an arrow is shot horizontally from the top of a tower at a speed of 15m/s and hits the ground with a speed of 25m/s. calculate the height of the tower

The two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.

A double replacement reaction have two ionic compounds that are exchanging anions or cations.

From the given options, we can choose the following based on their exchange of anions or cations.

Thus, the two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.

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Answer:

W = 1.875 J

Explanation:

For this exercise let's use the relationship between work and kinetic energy

          W = ΔK

The kinetic energy of rotational motion is

         K₀ = ½ I w²

we can assume that the box is small, so it can be treated as a point object, with moment of inertia

          I = m rₐ²

angular and linear velocity are related

          v = w r

          w = v / r

we substitute in the equation, for point A

         K₀ = ½ (m rₐ²) (v / rₐ)²

         K₀ = ½ m v²

For the final point B, as the system is isolated the angular momentum is conserved

initial        L₀ = Io wo

final          L_f = I_f w_f

                L₀ = L_f

                 I₀ w₀ = I_f w_f

                (m rₐ²) w₀ = (m  [tex]r_{b} ^2[/tex]) w_f

                 w_f = (rₐ/r_b)² w₀

with this value we find the final kinetic energy

         K_f = ½ I_f w_f²

         K_f = ½ (m [tex]r_{b}^2[/tex]) ( (rₐ / r_b)²  w₀) ²

         K_f = ½ m [tex]\frac{r_a^4}{r_b^2} \ w_o^2[/tex]

we substitute in the realcion of work

          W = K_f - K₀

          W = ½ m  [tex]( \( \frac {r_a^2 }{r_b} )^2[/tex] w₀² - ½ m v²

          W = ½ m  [tex]\frac{r_a^4}{r_b^2} ( \frac{v}{r_a} ) ^2[/tex] - ½ m v²

           W = ½ m [tex]\frac{r_a^2}{r_b^2} \ v^2[/tex] - ½ m v2

          W = ½ m v² (([tex]( \ (\frac{r_a}{r_b})^2 -1)[/tex]

let's calculate

           W = ½ ( [tex]\frac{8}{32}[/tex] ) 5 ((2/1)² -1)

           W = 0.625 (3)

           W = 1.875 J

Answer:

SO₂ + 2H₂ —> S + 2H₂O

The coefficients are: 1, 2, 1, 2

Explanation:

SO₂ + H₂ —> S + H₂O

The above equation can be balance as follow:

SO₂ + H₂ —> S + H₂O

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by writing 2 before H₂O as shown below:

SO₂ + H₂ —> S + 2H₂O

There are 2 atoms of H on the left side and 4 atoms the right side. It can be balance by writing 2 before H₂ as shown below:

SO₂ + 2H₂ —> S + 2H₂O

Now, the equation is balanced.

The coefficients are: 1, 2, 1, 2

Answer:

1 and 3

Explanation:

because they are going up from 0

Answer:

A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").

A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).

a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.

b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.

Answer:

A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").

A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).

a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.

b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.

Explanation:

The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.

The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;

ωf = ωi + 2αθ

where;

ωf = 0 + 2(2.0) x 6.3

ωf = 25.2 rad/s

Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.

​

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Answer: 5.0 rad/s

Explanation: Because that’s what khan said so try it out.

Answer:

In the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.

Explanation:

It is given that a student performs two types of experiment to see how change in its resistance while in the state of motion and in rest.

In the first experiment, an object is pushed with a force against a horizontal surface and the speed is measured using a sensor. Here, work is done against the inertia of the object as it is pushed from rest. So the mass is inertial mass.

In the second experiment, an object is pushed or thrown upwards with a force and speed is measured. Here, the mass is gravitational mass as the work done in the second experiment is against the gravity or against the weight of the object.

In the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.

As per the given problem, the student performs two types of experiment to see how change in its resistance while in the state of motion and in rest.

Thus, we can conclude that the in the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.

