When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant urinates at a remarkable rate of 0.0060 m3 (a bit over a gallon and a half) per second. Assume that the urine exits 1.0 m below the bladder and passes through the urethra, which we can model as a tube of diameter 8.0 cm and length 1.2 m. Assume that urine has the same density as water, and that viscosity can be ignored for this flow.
1) What is the speed of the flow? It is 1.2 m/s
2) If we assume that the liquid is at rest in the bladder (a reasonable assumption) and that the pressure where the urine exits is equal to atmospheric pressure, what does Bernoulli's equation give for the pressure in the bladder? (In fact, the pressure is higher than this; other factors are at work. But you can see that no increase in bladder pressure is needed!)

Answer:

A = [m/s]

B = [m/s²]

Explanation:

Assuming that V has SI units of m/s, then A and BT must also have units of m/s.

A = [m/s]

BT = [m/s]

Since T has SI units of s:

B [s] = [m/s]

B = [m/s²]

Answer:

Q1 acceleration = 16m/s²

Q2 Net force = 9N North

Explanation:

Q1 Using the formula F=ma

Q2 R = F1 + F2

Answer:

900 cm/s or 9 m/s.

Explanation:

Data obtained from the question include the following:

Length (L) = 30 cm

frequency (f) = 60 Hz

Velocity (v) =.?

Next, we shall determine the wavelength (λ).

This is illustrated below:

Since the wave have 4 node, the wavelength of the wave will be:

λ = 2L/4

Length (L) = 30 cm

wavelength (λ) =.?

λ = 2L/4

λ = 2×30/4

λ = 60/4

λ = 15 cm

Therefore, the wavelength (λ) is 15 cm

Now, we can obtain the speed of the wave as follow:

wavelength (λ) = 15 cm

frequency (f) = 60 Hz

Velocity (v) =.?

v = λf

v = 15 × 60

v = 900 cm/s

Thus, converting 900 cm/s to m/s

We have:

100 cm/s = 1 m/s

900 cm/s = 900/100 = 9 m/s

Therefore, the speed of the wave is 900 cm/s or 9 m/s.

Answer:

Explanation:

just use the gravational force equation which is G x m of earth x m of object divided by r squared (which is radius of earth)

The amount of lighting on a surface drops to (1/4) of the initial illumination when the distance between a light source and the surface it falls on is doubled.

The amount of light or luminous flux that falls on a surface is known as illumination. It is expressed as lux or lumens per square meter.

The square of the distance has an inverse relationship with the light intensity;

[tex]\rm I = \frac{1}{r^2}[/tex]

Where,

I is the light intensity

r is the distance

Let r is the distance and I is the sound intensity for case 2;

r' = 2r

[tex]\rm I' = \frac{1}{(2r)^2} \\\\ I' = \frac{1}{4r^2} \\\\ I' = \frac{I}{4}[/tex]

When we double the distance between a source of light and the surface on which it falls, the amount of illumination on the surface decreases to(1/2) of the original illumination.

Hence the value of the fraction is 1/4.

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Answer:

v = 46.67 km/h

Explanation:

We will use the following formula throughout this numerical:

s = vt

where,

s = distance covered

v = speed

t =  time taken

FOR FIRST 30 km:

s = 30 km

v = 30 km/h

t = t₃₀ = ?

Therefore,

30 km = (30 km/h)(t₃₀)

t₃₀ = (30 km)/(30 km/h)

t₃₀ = 1 h

FOR TOTAL 100 km:

s = 100 km

v = 40 km/h (Average Speed)

t = total time = ?

Therefore,

100 km = (40 km/h)(t)

t = (100 km)/(40 km/h)

t = 2.5 h

FOR LAST 70 km:

s = 70 km

t₇₀ = t - t₃₀ = 2.5 h - 1 h = 1.5 h

v = v₇₀ = ?

