A storage tank has the shape of an inverted circular cone with height 15 m and base radius of 5 m. It is filled with water to a height of 11 m. Find the work required to empty the tank by pumping all of the water to the top of the tank. (The density of water is 1000 kg/m3. Assume g = 9.8 m/s2 is the acceleration due to gravity. Round your answer to the nearest integer.) g

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Actually Welcome to the Concept of the Waves.

Wavelength is the distance between two consecutive crest or trough.

hence, here the distance is 10cm

So the wavelength is 10cm

===> 10 cm

Answer:

)Give the definition of poverty line as defined by the World Bank.

Given values are:

Speed,

Time,

As we know,

→ [tex]Acceleration = \frac{Speed}{Time}[/tex]

or,

→                   [tex]a = \frac{v}{t}[/tex]

By substituting the values,

                       [tex]= \frac{30}{3}[/tex]

                       [tex]= 10 \ m/s^2[/tex]

Thus the above solution is correct.    

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Answer:

it’s cosmopolitan

Explanation:

k12

Answer:

83.2 m/s to the West

Explanation:

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

Mass of plane = 180 Kg

Mass of pilot = 70 Kg

Momentum = 20800 Kg∙m/s West

Velocity =?

Next, we shall determine the total mass. This can be obtained as follow:

Mass of plane = 180 Kg

Mass of pilot = 70 Kg

Total mass =?

Total mass = Mass of plane + Mass of pilot

Total mass = 180 + 70

Total mass = 250 Kg

Finally, we shall determine the velocity. This can be obtained as follow:

Total mass = 250 Kg

Momentum = 20800 Kg∙m/s West

Velocity =?

Momentum = mass × Velocity

20800 = 250 × Velocity

Divide both side by 250

Velocity = 20800 / 250

Velocity = 83.2 m/s West

Thus, the velocity of both plane and pilot is 83.2 m/s to the West

Answer:

one sec let me think

Explanation:

(a)The average velocity of the ball over the following time intervals will be  [3,4] ft/sec.

(b)The instantaneous velocity at time t = 3 will be32 ft/sec.

(c)The instantaneous velocity at time t = 6 will be -64 ft/sec.

(d)The ball will hit the ground at 13.4 sec.

The change of displacement with respect to time is defined as the velocity.  velocity is a vector quantity. it is a time-based component.

The given data in the question will be ,

u is the initial velocity by which ball thrown=128 ft/sec.

V₃ is the instantaneous velocity at time t=3 sec.

V₆ is the instantaneous velocity at time t=6 sec.

t is the time when ball hits the ground=?

(a) Given equation for the displacement

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

Time when velocity is zero will be

[tex]\rm{ t=\frac{128}{32}[/tex]

[tex]\rm{ t=4 sec[/tex]

If the velocity got in the equation is 128 and 32 ft /sec. it can be only when the average velocity is [3,4] ft/sec .

Hence the average velocity obtained from the problem will be  [3,4] ft/sec

(b)  

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

At time( t=3 sec)

v(3) = 128-32×3

v(3) =32 m/sec.

Hence the instantaneous velocity at time t = 3 will be32 ft/sec.

(c)

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

At time( t=6 sec)

v(6) = 128-32×6

v(6) = -64 m/sec.

Hence the instantaneous velocity at time t = 6 will be -64 ft/sec.

(d)

According to Newtons third equation of motion we got

v=u+gt

If the body returens from a certain height at max height its velocity must be zero; ( u=0)

[tex]\rm t=\frac{(v-u)}{g} \\\\\ \rm t=\frac{(128-0)}{9.81}\\\\\rm t=13.04 sec.[/tex]

Hence the ball will hit the ground at 13.4 sec.

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Answer:

i need help with the same question

Explanation:

Answer:

[tex]2.4375\ \text{m/s}[/tex]

Explanation:

[tex]m_1[/tex] = Mass of cannon ball = 39 kg

[tex]m_2[/tex] = Mass of performer = 65 kg

[tex]u_1[/tex] = Initial horizontal component of cannon ball's velocity = 6.5 m/s

[tex]u_2[/tex] = Initial horizontal component of performer's velocity = 0

v = Velocity of combined mass

As the momentum of the system is conserved we have

[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{39\times 6.5+0}{39+65}\\\Rightarrow v=2.4375\ \text{m/s}[/tex]

The speed of the performer immediately after catching the cannon ball is [tex]2.4375\ \text{m/s}[/tex].

6 floors down from 12 floors underground = 18 floors underground.

