The average human walks at a speed of 5km per hour if your PE teacher asks you to walk for 30 minutes in gym class how far would you walk(km)?

Answers

Answer 1

Answer:

2.5 km

Explanation:

Answer 2

Answer:

2.5 km

Explanation:

Distance = speed x time

So =5 x 0.5


Related Questions

Place each description under the correct theory
Gravity is an attractive force.
Universal Law of Gravitation
General Theory of Relativity
Mass and distance affect force.
Time and space are absolute,
Time and space are relative.
Gravity is due to space-time curving.
Mass affects space-time curving.

Answers

Answer:

1) Law of Universal Gravitation     Gravity is an attractive force

5) General relativity               Gravity is due to the curvature of spacetime

Explanation:

In this exercise you are asked to relate the correct theory and its explanation

Theory Explanation

1) Law of Universal Gravitation              Gravity is an attractive force

2) Law of universal gravitation              Mass and distance affect force

3) Classical mechanics                           time and space are absolute

4) Special relativity                                 Time and space are relative

5) General relativity                                Gravity is due to the curvature of

                                                               spacetime

6) General relativity                                 Mass affects the curvature of space - time

Answer:

Explanation:

edge2022

A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm. Fwithout belt

Answers

This question is incomplete, the complete question is;

Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.  

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

Fwith belt =

2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

Answer:

1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

we solve for a

a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt =  10.3 KN

Therefore, Net force of the driver is 10.3 KN

2)

No sit belt,  

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

An empty parallel plate capacitor is connected between the terminals of a 18.8-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor

Answers

Answer:

[tex]p.d' = 37.6 V[/tex]

Explanation:

From the question we are told that:

Potential difference [tex]p.d=18.8V[/tex]

New Capacitor [tex]C_1=C_2/2[/tex]

Generally the equation for Capacitor capacitance is mathematically given by

[tex]C=\frac{eA}{d}[/tex]

Generally the equation for New p.d' is mathematically given by

 [tex]C_2V=C_1*p.d'[/tex]

  [tex]p.d' = 2V[/tex]

 [tex]p.d'= 2 * 18.8[/tex]

 [tex]p.d' = 37.6 V[/tex]

A ball rolled along a horizontal surface comes to rest in a distance of 72m in 6s. Its initial velocity
and deceleration are
and​

Answers

Answer:

1. Initial velocity = 24 m/s

2. Deceleration = –4 m/s²

Explanation:

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

Distance travelled (s) = 72 m

Time (t) = 6 s

Final velocity (v) = 0 m/s

1. Determination of the initial velocity.

Distance travelled (s) = 72 m

Time (t) = 6 s

Final velocity (v) = 0 m/s

Initial velocity (u) =?

s = (u + v)t / 2

72 = (u + 0) × 6 / 2

72 = u × 3

Divide both side by 3

u = 72 / 3

u = 24 m/s

2. Determination of the deceleration.

Time (t) = 6 s

Final velocity (v) = 0 m/s

Initial velocity (u) = 24 m/s

Deceleration (a) =?

v = u + at

0 = 24 + (a × 6)

0 = 24 + 6a

Collect like terms

0 – 24 = 6a

–24 = 6a

Divide both side by 6

a = –24 / 6

a = –4 m/s²

Cho điện cực dưới điện cực trên . Hàm thế biến thiên theo qui luật:
Xác định sự phân bố điện tích khối và .

Answers

Answer:

Quantum theory gives the concept of

An infinite plane lies in the yz-plane and it has a uniform surface charge density.
The electric field at a distance x from the plane
a.) decreases as 1/x^2
b.) increases linearly with x
c.) is undertermined
d.) decreases linearly with x
e.) is constant and does not depend on x

Answers

Answer:

So the correct answer is letter e)

Explanation:

The electric field of an infinite yz-plane with a uniform surface charge density  (σ) is given by:

[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]

Where ε₀ is the electric permitivity.

As we see, this electric field does not depend on distance, so the correct answer is letter e)

I hope it helps you!

Galaxies that are 400 million light years away have a red shift of 0.03 approximately. A radio wave coming from one of these galaxies has an observed wavelength of 125 meters. What is the emitted wavelength in meters and the observed frequency in Megahertz

Answers

Answer:

The correct answer is "121.36 meters and 2.40 MHz".

Explanation:

Given:

Red shift,

[tex]\frac{v}{c}=\frac{\lambda - \lambda_0}{\lambda_0} = 0.03[/tex]

Wavelength,

[tex]\lambda = 125 \ meters[/tex]

The observed frequency will be:

⇒ [tex]f = \frac{c}{\lambda}[/tex]

      [tex]=\frac{3\times 10^8}{125}[/tex]

      [tex]=24\times 10^5[/tex]

      [tex]= 2.40\times 10^6[/tex]

      [tex]=2.40 \ MHz[/tex]

hence,

The Emitted wavelength will be:

⇒ [tex]\frac{125-\lambda_0}{\lambda_0} =0.03[/tex]

   [tex]\frac{125}{\lambda_0}-1 =0.03[/tex]

         [tex]125=1.03 \ \lambda_0[/tex]

           [tex]\lambda_0=\frac{125}{1.03}[/tex]

                [tex]=121.36 \ m[/tex]

In a robotics circuit, a voltage source of 75V is supplying a current, I to a series circuit of 5
resistances. Resistance, R1 = 5 KΩ and R2 = 10 KΩ. The voltage drops across 3 black boxes of
resistances R3 , R4 and R5 are 15V, 20V and 25V respectively. The current through the black
box of resistance, R5 is measured as 1mA. Calculate the voltage V1 and V2 across the
resistance R1 and R2 using the Voltage Divider Rule.