Therefore,

70 km = v(1.5 h)

v = 70 km/1.5 h

v = 46.67 km/h

Answer:

[tex]D.\ 970\ N[/tex]

Explanation:

Given:

Mass of Piano: 99kg

Required:

Calculate its weight

The weight of an object is calculate as thus;

[tex]Weight = Mass\ (m) * Acceleration\ due\ to\ gravity\ (g)[/tex]

[tex]m = 99kg[/tex] and [tex]g = 9.8m/s^2[/tex]

The formula becomes

[tex]Weight = 99kg * 9.8m/s^2[/tex]

[tex]Weight = 970.2\ kgm/s^2[/tex]

[tex]Weight = 970.2\ N[/tex]

[tex]Weight = 970\ N[/tex] Approximated

Hence, the weight of the piano is 970N

Answer:

1666.7 ETB (birr)

833.3 ETB (birr)

2083.3 ETB (birr)

Explanation:

The first worker

5000*1/3=1666.7

The second worker

5000*1/6=833.3

The third worker

5000*5/12=2083.3

Hope this helps :) ❤❤❤

Answer:

Explanation:

Load ( L ) = 500 N

Effort ( E ) = 200 N

Load distance ( LD ) = 3 m

Effort distance ( ED ) = ?

Now, Let's find the Effort distance ( ED )

We know that,

Output work = Input work

i.e L × LD = E × ED

plug the values

[tex]500 \times 3 = 200 \times ED[/tex]

multiply the numbers

[tex]1500 = 200 \times ED[/tex]

Swipe the sides of the equation

[tex]200 \: ED \: = 500[/tex]

Divide both sides of the equation by 200

[tex] \frac{200 \: ED}{200} = \frac{500}{200} [/tex]

Calculate

[tex]ED\: = 2.5 \: m[/tex]

Hope this helps..

best regards!!

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

Answer:

The teacher should swim in a direction 29.24° North of East

Explanation:

Given that the there is a water  current across the lake, and the physics teacher intends to swim directly across the lake, the direction the physics teacher will have to swim is found as follows;

The speed of the water current is given by the speed of the floating leaf traveling with the water current  

Distance traveled by the leaf = 4.2 m South

Time of travel of the leaf = 5.0 s

Speed of leaf = 4.2/5 = 0.84 m/s = Speed of the water current

Swimming peed of the teacher, v = 1.9 m/s

To swim directly across the lake, the teacher has to swim slightly in the opposite direction of the water current, the y-component of the teacher's swimming speed should be equal to and opposite that of the speed of the water current.

Y-component of v = v×sin(θ), where θ is the angle of the direction, the teacher should swim

Therefore;

1.9 × sin(θ) = 0.84

sin(θ) = 0.84/1.9 = 0.44

θ = 26.24°

That is the teacher should swim in a direction 29.24° North of East.

To cross the lake the teacher has to swim in a direction 29.24° North of the East

The speed of the water current can be derived from the speed of the floating leaf :

The distance traveled by the leaf L = 4.2 m South

Time taken T = 5s

So, the speed of the leaf is:

u = 4.2/5

u = 0.84 m/s South

So, the speed of the current is 0.84 m/s South

Now, it is given that the speed of the teacher is, v = 1.9 m/s East

To cross the lake the speed of the teacher must be in a Northeast direction so that the North component of the speed of the teacher cancels out the speed of the current which is directed towards the South.

Let, the speed of the teacher makes an angle of θ from the EAST.

So, the North component is given by:

v(north) = vsinθ

it must be equal to the speed of the current:

vsinθ = u

1.9 × sinθ = 0.84

sinθ = 0.84/1.9

sinθ = 0.44

θ = 26.24°

The teacher should swim in a direction 29.24° North of East.

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Answer:

W =1562.53 N

Explanation:

It is given that,

Radius of the aluminium ball, r = 24 cm = 0.24 m

The density of Aluminium, [tex]d=2698.4\ kg/m^3[/tex]

We need to find the thrust and the force. The mass of the liquid displaced is given by :

[tex]m=dV[/tex]

V is volume

Weight of the displaced liquid

W = mg

[tex]W=dVg[/tex]

So,

[tex]W=dg\times \dfrac{4}{3}\pi r^3\\\\W=2698.4\times 10\times \dfrac{4}{3}\times \pi \times (0.24)^3\\\\W=1562.53\ N[/tex]

So, the thrust and the force is 1562.53 N.

Answer:

[tex]E=6.01\times 10^{-3}\ J[/tex]

Explanation:

It is given that,

Side of a square loop, l = 24 cm = 0.24 m

Resistance of loop, R = 6.1 ohms

Initial magnetic field is 0.665 T and final magnetic field is 0 as the field is removed in 40 ms

We need to find the electrical energy dissipated in this process.

