What volume of .2500 m cobalt iii chloride is required to react completely with 25 ml of .0315 m calcium hydroxide?

The stone was originally thrown from a height of 234 feet. This solution was obtained using the physics kinematic equation for vertical motion.

This problem can be solved using the kinematic equation:

Δy = V₀t + 1/2gt²

, where:

From the statement we know: Δy = 10 ft - 90 ft = -80 ft after 2 seconds. We substitute these values and solve for the initial position, y₀. The equation becomes:

-80 = 4*2 + 1/2*(-32)*2²

. Solving gives us -80 = 8 - 64, so

y₀ = 80 + 64 + 90 = 234 ft

. Therefore, the stone was thrown from a height of 234 feet.

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Using the equations of motion under gravity, it's determined that the stone was thrown from a height of 162 feet considering it was thrown down with an initial velocity and passed specific heights in its journey.

The question involves finding the original height from which a stone was thrown, given that it was thrown down with an initial speed and passes specific points in its descent. To solve this problem, we can use the equations of motion under the influence of gravity. The formula that relates the initial velocity (u), the acceleration due to gravity (g = 32 feet/second2 downward), the time taken (t), and the displacement (s) is s = ut + (1/2)gt2.

In this instance, the stone is observed to move from 90 feet above the ground to 10 feet above the ground in 2 seconds, with an initial downward speed of 4 feet/second. The drop in height (displacement) is 80 feet (90 - 10). We can insert these values into the formula to find the initial height: Let H be the initial height, the equation becomes H - 90 = 4(2) + (1/2)(32)(22). Simplifying, we find that H - 90 = 8 + 64, which resolves to H = 162 feet.

Hence, the stone was thrown from a height of 162 feet.

Answer:

Creation and intelligent design

Explanation:

First let us calculate the acceleration.

v1 = v0 + a t1

where v1 is final velocity, v0 is initial velocity, a is acceleration and t is time

Calculating for a:

14 m/s = 7 m/s + a * 8 s

a = 0.875 m/s^2

Therefore the final speed is calculated similarly:

v2 = v1 + a t2

v2 = 14 m/s + (0.875 m/s^2) * 16 s

v2 = 28 m/s

To find the bird's instantaneous velocity at t = 8.00 s, the derivative of the position function x(t) is calculated to get the velocity function v(t), and then t = 8.00 s is substituted into this function to obtain the velocity at that instant.

The question asks for the instantaneous velocity of a bird at a specific time given its position as a function of time, x(t). To find the instantaneous velocity, we need to take the derivative of the position function with respect to time. Given the position function x(t) = 30.0 m + (11.7 m/s)t - (0.0450 m/s3)t3, the derivative of this function will give us the velocity function v(t).

Therefore, v(t) = dx(t)/dt = 11.7 m/s - 3*(0.0450 m/s3)*t2. Plugging in t = 8.00 s into this velocity function, we get v(8.00 s) = 11.7 m/s - 3*(0.0450 m/s3)*(8.00 s)2, which we can calculate to find the instantaneous velocity at t = 8.00 seconds.

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The auditory cortex is located in temporal lobe of the brain.

The  auditory cortex is the part of the temporal lobe that processes hear-able data in people and numerous different vertebrates.

It is a piece of the  auditory framework, carrying out fundamental and higher roles in hearing. For example, potential relations to language switching.

The auditory cortex is situated on the predominant transient gyrus in the fleeting curve and gets highlight point input from the ventral division of the average geniculate complex.  Consequently, it contains an exact tonotopic map.

It is found generally at the upper sides of the fleeting curves in people, bending down and onto the average surface, on the prevalent transient plane, inside the sidelong sulcus and containing portions of the cross over worldly gyri, and the unrivaled worldly gyrus, including the planum polare and planum temporale.

Thus,  auditory cortex is located in temporal lobe of the brain.

Learn more about auditory cortex.

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Answer:

buzzer

Explanation:

Answer:

D. Do you think we should ignore our constitutional rights and let the government take citizens' guns away?

Explanation:

A question wording biased is what happens when the question states directly a point of biew and suggests the interviewed a certain answer that is clear once you´ve heard the question, in this case it is obvious that the question is against the ban on guns, because it is already judging any decision that the congress could make on it and suggesting a point of view to the interviewed.

Angular speed is  0.188 rad/s ,

Tangential speed is 18.8 ft/s ,

Time for one revolution is 33.4 s.

Given :

Ferris wheel of radius 100 ft.

The lowest point of the ride is 3 feet above ground level.

Solution :

Refer the attached diagram for better understanding.