6 degrees colder than 12 degrees below zero  =  18 degrees below zero

6 brown cows taken away from -12 brown cows = -18 brown cows

6 cars sold from a dealer that 12 cars were stolen from = 18 cars gone

6xy taken away from -12xy  =  -18xy

The weight is measured in a unit called Newton. The newton is a standard unit of weight measurements.

The International System of Units (SI) uses Newtons (N) as the unit of weight measurement. Weight is frequently expressed in pounds (lb) or ounces (oz) in some nations that employ the Imperial system. However, Newton is the accepted weight measurement unit in scientific and international contexts.

The International System of Units (SI) uses Newton as the primary unit of weight measurement. Sir Isaac Newton, a great physicist who significantly influenced our understanding of classical mechanics, is honored by having his name attached to it. Newton is the force needed to accelerate a mass of one kilogram by one meter per second squared in the SI system.

Hence, the weight is measured in a unit called Newton. The newton is a standard unit of weight measurements.

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Answer:

(a) The change in momentum is 12.04 kg-m/s

(b) The force exerted by the bat is 1003.33 N

Explanation:

Given that,

The mass of a ball, m = 0.14 kg

Initial speed of the ball, u = 40 m/s

Final speed of the ball, v = -46 m/s

(a) The change in momentum of the ball during the collision with the bat is given by :

[tex]\Delta p=m(v-u)\\\\=0.14(-46-40)\\\\=-12.04\ kg-m/s[/tex]

(b) Time for collision, t = 0.012 s

Now the force can be calculated as follows :

[tex]F=\dfrac{\Delta p}{t}\\\\F=\dfrac{12.04}{0.012}\\\\=1003.33\ N[/tex]

Hence, this is the required solution.

Answer:

a. = 12.04 kg*m/s

b. = 1,003.3N

Explanation:

The answer above is correct.

Answer:

160.43 meters

Explanation:

T=(2*pi*r)/v

840=(2*pi*r)/1.2

1008=2*pi*r

504=pi*r

160.43=radius

The distance around the circular track is equal to its circumference. If it takes840 s  to complete the path with a velocity of 1.20 m/s then the radius of the path will be 160 m.

Velocity is a physical quantity measuring the distance traveled by unit time. It is a vector quantity thus, characterized by its magnitude and direction. Velocity is usually expressed in m/s.

Velocity is the ratio of distance to the time taken to complete the travel. Thus, greater distance within short time indicates greater velocity.

Given that the time take to complete the circular path = 840 s

velocity = 1.20 m/s

distance = circumference of the circular path = 2πr.

distance = time × velocity

2πr = 840 s × 1.20 m/s

      = 1008 m.

Therefore, the radius r of the circular path  = 1008 m/ 2 π = 160 m.

Hence, the radius of the circular path 160 m.

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Answer:

t_pass = 2.34 m

t_stop = 4.68 s

Thus, for the car passing at constant speed the pedestrian will have to wait less.

Explanation:

If the car is moving with constant speed, then the time taken by it will be given as:

[tex]t_{pass} = \frac{D}{v}[/tex]

where,

t_pass = time taken = ?

D = Distance covered = 23 m

v = constant speed = (22 mi/h)(1609.34 m/1 mi)(1 h/3600 s) = 9.84 m/s

Therefore,

[tex]t_{pass} = \frac{23\ m}{9.84\ m/s} \\[/tex]

t_pass = 2.34 m

Now, for the time to stop the car, we will use third equation of motion to get the acceleration first:

[tex]2as = v_{f}^{2} - v_{i}^2\\a = \frac{v_{f}^{2} - v_{i}^2}{2D}\\\\a = \frac{(0\ m/s)^{2}-(9.84\ m/s)^2}{(2)(23\ m)}\\\\a = -2.1\ m/s^2[/tex]

Now, for the passing time we use first equation of motion:

[tex]v_{f} = v_{i} + at_{stop}\\t_{stop} = \frac{v_{f}-v_{i}}{a}\\\\t_{stop} = \frac{0\ m/s - 9.84\ m/s}{-2.1\ m/s^2}[/tex]

t_stop = 4.68 s

Constant velocity is the velocity which covers the same distance for each interval of the time.

The time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.

Constant velocity is the velocity which covers the same distance for each interval of the time.

It can be given as,

[tex]v=\dfrac{x}{t}[/tex]

As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.

Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.

Thus amount of time, [tex]t_{pass}[/tex] is,

[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]

As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.

Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.

Thus amount of time, [tex]t_{pass}[/tex] is,

[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]

According to the third equation of the motion acceleration can be given as,

[tex]v^2-u^2=2ax\\a=\dfrac{v^2-u^2}{2x}\\a=\dfrac{0^2-9.84^2}{2\times 23}\\a=-2.1 \rm \; m/s^2[/tex]

Now, use the first equation of motion, to get the required time,

[tex]v=u+at\\0=9.84+(-2.1)t\\t=4.68\rm \; s[/tex]

Therefore, the time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.