Due to change in magnetic field, the loop will induce a voltage. The induced voltage is given by :

[tex]V=-\dfrac{dB}{dt}\\\\V=\dfrac{BA}{t}[/tex]

If I is induced current then,

[tex]V=IR[/tex]

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

Power is given by voltage times current. So,

[tex]P=\dfrac{(BA)^2}{(t^2R)}[/tex]

Now, energy is given by the product of power and time. So,

[tex]E=\dfrac{(BA)^2}{(t^2R)}\times t\\\\E=\dfrac{(BA)^2}{(tR)}[/tex]

Now putting all the values in above formula. So,

[tex]E=\dfrac{(0.665\times (0.24)^2)^2}{(40\times 10^{-3}\times 6.1)}\\\\E=6.01\times 10^{-3}\ J[/tex]

So, the electrical energy of [tex]6.01\times 10^{-3}\ J[/tex] is dissipated in this process.

The electrical energy dissipated throughout this process will be "6.01 × 10⁻³ J".

According to the question,

Square loop's side, l = 24 cm or,

                                  = 0.24 m

Loop's resistance, R = 6.1 ohms

Initial magnetic field = 0.665 T

Final magnetic field = 0

We know the relation,

→ V = - [tex]\frac{dB}{dt}[/tex]

      = [tex]\frac{BA}{t}[/tex]

Also we know,

Current, V = IR

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

                 = [tex]\frac{BA}{tR}[/tex]

Now, Energy, E = [tex]\frac{(BA)^2}{t^2R}[/tex] × t or,

                          = [tex]\frac{(BA)^2}{tR}[/tex]

By substituting the values,

                          = [tex]\frac{(0.665\times (0.24)^2)^2}{40\times 10^{-3}\times 6.1}[/tex]

                          = 6.01 × 10⁻³ J

Thus the response above is correct.                                  

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Answer:

8.8 mT

Explanation:

Current through the conductor = 12.5 A

Magnetic force per unit length on the wire = 0.110 N/m

Recall that the magnetic force per unit length on a current carrying conductor is in a uniform magnetic field is

[tex]\frac{F}{l}= IBsin\alpha[/tex]

where B is the magnetic field magnitude

[tex]I[/tex] is the current in the conductor

α is the angle the conductor makes with the  magnetic field ( since it is perpendicular in this case, α is 90°)

imputing values into the equation, we'll have

0.11 = 12.5 x B x sin 90°

but sin 90° = 1, therefore, we have

0.11 = 12.5B

B = 0.11/12 = 8.8 x 10^-3 T

or rather = 8.8 mT

Answer:

According to super hero logic , nothing will happen to him.

But according to science , yes he will get current shock but good news is that he wouldn't get elected until he is in contact with the wires.

He may / may not be affected but his suit will be damaged for sure as it is made of metal.

HOPE THIS HLEP AND PLSSSSS MARK AS BRAINLIEST AND THNXX :)

Explanation:

Given,

I = 5 A

V = 240 V

T = 25 mins = 1500 sec

Now,

Energy dissipated = IVT= 5×240×1500 = 1800000 J

Explanation:

m = kg. v=m/s. g=m/s^2. h= m

>>1/2mv^2=mgh

>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m

>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m>>kg m^2/s^2=kg m^2/s^2 the fraction 1/2 won't be able to make any changes to to the dimensional expression of energy i.e half of energy is still energy therefore you can neglect the number .

>>kg m^2/s^2=kg m^2/s^2

>>J= J

Answer:

210 m

Explanation:

The speed of train 2 relative to train 1 is 15 m/s + 20 m/s = 35 m/s.