From the diagram we know that,

[tex]\rm y = 100-(44-3)=59 \; ft[/tex]

y = 59 ft

Now applying pythagorean theorem,

[tex]x^2 + 59 ^2= 100^2[/tex]

[tex]x=\sqrt{100^2-59^2}[/tex]

[tex]\rm x = 80.7403\;ft[/tex]

Now to calculate angle [tex]\theta\\[/tex],

[tex]\rm cos\theta = \dfrac{59}{100}=0.59[/tex]

[tex]\rm \theta = 53.84^\circ=0.9397\;radians[/tex]

Now the arc length pq is given by,

[tex]\rm S =pq=0.9397\times100[/tex]

[tex]\rm S = 93.97\; ft[/tex]

Now the angular velocity is given by,

[tex]\omega = \dfrac {0.9397}{5}[/tex]

[tex]\rm \omega = 0.188\;rad/sec[/tex]

Now the tangential velocity is given by,

[tex]\rm v={100}\times{0.188}[/tex]

[tex]\rm v = 18.8\;ft/sec[/tex]

Now the time for a single revolution is given by,

[tex]\rm T = \dfrac{2\pi}{0.188}[/tex]

[tex]\rm T= 33.4\; sec[/tex].

Therefore, angular speed is  0.188 rad/sec , tangential speed is 18.8 ft/s ec and time for one revolution is 33.4 sec.

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Answer: A

Explanation:

Answer:

False

Explanation:

Activation energy is simply the initial energy input which is needed to proceed a chemical reaction. The source of this energy is heat, which is obtained when reactant molecules absorb thermal energy from their surroundings. This thermal energy provides the kinetic energy of moving molecules, by speeding up the motion of the reactant molecules.

Therefore, the correct option is false.

We can define density as per unit volume.

The SI unit to measure the density of solid, liquid and gas is kilogram per cubic metre (kg/m3) and in the centimetre–gram–second system of units (cgs unit) is gram per cubic centimetre (g/cm3). Gas density is very dependent of pressure and temperature whereas the density of solids and liquids is not so dependent, that is why the gas density is given at a standard temperature and pressure.

Chemists typically use g/mL or g/cm³ for liquids and solids, and g/L for gases. Differences arise due to the varying densities and volumes of these states of matter.

Chemists use different units for density based on the state of matter being measured. For liquids and solids, the density is usually expressed in grams per milliliter (g/mL) or grams per cubic centimeter (g/cm³). This is because liquids and solids generally have higher densities and smaller volumes compared to gases, making these units convenient for laboratory measurements and calculations.

For gases, the density is typically expressed in grams per liter (g/L). Gases have much lower densities and occupy larger volumes compared to liquids and solids, so g/L is more appropriate. The larger unit (liter) helps to manage the lower mass of gases, ensuring the numerical values are easy to work with.

The choice of units helps chemists maintain consistency and accuracy when measuring and comparing densities across different substances and states of matter. Using units that reflect the typical scales of measurement for each state ensures that the values are neither too large nor too small, facilitating easier interpretation and communication of data.

The cannonball takes approximately 4.40 seconds to hit the wall and strikes it at a height of around 157.15 meters.

Let's solve this step by step:

Step 1: Calculate the time of flight

To find how long it takes for the cannonball to hit the wall, we need to consider the horizontal motion.

So, it takes approximately 4.40 seconds for the ball to hit the wall.

Step 2: Calculate the height at which the ball hits the wall

For vertical motion:

Vertical displacement (y) after time t: [tex]y = u_y \times t - 0.5 \times g \times t^2[/tex] (where g is the acceleration due to gravity, approximately 9.8 m/s²)

Therefore, the ball hits the wall at a height of approximately 157.15 meters.

The complete question is as follows:

A cannon elevated at 40 degrees is fired at a wall 300 m away on level ground. The initial speed of the cannonball is 89 m/sec.

1) how long does it take for the ball to hit the wall?

2) at what hight does the ball hit the wall?

Final answer:

To find the period of simple harmonic motion for the given mass and spring constant, first convert units, then apply the period formula to get approximately 0.6283 seconds.

Explanation:

The period of simple harmonic motion for a mass-spring system can be calculated using the formula for the period T of a simple harmonic oscillator:

T = 2π√(m/k)

where m is the mass in kilograms, k is the spring constant in newtons per meter (N/m), and π is approximately 3.1416. To use this formula, we must convert the mass from pounds to kilograms and the spring constant from pounds per foot to newtons per meter. The period T is the time it takes for one complete cycle of oscillation.

