A body is dropped from the roof of a 20 m high building by how much:
Does it take to reach the ground? How fast does it hit the ground?

Answer:

Length, l = 0.126 meters.

Explanation:

Given the following data;

Period = 6.98s

Acceleration due to gravity, g = 9.8m/s²

To find the length, l;

[tex] Period, T = 2 \pi \sqrt {lg} [/tex]

Substituting into the equation, we have;

[tex] 6.98 = 2*3.142 \sqrt {l*9.8} [/tex]

[tex] 6.98 = 6.284 \sqrt {9.8l} [/tex]

[tex] \frac {6.98}{6.284} = \sqrt {9.8l} [/tex]

[tex] 1.1108 = \sqrt {9.8l} [/tex]

Taking the square of both sides

[tex] 1.1108^{2} = 9.8l [/tex]

[tex] 1.2339 = 9.8l [/tex]

[tex] l = \frac {1.2339}{9.8} [/tex]

Length, l = 0.126m.

Answer:

amplitude is the answer

Answer:

The answer is a 4.2m

Explanation:

Given data

Please see attached the rough drawing for your reference.

From the drawing, you ran 18m west and 2.4m south

The displacement is

= 1.8+2.4

=4.2m

Answer:

27872.2 miles per hour

Explanation:

Given that :

Radius of Jupiter is approximately = 44,360 miles

Revolution is 10 hours ;

Jupiter makes one revolution in 10 hours :

Using the relation to obtain the velocity :

V = re

r = radius

w = 2π/T

Hence,

V = r * 2π/ T

V =44360 * 2 * π/10

V = 88720 * π/10

V = 278722.10 / 10

V = 27872.210

V = 27872.2 miles per hour

Solution :

The volume rate of flow is given by : R = 5.0 L/s

                                                                 [tex]$ = 5.0 \times 10^{-3} \ m^3/s$[/tex]

The radius of the pipe, [tex]$r_1= 5 \times 10^{-2} \ m$[/tex]

∴ [tex]$ 5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]

then, [tex]$v_1 = \frac{5.0 \times 10^{-3}}{(3.14)(5 \times 10^{-2})^2}$[/tex]

             = 0.637 meter per second

Then the speed of the water at wider section,

[tex]$R=A_1v_1$[/tex]

Similarly, the speed of water at narrow pipe.

The radius of the [tex]$r_2 = 2.5 \times 10^{-2}$[/tex] m

[tex]$5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]

then, [tex]$v_2 = \frac{5.0 \times 10^{-3}}{(3.14)(2.5 \times 10^{-2})^2}$[/tex]

             = 2.55 meter per sec

Now from Bernoulli's theorem,

[tex]$P_1 + \frac{1}{2} \rho v_1^2 =P_2 + \frac{1}{2} \rho v_2^2 $[/tex]

[tex]$P_1 = P_2 + \frac{1}{2} \rho (v_2^2 - v_1^2)$[/tex]

    [tex]$= 50 \kPa + (0.5)(10^3)[(2.55)^2-(0.637)^2]$[/tex]

    = 50 kPa + 3.05 kPa

    = 53.05 kPa

or 53000 Pa

This question involves the concepts of Bernoulli's Theorem and Volumetric Flowrate.

The pressure reading in the wide section is "53.05 KPa".

First, we will use the volumetric flow rate to find the velocities of the water at wide and narrow sections.

[tex]V = A_1v_1[/tex]

where,

V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s

r₁ = radius of narrow section = 5 cm/2 = 2.5 cm = 0.025 m

A₁ = Area of narrow section = πr₁² = π(0.025 m)²

v₁ = velocity at narrow section = ?

Therefore,

[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.025\ m)^2](v_1)\\\\v_1=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.025\ m)^2}\\\\v_1=2.55\ m/s\\[/tex]

Similarly,

[tex]V = A_2v_2[/tex]

where,

V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s

r₂ = radius of wide section = 10 cm/2 = 5 cm = 0.05 m

A₂ = Area of wide section = πr₁² = π(0.05 m)²

v₂ = velocity at wide section = ?