It takes 6 seconds for the train to pass, so the length of the train is:

(35 m/s) (6 s) = 210 m

Answer:

First, let's think in the vertical problem:

The acceleration will be the gravitational acceleration:

g = 9.8 m/s^2

a = -9.8 m/s^2

For the velocity, we integrate over time:

v(t) = (-9.8 m/s^2)*t + v0

Where v0 is the initial velocity, in this case  v0 = 30m/s.

v(t) =  (-9.8 m/s^2)*t + 30m/s

Now, for the position we integrate again over time, and get:

P(t) = (1/2)*(-9.8 m/s^2)*t^2 + 30m/s*t + p0

Where p0 is the initial position, as the ball is launched from the ground, we can use p0 = 0m

p(t) = (-4.9m/s^2)*t^2 + 30m/s*t

Now, the maximum vertical height is reached when:

v(t) = 0m/s = -9.8m/s^2*t + 30m/s

t = 30m/s/9.8m/s^2 = 3.06s

Now we can evaluate the vertical position in t = 3.06s

p(3.06s) = (-4.9m/s^2)*(3.06)^2 + 30m/s*3.06 = 62m

So, rounding down, the correct option is: C. 60 m

Answer:

(a) Error: Other's ; Correction: Others'

(b) Error: to ; Correction: for

(c) Error: with ; Correction: of

(d) Error: prepares ; Correction: prepare

Explanation:

a)  The error is in the word "other's" as the position of apostrophe is wrong, so the correct word will be "others'", it shows plural nouns.

b)  The error is in the word "to", so the correct word will be "for" as for is use to talk about a purpose.

c)  The error is in the word "with"  and the correct word will be "of" as of indicates relationships between other words including things that made of other things.

d)  The error is in the word "preapres" and the correct word will be "prepare".

Answer:

λ₀ = 2 d n

Explanation:

A soap film is a layer where the lus is reflected on the surface and on the inside of the film, these two reflected rays can interfere with each other either constructively or destructively.

Let's analyze the general conditions of this interference,

* When the ray of light reaches the surface of the film it is reflected, as the index of refraction of the air is less than the index of the film, the reflected ray has a phase change of 180º

* When the ray penetrates the film, its wavelength changes due to the refractive index of the film.

          λ = λ₀ / n

where lick is the wavelength in the vacuum or air and n index of refraction of the film, in general this interference is observed perpendicular to the film, so the sine veils 1. the expression for constructive interference taking in what previous remains

         2d = (m + ½) λ

the expression for destructive interference remains

         2d = m λ

          2d = m λ₀ / n

When the film is placed on a glass plate whose index of refraction is greater than the index of refraction of the film, in the reflection in the lower part of the film another phase difference of 180º is created, for which we have a difference of total phase of 180 +180 = 360º, which is equivalent to no phase difference, therefore the two previous equations are interchanged.

Therefore where we had destructive interference now a cosntructive interference happens we can see the reflected light.

Find us the wavelength that this constructive interference creates

           2d n = m λ₀

           λ₀ = 2 d n / m

To find the minimum wavelength, suppose we observe the first interference pattern m = 1

         λ₀ = 2 d n

where d is the thickness of the film and n the index of refraction of the same

Answer:

D. Mars is a long distance away

Answer:

1.25x10^-19N

Explanation:see attached file pls

Answer:

1) a.  52.41 m/s

b. The skier will be going 15.35 m/s slower

2)  103.68 m

3) 35,127 J

4) a.  88.825 kJ

(b) 16.36 %

5) 3,071.12 J

Explanation:

1) a. The given height of the hill, h = 140.0 m

The mass of the skier at the top of the hill, m = 85.0 kg

The acceleration due to gravity, g = 9.81 m/s²

The initial potential energy, P.E of the skier = m×g×h = 85.0×140.0×9.81 = 116739 J

From the principle of conservation of energy, we have;

The potential energy, P.E. lost = The gain in kinetic energy, K.E.

m×g×h = 1/2×m×v²

116739 J = 1/2×85.0×v²

v² = 116739/(1/2*85.0)= 2746.8 m²/s²

v = √(2746.8 m²/s²) = 52.41 m/s

b. From 70 m up, we have;

The initial potential energy, P.E., of the skier is now = 85.0×70×9.81 = 58,369.5 J

The potential energy, P.E. lost = The gain in kinetic energy, K.E.