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Answer:

[tex]36.44\ \text{kg m/s}\hat{i}[/tex]

Explanation:

I = Moment of inertia of sphere = [tex]\dfrac{2}{5}MR^2[/tex]

M = Mass of sphere = 29 kg

R = Radius of sphere = 0.5 m

T = Time taken for one revolution = 0.5 s

[tex]\omega[/tex] = Angular velocity = [tex]\dfrac{2\pi}{T}[/tex]

[tex]L=I\omega\\\Rightarrow L=\dfrac{2}{5}MR^2\dfrac{2\pi}{T}\\\Rightarrow L=\dfrac{4MR^2\pi}{5T}\\\Rightarrow L=\dfrac{4\times 29\times 0.5^2\pi}{5\times 0.5}\\\Rightarrow L=36.44\ \text{kg m/s}[/tex]

The rotational angular momentum of the sphere is [tex]36.44\ \text{kg m/s}\hat{i}[/tex].

Answer:

speed: 35m/s

direction: left

Explanation:

Assuming the right side is the positive direction:

before explosion:

P = mv = 0

after explosion:

P' = 15P + 5P

(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')

P' = 0.75mv1' + 0.25mv2'

P' = (15kg)v' + (5kg)(105m/s)

P' = 525kg/m/s + (15kg)v1'

P = P'

525kg/m/s + (15kg)v1' = 0

(15kg)v1' = -525kg/m/s

v1' = -35m/s

speed = |-35| = 35m/s

direction is to the left since the right side is the positive direction.

Answer:

The moon does not stay at the same distance of the earth because the ortbit of the moon is slightly elliptical. If earth is not tilted at an angle of 66.5°, there will be no change in the season and the earth will have equal length of days and night.

Explanation:

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Answer:

Get to work.

Explanation:

This is not Brainly material. This is a project that you need to get started on and do your best on. It always looks worst before you get started on it. Just sink your teeth in and you'll be finishing it up before you know it.

Answer:

careers related to the promotion or treatment of reproductive health and choose one career to focus on. You are going to identify the key factors of the educational path that you would need to take to work in this career, the annual salary that you would earn, the schedule that you may find yourself working, physical requirements for the job, and how long it will take you to achieve this goal.

Explanation:

Answer:

B. 3.1 × 10^5 V

Explanation:

Answer:

B

Explanation:

e2021

Answer:

B. it increases

Explanation:

As shown in the table provided, the speed of sound in water (1493 m/s) is greater than the speed of sound in air (346 m/s).

Answer:

B is the correct answer.

Explanation:

Answer:

I guess it is kinetic energy

Answer:

kinetic energy because my dog told me

Answer:

C. Constant

Explanation:

The total energy of the cannonball remains constant as it travels through the air.

Answer:

Explanation:

hi my name is Ava

Answer:

a. i. 3.43 m/s ii. 2.8 m/s

b. The thin-walled cylinder

Explanation:

a. Find translational speed of each cylinder upon reaching the bottom

The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy

So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.

i. translational speed of thin-walled cylinder upon reaching the bottom

So, For the thin-walled cylinder, I = mr², we find its translational velocity, v

So, mgh = 1/2mv² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²)(v/r)²  

mgh = 1/2mv² + 1/2mv²

mgh = mv²

v² = gh

v = √gh

v = √(9.8 m/s² × 1.2 m)

v = √(11.76 m²/s²)

v = 3.43 m/s

ii. translational speed of solid cylinder upon reaching the bottom

So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'

So, mgh = 1/2mv'² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²  

mgh = 1/2mv'² + mv'²

mgh = 3mv'²/2

v'² = 2gh/3

v' = √(2gh/3)

v' = √(2 × 9.8 m/s² × 1.2 m/3)

v' = √(23.52 m²/s²/3)

v' = √(7.84 m²/s²)

v' = 2.8 m/s

b. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Since v = 3.43 m/s > v'= 2.8 m/s,

the thin-walled cylinder has the greatest translational speed upon reaching the bottom.

Answer:

The example of the center of the gravity is the middle of a seesaw

Explanation:

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Answer:

8 kV

Explanation:

Here is the complete question

Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?