Therefore,

[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.05\ m)^2](v_2)\\\\v_2=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.05\ m)^2}\\\\v_2=0.64\ m/s\\[/tex]

Now, we will use Bernoulli's Theorem to find out the pressure wide section.

[tex]P_1 + \frac{1}{2}\rho v_1^2=P_2 + \frac{1}{2}\rho v_2^2[/tex]

where,

[tex]\rho[/tex] = density of water = 1000 kg/m³

P₁ = pressure in narrow section = 50 KPa = 50000 Pa

P₂ = pressure in wide section = ?

Therefore,

[tex]50000\ Pa + \frac{1}{2}(1000\ kg/m^3)(2.55\ m/s)^2=P_2 + \frac{1}{2}(1000\ kg/m^3)(0.64\ m/s)^2[/tex]

P₂ = 50000 Pa + 3251.25 Pa - 204.8 Pa

P₂ = 53046.45 Pa = 53.05 KPa

Learn more about Bernoulli's Theorem here:

https://brainly.com/question/13098748?referrer=searchResults

The attached picture shows Bernoulli's Theorem.

Answer:

3

Explanation:

my family i hope thinks of me.  And I don't have friends for them to think of me.

Answer:

Endothermic, because the reactants are lower in energy (C)

Explanation:

From the graph, you can see the energy of the products is higher than the energy of the reactants. If you recall that  when the enthalpy change  Eproducts is gretater than  Ereactants, the reaction is said to be endothermic.

Answer:

I hear points of low volume sound and points of high volume of sound.

Explanation:

This is because, since the two sources of sound have the same frequency and are separated by a distance, d = 10 mm, there would be successive points of constructive and destructive interference.

Since their frequencies are similar, we should have beats of high and low frequency.

So, at points of low frequency, the amplitude of the wave is smallest and there is destructive interference. The frequency at this point is the difference between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, f - f' = 400 Hz - 400 Hz = 0 Hz. So, the volume of the sound is low(zero) at these points.

Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.

So, as you wander around the room, I should hear points of high and low sound across the room.

Answer:

A. Doppler ultrasound collects data from moving objects

Explanation:

Did the test !!

Answer:A. Doppler ultrasound collects data from moving objects.

Explanation: just got it right

on my test

Answer:

Hello the diagram related to your question is attached below

answer: a) 851 m/s

             b)  8506.1 secs

Explanation:

calculate the periodic time of the satellite using the equation below

t = [tex]\frac{2\pi }{R} \sqrt{\frac{(R+h)^{3} }{g} }[/tex]  --  ( 1 )

where ; R = 6370 km

h = 500 km

g = 9.81 m/s^2

input given values into equation 1

t = 5670.75 secs

next calculate the periodic time taken by the space craft  

a) determine the increase in speed

V = v - [tex]\sqrt{\frac{gR^2}{R + h} }[/tex]  

where ; v = 8463 m/s , R = 6370 km, h = 500 km

V = 851 m/s

b) Determine the periodic time for the elliptic orbit

τ = [tex]\frac{3t}{2}[/tex]

 = [tex]\frac{3*5670.76}{2}[/tex]  =  8506.1 secs

attached below is the remaining part of the detailed solution

Answer:

Option D

2Na + Cl₂ —> 2NaCl

Explanation:

We'll begin by stating the law of conservation of matter.

The law of conservation of matter states that matter can neither be created nor destroyed during a chemical reaction but can be transferred from one form to another.

For an equation to comply with the law of conservation of matter, the number of atoms of each element must be the same on both side of the equation. This simply means that the equation must be balanced!

NOTE: An unbalanced equation simply means matter has been created or destroyed.

Now, we shall determine which equation is balanced. This can be obtained as follow:

For Option A

Na + Cl₂ —> 2NaCl

Reactant:

Na = 1

Cl = 2

Product:

Na = 2

Cl = 2

Thus, the equation is not balanced!

For Option B

2Na + 2Cl₂ —> 2NaCl

Reactant:

Na = 2

Cl = 4

Product:

Na = 2

Cl = 2

Thus, the equation is not balanced!

For Option C

2Na + Cl₂ —> NaCl

Reactant:

Na = 2

Cl = 2

Product:

Na = 1

Cl = 1

Thus, the equation is not balanced!