First, convert 16 pounds to kilograms (1 pound is approximately 0.453592 kilograms):

16 pounds × 0.453592 = 7.257 kilograms

Now, convert the spring constant from lb/ft to N/m (1 lb/ft is approximately 14.5939 N/m):

49 lb/ft × 14.5939 = 715.6011 N/m

Using the conversion values:

T = 2π√(7.257 kg / 715.6011 N/m) = 2π√(0.010139 kg/N·m)

Perform the calculation to determine the period:

T = 2π√(0.010139) ≈ 2π√(0.01) ≈ 2π(0.1) ≈ 0.6283 seconds

Thus, the period of simple harmonic motion for the given mass-spring system is approximately 0.6283 seconds.

(a). Position of sandbag at time [tex]1.05\text{ s}[/tex] after its release is [tex]\boxed{39.84\text{ m}}[/tex] above the ground.

(b). Velocity of the sandbag after time [tex]1.05\text{ s}[/tex] is [tex]\boxed{5.3\text{ m/s}}[/tex].

(c). The time taken after release the bag to strike the ground is [tex]\boxed{3.41\text{ s}}[/tex].

Further explanation:

Here, all the actions performed is under free fall. So, we will use the kinematic equations of motion for free falling body.

Given:

The velocity of rising of hot air balloon is [tex]5\text{ m/s}[/tex].

Height of hot air balloon when sandbag released is [tex]40\text{ m}[/tex].

Calculation:

Part (a)

Position of sandbag at time [tex]1.05\text{ s}[/tex] after its release.

When sandbag released the hot air balloon was rising up with the velocity of [tex]5\text{ m/s}[/tex].

So, initial velocity of sandbag will be [tex]5\text{ m/s}[/tex] in upward direction.

So, the time taken by the sand bag to reach at its top position is given by,

[tex]\boxed{v = u - gt}[/tex]                                                     …… (1)

Here, [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity, [tex]g[/tex] is the acceleration due to gravity and negative sign is due upward motion of sandbag, [tex]t[/tex] is the time required to reach at top position.

Substitute values for v and u in equation (1).

[tex]\begin{aligned}0&=5-9.8t\\9.8t&=5\\t&=0.51\text{ s}\\\end{aligned}[/tex]

So, the distance travel by sandbag to top position can be calculated as,

[tex]\boxed{{v^2}={u^2}-2g{s_1}}[/tex]

Substitute values for [tex]v[/tex] and u in above equation.

[tex]\begin{aligned}{0^2}&={5^2}-2\times9.8\times{s_1}\\19.6{s_1}&=25\\{s_1}&=1.27\text{ m}\\\end{aligned}[/tex]

After that sandbag will start falling.

The time remain from the given time is,

[tex]\begin{aligned}{t_1}&=1.05-0.51\\{t_1}&=0.54\text{ s}\\\end{aligned}[/tex]

The distance travel by sandbag in [tex]0.54\text{ s}[/tex] in downward direction can be calculated as,

Substitute [tex]0[/tex] for [tex]u[/tex] and [tex]0.54\text{ s}[/tex] for [tex]t[/tex] in above equation.

[tex]\begin{aligned}{s_2}&=0\times0.54+\frac{1}{2}\times9.8{\left({0.54}\right)^2}\\&=1.43\text{ m}\\\end{aligned}[/tex]

So, the position of the sandbag after [tex]1.05\text{ s}[/tex] from the ground can be calculated as,

[tex]\begin{aligned}h&=40+{s_1}-{s_2}\\&=40+1.27-1.43\\&=39.84\text{ m}\\\end{aligned}[/tex]

Part (b)

Velocity of the sandbag after time [tex]1.05\text{ s}[/tex].

The velocity of the sandbag after time [tex]t[/tex] can be calculated as,

[tex]\boxed{v=u+gt}[/tex]

Substitute the values for [tex]u[/tex] and t in above equation.

[tex]\begin{aligned}v&=0+9.8\times0.54\\&=5.3\text{ m/s}\\\end{aligned}[/tex]

Thus, the velocity of the sandbag after time [tex]1.05\text{ s}[/tex] is [tex]\boxed{5.3\text{ m/s}}[/tex].

Part (c)

The time taken after release the bag to strike the ground.

The total distance the top most position of the bag and the ground is,

[tex]\begin{aligned}S&=40+{s_1}\\&=40+1.27\\&=41.27\text{ m}\\\end{aligned}[/tex]

Now, time taken by the bag to strike the ground from its top most position,

[tex]\boxed{S=ut+\dfrac{1}{2}g{t_2}^2}[/tex]

Substitute [tex]41.27{\text{ m}}[/tex] for [tex]S[/tex] and [tex]0[/tex] for [tex]u[/tex] in above equation.