Solution

Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V

So, Q = CV

= 500 × 10⁻⁶ F × 800 V

= 400000 × 10⁻⁶ C

= 0.4 C

Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV

V = Q/C

= 0.4 C/500 × 10⁻⁶ F

= 0.0008 × 10⁶ V

= 800 V

The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)

= 10 × 800 V

= 8000 V

= 8 kV

Answer:

The third approach to cross-cultural studies of personality is the combined approach, which serves as a bridge between Western and indigenous psychology as a way of understanding both universal and cultural variations in personality

Explanation:

Answer:

The third approach to cross-cultural studies of personality is the combined approach, which serves as a bridge between Western and indigenous psychology as a way of understanding both universal and cultural variations in personality

Answer:

a) The minimum thickness of the oil slick at the spot is 313 nm

b) the minimum thickness be now will be 125 nm

Explanation:  

Given the data in the question;

a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

t[tex]_{min[/tex] = λ/2n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.20

we substitute

t[tex]_{min[/tex] = 750 / 2(1.20)

t[tex]_{min[/tex] = 750 / 2.4

t[tex]_{min[/tex] = 312.5 ≈ 313 nm

Therefore, The minimum thickness of the oil slick at the spot is 313 nm

b)

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

minimum thickness of the oil slick at the spot will be;

t[tex]_{min[/tex] = λ/4n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.50

we substitute

t[tex]_{min[/tex] = 750 / 4(1.50)

t[tex]_{min[/tex] = 750 / 6

t[tex]_{min[/tex] = 125 nm

Therefore, the minimum thickness be now will be 125 nm

Answer:

1.9m/s²

Explanation:

Use the equation v=u+at, where v is the final speed, u is the initial speed, a is the acceleration and t is the time.

v=u+at

15.3=0+a(8)

a=15.3/8

a= 1.9125 m/s²

Answer:

The final speed will be "[tex]1.185\times 10^7 \ m/sec[/tex]".

Explanation:

The given values are:

Potential difference,

Δv = 400 v

Radius,

r = 0.5580 cm

As we know,

⇒  [tex]W=e \Delta v[/tex]

and,

⇒  [tex]\frac{1}{2}mv^2=e \Delta v[/tex]

then,

⇒  [tex]v=\sqrt{\frac{2e \Delta v}{m} }[/tex]

On substituting the values, we get

⇒     [tex]=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }[/tex]

⇒     [tex]=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}[/tex]

⇒     [tex]=1.185\times 10^7 \ m/sec[/tex]

Answer:

0.03 A

Explanation:

From the question given above, the following data were obtained:

Voltage (V) = 12 V

Resistor (R) = 470 Ω

Current (I) =?

From ohm's law, the voltage, current and resistor are related by the following formula:

Voltage = current × resistor

V = IR

With the above formula, we can obtain the current in the circuit as follow:

Voltage (V) = 12 V

Resistor (R) = 470 Ω

Current (I) =?

V = IR

12 = I × 470

Divide both side by 470

I = 12 / 470

I = 0.03 A

Thus, the current in the circuit is 0.03 A

Answer:

0.03 A

Explanation:

Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit

0.03 A

A car making this turn is pulled downward by its own weight, and pushed up by the road at an angle of 45°, so by Newton's second law,

• the net horizontal force on the car is

∑ F = N cos(45°) = m a = m v ² / R

• the net vertical force on the car is

∑ F = N sin(45°) - m g = 0

where

• N = magnitude of the normal force

• m = mass of the car

• a = v ² / R = centripetal acceleration of the car

• v = tangential speed of the car

• R = 100 m = radius of curvature

• g = 9.8 m/s² = acceleration due to gravity

From the net vertical force equation, we get

N = m g / sin(45°)

and substituting this into the net horizontal force equation and solving for v gives

(m g / sin(45°)) cos(45°) = m v ² / R

v = √(R g cos(45°) / sin(45°)) ≈ 31 m/s

We have that A highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of

V=32m/s

From the question we are told

a highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of:

Generally the equation for the Velocity is mathematically given as

[tex]V=\sqrt{rgtan\theta}[/tex]

Therefore

[tex]V=\sqrt{rgtan\theta}\\\\V=\sqrt{100*9.8*tan45}\\\\V=32m/s[/tex]

Therefore

A highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of

V=32m/s

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Answer:

Explanation:

Moment of inertia I = M k² , where M is mass and k is radius of gyration .

Putting the given values in the equation

5000 = 200 x k²

k² = 25

k = 5 cm .

Radius of gyration is 5 cm .

Answer:

https://wtamu.edu/~cbaird/sq/2015/02/26/when-i-sit-by-a-campfire-how-does-its-hot-air-heat-me/#:~:text=When%20you%20sit%20by%20a,It%20comes%20from%20thermal%20radiation.&text=Since%20air%20is%20a%20good,of%20pockets%20of%20heated%20fluid.

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