For Option D

2Na + Cl₂ —> 2NaCl

Reactant:

Na = 2

Cl = 2

Product:

Na = 2

Cl = 2

Thus, the equation is balanced!

From the above illustrations, only option D has a balanced equation. Thus, option D illustrate the law of conservation of matter.

Answer:

14.1seconds approx

Explanation:

Given data

Distance= 59.1m

Your velocity= 2.35m/s

Walkway velocity= 1.85m/s

Total velocity= 2.35+1.85= 4.2m/s

We know that

Velocity= distance/time

time= distance/velocity

substitute

time= 59.1/4.2

time= 14.07

time=14.1seconds approx

Hence the time is 14.1seconds approx

Answer:

d = 29.89 m

Explanation:

To solve this, we need to separate this problem in two parts.

One part would be the the time taken by the rock to actually hit the bottom, and the other part would be the time taken by the sound to reach the inspector.

Joining these two times we have:

t = t₁ + t₂    (1)

This time is 2.56 s.

Now, as we are asked to determine the distance from the top floor to the bottom, and we have two times taken in different ways, one by sound and the other the actual, we can say the same thing on distance, we need a distance relationed to the time taken by rock to hit the bottom, and the other distance relationet to the time taken by sound to reach the inspector.

Doing this we have that the distance traveled by the rock is:

y₁ = gt²/2

y₁ = 9.8t²/2 = 4.9t₁²    (2)

Now, the distance traveled by sound would be:

y₂ = v * t₂ = 336t₂   (3)

Remember that the speed of the sound is 336 m/s

From this last expression (3), we can actually write t₂ in function of t₁, using (1):

2.56 = t₁ + t₂

t₂ = 2.56 - t₁    (4)

Replacing (4) in (3):

y₂ = 336(2.56 - t₁)    (5)

Now that we have y₁ and y₂, we can equal (2) and (5), both expressions to get the value of t₁, and then, calculate the distance:

4.9t₁² = 336(2.56 - t₁)

4.9t₁² = 860.16 - 336t₁

4.9t₁² + 336t₁ - 860.16 = 0

Using the quadractic formula, we can calculate t₁:

t₁ = -336 ±√(336)² + 4*4.9*860.16 / (2*4.9)

t₁ = -336 ±√129,7555.136 / 9.8

t₁ = -336 ± 360.21 / 9.8     Using only the positive value we have:

t₁ = 2.47 s

This means that the rock hits the bottom in 2.47 s, and the remaining 0.09 s belongs to the time taken by sound. (2.47 + 0.09 = 2.56 s)

With this, we can calculate the distance of the rock using expression (2):

y₁ = 4.9 * (2.47)²

Hope this helps

Answer:The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would..

Explanation: It would decrease.

The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would decrease because the acceleration due by gravity on the moon is less than the acceleration due to gravity on the earth, therefore the correct answer is option C.

It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another.

In comparison to the Earth, the Moon is less massive. Your weight would drop if you traveled to the moon because the acceleration caused by gravity there is lower than that caused by gravity here on Earth.

As a result of the less gravity on the moon, the weight would decrease.

Thus, the correct answer is option C.

To learn more about gravity from here, refer to the link;

brainly.com/question/4014727

#SPJ2

Answer:

120.237 seconds

Explanation:

Given that:

V = 144 V

I = 10.5 A

H = 60.6 kJ

Using the formula:

H = I²RT

From H = I²RT; making T the subject, we have:

[tex]T = \dfrac{H}{I^2R}[/tex]

where;

[tex]R = \dfrac{V}{I}[/tex]

∴

[tex]T = \dfrac{H}{V \times I}[/tex]

[tex]T = \dfrac{60.6 \times 10^3 }{144 \times 10.5}[/tex]

T = 40.079

[tex]T_{neq} = 3T[/tex]

[tex]T_{neq} =3 \times 40.079[/tex]

[tex]\mathbf{T_{neq} =120.237 \ sec}[/tex]

The car's rate of  acceleration : a = 2.04 m/s²

Given

speed = 110 km/hr

time = 15 s

Required

The acceleration

Solution

110 km/hr⇒30.56 m/s

Acceleration is the change in velocity over time

a = Δv : Δt

Input the value :

a = 30.56 m/s : 15 s

a = 2.04 m/s²

Answer:

nique ta mama  

Explanation:

Answer:

Only the car transforms electrical energy into more than one form of energy.