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Answer:

c

Explanation:

Answer:

rotates

Explanation:

I'm so bored

yrfgggghhgghhyuj

Answer:

what situation?

Explanation:

Answer:

An object is considered to be in motion only when its position changes over time with reference to a point which will be known as orgin

Answer:

im not sure maybe viberation

Explanation:

Answer:

X-ray waves are transmitted by ....... body tissues with very little absorption

Hope it helps

Please mark me as the brainliest

Thank you

Answer:

λ = 5.434 x 10⁻⁷ m = 543.4 nm

Explanation:

To solve this problem we can use the formula provided by Young's Double Slit experiment:

[tex]\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta xd}{L}[/tex]

where,

λ = wavelength of light = ?

Δx = distance between adjacent bright fringes = 1.6 cm = 0.016 m

d = slit separation = 0.18 mm = 0.00018 m

L = Distance between slits and screen = 5.3 m

Therefore,

[tex]\lambda = \frac{(0.016\ m)(0.00018\ m)}{5.3\ m}[/tex]

λ = 5.434 x 10⁻⁷ m = 543.4 nm

Answer:

24.5 m/s

Explanation:

KE=1/2mv^2

15000=1/2(50)v^2

30000=(50)v^2

600=v^2

sqrt600=v

v=24.5 m/s!!

Answer:

a)       [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] =  [tex]\frac{d^2 \theta}{d t^2}[/tex]

b)     f = 2π  [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]  

Explanation:

a) To have the equations of motion, let's use Newton's second law.

Let's set a reference system where the x-axis is parallel to the path and the y-axis is in the direction of tension of the rope.

For this reference system the tension is in the direction of the y axis, we must decompose the weight and the electrical force.

Let's use trigonometry for the weight that is in the vertical direction down

             sin θ = Wₓ / W

             cos θ = W_y / w

             Wₓ = W sin θ

             W_y = W cos θ

we repeat for the electric force that is vertical upwards

              F_{ex} = F_e sin θ

              F_{ey} = F_e cos θ

the electric force is

               F_e = q E

where the field created by an infinite plate is

               E = [tex]\frac{ \sigma}{2 \epsilon_o}[/tex]  

let's write Newton's second law

Y  axis  

           T - W_y = 0

            T = W cos θ

X axis

            F_{ex} - Wₓ = m a                   (1)

we use that the acceleration is related to the position

            a = dv / dt

            v = dx / dt

where x is the displacement in the arc of the curve

substituting

            a = d² x /dt²

we substitute in 1

           q E sin θ - mg sin θ = m [tex]\frac{d^2 x}{dt^2}[/tex]

we have angular (tea) and linear (x) variables, if we remember that angles must be measured in radians

           θ = x / R

           x = R θ

we substitute

           sin θ (q E - mg) = m \frac{d^2 R \ theta}{dt^2}  

          [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] =  [tex]\frac{d^2 \theta}{d t^2}[/tex]

this is the equation of motion of the system

b) for small oscillations

         sin θ = θ

therefore the solution is simple harmonic

      θ = θ₀ cos (wt + Ф)

if derived twice, we substitute

- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o}  ) \frac{\theta}{R } θ₀ cos (wt + Ф) = -w² θ₀ cos (wt + Ф)

       w² =  [tex]\frac{g}{R}[/tex] - [tex]\frac{q}{m} \frac{ \sigma }{2 \epsilon_o} \frac{1}{R}[/tex]  

angular velocity is related to frequency

       w = 2π f

        f = 2π / w

        f = 2π/w

        f = 2π  [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]  

Answer:

2.2 ma

Explanation:

Given :

Length of the rope = L

Mass of the rope = m

Mass of the object pulled by the rope = M

M = 1.5 m

So, L [tex]$\rightarrow$[/tex] m

For unit length [tex]$\rightarrow \frac{m}{L}$[/tex]

∴ 0.3 L = [tex]$0.3 \ L \left(\frac{m}{L}\right)$[/tex]

            = 0.3 m

And for remaining 0.7 L =  [tex]$0.7 \ L \left(\frac{m}{L}\right)$[/tex]

            = 0.7 m

By Newtons law of motion,

F - T = ( 0.3 m) a .........(1)

T = ( M + 0.7 m) a

T = ( 1.5 m + 0.7 m) a

T = ( 2.2 m ) a  ..............(2)

So from equation (1) and (2), we have

Tension on the rope

T = 2.2 ma