58,369.5 J = 1/2×85.0×v²

v² = 58,369.5/(1/2*85.0) = 1373.4 m²/s²

v = 37.06 m/s

The skier will be going 52.41 - 37.06 = 15.35 m/s slower

The skier will be going 15.35 m/s slower

2) From the principle of conservation of energy, the amount of work done (energy used) = The (potential) energy gained by the load

The amount of work done by the electric hoist = 356,000 J

The mass of the load = 350.0 kg

The height to which the load is raised = h

The potential energy gained by the load = m×g×h = 350.0×9.81×h

356,000 J = 350.0×9.81×h

h = 356,000/(350.0*9.81) = 103.68 m

The height to which the load is lifted= 103.68 m

3) The initial potential energy of the roller coaster cart = 600*35.0*9.81 = 206010 J

The final potential energy = 600*28.0*9.81= 164808 J

The velocity at point 3  = 4.5 m/s

The kinetic energy at point 3 = 1/2*600*4.5^2 = 6075 J

The total energy at point 3 = 164808 + 6075 = 170,883 J

The energy loss = The initial potential energy at point 1 - Total energy at point 3

The energy loss = 206010 - 170,883 = 35,127 J

The heat energy due to friction that must have been produced between points 1 and 3 = 35,127 J

4) a. The heat energy absorbed = mass × specific heat capacity for water, [tex]C_{water}[/tex] × Temperature change

The mass of the water = 2.5×10² g = 0.25 kg

[tex]C_{water}[/tex] = 4,180 J/(kg·°C)

Initial temperature = 10.0°C

Final temperature = 95°C

The temperature change = 95.0°C - 10.0°C = 85.0°C

The heat energy absorbed = 0.25*4,180* 85 = 88,825 J = 88.825 kJ

(b) The percentage efficiency = (Heat absorbed/(Heat supplied)) × 100

The heat supplied = 543 kJ

The efficiency = (88.825/543) × 100  = 16.36 %

5) The mass of the box = 115 kg

Force acting on the rope = 255 N

The angle of inclination of the force to the horizontal = 24.5°

The distance the box is displaced = 15.0 m to the right

The work done = Force applied × distance moved in the direction of the force

The work done = Force applied × distance moved in the direction of the force

Given that the load moves a distance 15.0 m to the right,we have;

The component of the force acting in the direction of the movement of the load (to the right) is 225 × cos(24.5°) =  204.74 N

The work done = 204.7*15 = 3071.12 J

The amount of work done  = 3,071.12 J

Explanation:

The kinetic energy of an object is given by :

[tex]E=\dfrac{1}{2}mv^2[/tex]

m is mass of the object

v is the speed of the object

It is clear from the above relation that the kinetic energy of the object is directly proportional to the product of mass and velocity.

The hypothesis can be :

If the speed of the object increased then its kinetic energy increase because kinetic energy is directly proportional to ist speed.

The kinetic energy of an object is given by :

m is mass of the object

v is the speed of the object

It is clear from the above relation that the kinetic energy of the object is directly proportional to the product of mass and velocity.

The hypothesis can be :

If the speed of the object increased then its kinetic energy increase because kinetic energy is directly proportional to its speed.

Answer:

625 cm²

Explanation:

The pressure is the same on both pistons.

F/A = F/A

200 N / 25 cm² = 5000 N / A

A = 625 cm²

Answer:

θ = 41.8º

Explanation:

This is an internal total reflection exercise, the equation that describes this process is

         sin θ = n₂ / n₁

where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂

        θ = sin⁻¹ n₂ / n₁

let's calculate

       θ = sin⁻¹ (1.00 / 1.50)

       θ = 41.8º

Answer:

1: the refracted angle in the first face is equal to the incident angle that is 60degrees

2. Emergence Angle is 42degrees

Explanation:

Pls see attached file

Explanation:

A theory is a way of explaining an object or event using a set of facts.

Answer:

(a) 1900 kg m²

(b) 8950 Nm

Explanation:

(a) The moment of inertia of a rod about its end is I = ⅓mL².

For 3 rods of mass m = 135 kg and length L = 3.75 m, the total moment of inertia is:

I = 3 (⅓ (135 kg) (3.75 m)²)

I = 1900 kg m²

(b) Net torque = moment of inertia × angular acceleration

∑τ = Iα

First, find the angular acceleration.

ω₀ = 0 rad/s

ω = 6.0 rev/s (2π rad/rev) = 37.7 rad/s

t = 8.0 s

α = (37.7 rad/s − 0 rad/s) / 8.0 rad/s = 4.71 rad/s²

∑τ = Iα

∑τ = (1900 kg m²) (4.71 rad/s²)

∑τ = 8950 kg m² / s²

∑τ = 8950 Nm