Explanation:

The motion of the car is mechanical energy but it can also transform into another energy witch is electrical energy

Answer:

m = 4.11 grams

Explanation:

Given that,

The specific heat of Iron, c = 0.45 J/g°C

Heat removed, [tex]Q=1.16\times 10^2\ J[/tex]

Initial temperature, [tex]T_i=89^{\circ} C[/tex]

Final temperature, [tex]T_f=26.41^{\circ} C[/tex]

We need to find the mass of the iron. We know that the heat removed in terms of specific heat is given by :

[tex]Q=mc\Delta T\\\\m=\dfrac{Q}{c\Delta T}\\\\m=\dfrac{-1.16\times 10^2}{0.45\times (26.41-89)}\\\\m=4.11\ g[/tex]

So, the mass of the iron is 4.11 grams.

Answer:

PRECAUTIONS

-The refracting faces of the glass prism should be smooth, transparent and without any air bubble or broken edge. ...

-Use a sharp pencil to draw boundary of the prism and rays of light.

-The alpins should have sharp tip and should be fixed exactly vertical to the plane of the paper.

Explanation:

Please give thanks to all my answers and please mark as brilliant and please follow me

Answer:

10 J.

Explanation:

Given that,

Net force acting on the rake, F = 2 N

Distance moved by the rake, d = 5 m

We need to find the kinetic energy gained by the rake. We know that,

Kinetic energy = work done

So,

K = F×d

K = 2 N × 5 m

K = 10 J

So, 10 J of kinetic energy is gained by the rake.

Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.

How much kinetic energy does the rake gain?

Answer: 10 J

Answer:

Vf = 29.4 m/s

h = 44.1 m

Explanation:

Data:

==================================================================

Final Velocity

Use formula:

Replace:

Multiply:

==================================================================

Height

Use formula:

Replace:

Multiply time squared:

Simplify the s², and multiply in the numerator:

It divides:

What is the velocity when falling to the ground?

The final velocity is 29.4 meters per seconds.

How high is the building?

The height of the building is 44.1 meters.

C

It is right because I took this and I got this answer correct

Answer:

It is a screw.

Explanation:

Answer:

A)  Δf = - 49 m, B)  v (7)  = -56 m / s, a = - 8 m / s², C)  t = 0.866 s

Explanation:

A) In this exercise ask to find the displacement and the average velocity, give the function of the movement

           f (t) = - 4t² +3

and the range of motion 0≤ t ≤ 7

the displacement is

for t = 0

           f (0) = 3

for t = 7 s

           f (7) = - 4 7² +3

           f (7) = -46 m

the total displacement is

           Δf = f (7) - f (0)

           Δf = -46 - 3

            Δf = - 49 m

the average speed is defined as the displacement between the time interval

           v = Df / Dt

           v = -49 / 7

           v = - 7 m / s

B) the speed and acceleration of the end points of the motion

the speed of defined by

          v = [tex]\frac{dx}{dt}[/tex]

in this case

         v = [tex]\frac{df}{dt}[/tex]

         v = -8t

let's calculate

          v (7) = -8 7

          v (7)  = -56 m / s

acceleration is defined by

          a = [tex]\frac{dv}{dt}[/tex]

          a = - 8 m / s²

acceleration is constant throughout the movement

C) the point where the direction changes.

This point is a point where the position goes from positive to negative, the point f = 0

        0 = -4t² +3

        t = √¾

        t = 0.866 s

Answer:

0.392

Explanation:

Mm = 0.11Me

Rm = 0.53Re

g = GM / r^2

G = 6.67 * 10^-11

gmars = (G * 0.11Me) / (0.53Re)^2

Recall:

gearth = GMe /Re^2

Hence, gmars in terms of gearth equals

gmars = gearth * (0.11 / 0.53^2)

gmars = gearth * 0.3915984

gmars = 0.392gearth