kHz, therefore the correct answer is option A.

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If a 5kg mass starts from rest at xo = -1 and moves under the action of a variable force F(x) = √1-x² to point xf = 1. Then the total work done by the force is equal to π/2 + 1.

To calculate the total work done by the force in this scenario, we can use the formula for work:

Work = ∫F(x) dx

where F(x) is the force as a function of position and dx represents an infinitesimal displacement.

In this case, the force is given by F(x) = √(1 - x²), and we need to find the total work done as the object moves from xo = -1 to xf = 1.

Let's break down the calculation step by step:

Write the integral for work:

Work = ∫F(x) dx

Substitute the given force:

Work = ∫√(1 - x²) dx

Integrate with respect to x:

To integrate the square root of (1 - x²), we use the trigonometric substitution. Let's substitute x = sin(θ) and dx = cos(θ) dθ.

Work = ∫√(1 - sin²(θ)) cos(θ) dθ

Simplify the integrand:

Using the trigonometric identity sin²(θ) + cos²(θ) = 1, we can rewrite the integrand as cos²(θ).

Work = ∫cos²(θ) dθ

Apply the power-reducing formula:

The power-reducing formula states that cos²(θ) = (1 + cos(2θ)) / 2. We can use this formula to simplify the integrand further.

Work = ∫(1 + cos(2θ))/2 dθ

Integrate the terms separately:

Work = (1/2) ∫dθ + (1/2) ∫cos(2θ) dθ

The first integral, ∫dθ, is simply θ, and the second integral, ∫cos(2θ) dθ, can be calculated as sin(2θ)/2.

Work = (1/2) θ + (1/2) (sin(2θ)/2) + C

Evaluate the integral limits:

To find the total work done, we need to evaluate the integral at the upper and lower limits of integration.

At xf = 1, the angle θ is π/2, and at xo = -1, the angle θ is -π/2.

Work = (1/2) (π/2) + (1/2) (sin(2(π/2))/2) - [(1/2) (-π/2) + (1/2) (sin(2(-π/2))/2)]

Simplifying further:

Work = π/4 + (1/2) - (-π/4 + (1/2))

Work = π/4 + 1/2 + π/4 + 1/2

Work = π/2 + 1

Therefore, the total work done by the force is equal to π/2 + 1.

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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:

It accelerate but will not spin.

Explanation:

If an irregular shaped object is dropped from rest without feeling any form of air resistance it will accelerate without spinning and this is due to the fact that there is no Torque around  the center of gravity