[tex]\begin{aligned}41.27&=0\timest+\dfrac{1}{2}\times9.8{t_2}^2\\41.27&=4.9{t_2}^2\\{t_2}^2&=\dfrac{{41.27}}{{4.9}}\\&=2.9{\text{ s}}\\\end{aligned}[/tex]

Now, the total time taken by bag to strike the ground from the instant of release is,

[tex]\begin{aligned}T&={t_2}+t\\&=2.9+0.51\\&=3.41\text{ s}\\\end{aligned}[/tex]

Thus, the time taken after release the bag to strike the ground is [tex]\boxed{3.41\text{ s}}[/tex].

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Answer detail:

Grade: Senior School

Subject: Physics

Chapter: Kinematics

Keywords:

Hot air balloon, constant velocity, height of, position of sandbag, velocity of sandbag, total time, rising up, 5m/s, 40m, 1.05 s, displacement, balloonist, strike the ground.

The speed of the rain after accelerating for addition [tex]7\text{ s}[/tex] will be [tex]\boxed{11.62\text{ m/s}}[/tex].

Explanation:

Given:

The speed of the train after [tex]5.1\text{ s}[/tex] is [tex]4.9\text{ m/s}[/tex].

The initial speed of the train is [tex]0\text{ m/s}[/tex].

Concept:

The acceleration of a body is defined as the rate at which the velocity of the body in motion changes every second. If the acceleration of the body is in the direction of motion of the body, the body will be accelerated.

As the train stars from rest and accelerates away from the station, the speed of the train will increase according to the first equation of motion.

The expression for the first equation of motion for the motion of the train is:

[tex]\boxed{v_f=v_i+at}[/tex]

Here, [tex]v_f[/tex] is the final speed of the train, [tex]v_i[/tex] is the initial speed of the train, [tex]a[/tex] is the acceleration of the train and [tex]t[/tex] is the time taken by the train.

Substitute the values of velocity for first [tex]5.1\text{ s}[/tex] of motion of the train.

[tex]\begin{aligned}4.9&=0+a.(5.1)\\a&=\dfrac{4.9}{5.1}\text{ m/s}^2\\&=0.96\text{ m/s}^2\end{aligned}[/tex]

Now, the for the speed of the train after it travels for addition [tex]7.0\text{ s}[/tex] or a total of [tex]12.1\text{ s}[/tex].

[tex]\begin{aligned}v_f&=0+(0.96)(12.1)\text{ m/s}\\v_f&=11.62\text{ m/s}\end{aligned}[/tex]

Thus, The speed of the rain after accelerating for addition [tex]7\text{ s}[/tex] will be [tex]\boxed{11.62\text{ m/s}}[/tex].

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Answer Details:

Grade: Senior School

Subject: Physics

Chapter: Laws of motion

Keywords:

train, accelerates, constant, rest, equation of motion, initial, final, velocity, time taken, addition 7 seconds, acceleration.

The height of the building is 12.898 meters.

First, we calculate the time of flight using the horizontal distance and the horizontal velocity:

[tex]\[ t = \frac{d}{v_x} \][/tex]

where [tex]\( t \)[/tex] is the time of flight, [tex]\( d \)[/tex] is the horizontal distance (36.0 m) and is the horizontal velocity (22.2 m/s). Plugging in the values:

[tex]\[ t = \frac{36.0 \text{ m}}{22.2 \text{ m/s}} \][/tex]

[tex]\[ t = 1.6216 \text{ s} \][/tex]

Now, we use the time of flight to find the height of the building using the vertical motion equation:

[tex]\[ h = \frac{1}{2} g t^2 \][/tex]

where [tex]\( h \)[/tex] is the height of the building, [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.81 \text{ m/s}^2 \)[/tex]), and [tex]\( t \)[/tex] is the time of flight we just calculated. Plugging in the values:

[tex]\[ h = \frac{1}{2} \times 9.81 \text{ m/s}^2 \times (1.6216 \text{ s})^2 \][/tex]

[tex]\[ h = \frac{1}{2} \times 9.81 \text{ m/s}^2 \times 2.6297 \text{ s}^2 \][/tex]

[tex]\[ h = 4.905 \text{ m/s}^2 \times 2.6297 \text{ s}^2 \][/tex]

[tex]\[ h = 12.898 \text{ m} \][/tex]