Answers

Answer:

In the given circuit, R

2

,R

6

and R

4

are in series. So,

R

1

=7+5+12=24Ω

Now R

1

and R

5

are in parallel. So,

R

2

1

=

8

1

+

24

1

=

24

3+1

=

24

4

=

6

1

R

2

=6ohm.

Now R

2

,R

1

and R

3

are in series. So,

R=R

2

+R

1

+R

3

=6+3+2=11ohm.

We know i=

R+r

E

=

11+1

6

=

12

6

=

2

1

i=0.5amp.

A person pushes horizontally on a heavy box and slides it across the level floor at constant velocity. The person pushes with a 60.0 N force for the first 16.4 m at which time he begins to tire. The force he exerts then starts to decrease linearly from 60.0 N to 0.00 N across the remaining 6.88 m. How much total work did the person do on the box

Answers

Over the first 16.4 m, the person performs

W = (60.0 N) (16.4 m) = 984 J

of work.

Over the remaining 6.88 m, they perform a varying amount of work according to

F(x) ≈ 60.0 N + (-8.72 N/m) x

where x is in meters. (-8.72 is the slope of the line segment connecting the points (0, 60.0) and (6.88, 0).) The work done over this interval can be obtained by integrating F(x) over the interval [0, 6.88 m] :

W = ∫₀⁶˙⁸⁸ F(x) dx ≈ 206.4 J

(Alternatively, you can plot F(x) and see that it's a triangle with base 6.88 m and height 60.0 N, so the work done is the same, 1/2 (6.88 m) (60.0 N) = 206.4 J.)

So the total work performed by the person on the box is

984 J + 206.4 J = 1190.4 J ≈ 1190 J

The thermal efficiency (in %) of a system that undergoes a power cycle while receiving 1000 kJ of energy by heat transfer from a hot reservoir at 1000 K and discharging 500 kJ of energy by heat transfer to a cold reservoir at 400 K is:

Answers

Answer:

η = 0.5 = 50%

Explanation:

The efficiency of the power cycle is given by the following formula:

[tex]\eta = \frac{W}{Q_1}\\\\\eta = \frac{Q_1-Q_2}{Q_1}[/tex]

where,

where,

η = efficiency = ?

Q₁ = heat received from hot reservoir = 1000 KJ

Q₂ = heat discharged to cold reservoir = 500 KJ

Therefore,

[tex]\eta = \frac{1000\ KJ-500\ KJ}{1000\ KJ}[/tex]

η = 0.5 = 50%

4
Type the correct answer in the box. Use numerals instead of words.
Anne has a sample of a substance. Its volume is 20 cm and its mass is 100 grams. What is the sample's density?
The sample's density is
g/cm?
Reset
Next

Answers

Answer:

5g/cm

Explanation:

denisty=mass/volume

100/20

5g/cm

True or false: Increasing the Young’s modulus of a beam in bending will cause it to deflect less.

Answers

Answer:

false?

Explanation:

The higher the modulus, the more stress is needed to create the same amount of strain; an idealized rigid body would have an infinite Young's modulus.

Answer:

I think the answer is False.

3. If you change the resistance of the resistor:
a. How does the current through the circuit change? (answer, explain, evidence)
b. How does the voltage of the battery change? (answer, explain, evidence)

Answers

Answer:

Explanation:

Changing the resistance of a resistor means the resistance is either increased or decreased.

a. When the resistance of the resistor is increased, the value of current flowing through the circuit decreases.

Example: given voltage of 6V, and a resistance of 30 Ohm's. The value of current flowing in the circuit is;

V = IR

6 = I x 30

I = 0.2 A

If the resistance is changed to 50 Ohm's, then:

I = 0.12 A

(ii) When the resistance of the resistor is decreased, the value of the current flowing through the circuit increases.

In the previous example, if the resistance is changed to 5 Ohm's, then:

V = IR

6 = I x 5

I = 1.2 A

(b) The voltage of the battery does not change since it is directly proportional to the current flowing through the circuit. Consider the examples stated above.

A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m) which has its other end fixed. The block passes through the equilibrium position with a speed of 2.6 m/s and first comes to rest at a displacement of 0.20 m from equilibrium. What is the coefficient of kinetic friction between the block and the horizontal surface?

Answers

Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.

There are two forces performing work on the block: the restoring force of the spring and kinetic friction.

By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:

[tex]W_{\rm total}=\Delta K[/tex]

or

[tex]W_{\rm friction}+W_{\rm spring}=0-K=-K[/tex]

where K is the block's kinetic energy at the equilibrium point,

[tex]K=\dfrac12\left(2.0\,\mathrm{kg}\right)\left(2.6\dfrac{\rm m}{\rm s}\right)^2=6.76\,\mathrm J[/tex]

Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is

[tex]W_{\rm spring}=-\dfrac12\left(250\dfrac{\rm N}{\rm m}\right)(0.20\,\mathrm m)^2=-5.00\,\mathrm J[/tex]

Compute the work performed by friction:

[tex]W_{\rm friction}-5.00\,\mathrm J=-6.76\,\mathrm J \implies W_{\rm friction}=-1.76\,\mathrm J[/tex]

By Newton's second law, the net vertical force on the block is

F = n - mg = 0   ==>   n = mg

where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.

So we have

[tex]W_{\rm friction}=-f(0.20\,\mathrm m)[/tex]

[tex]\implies -1.76\,\mathrm J=-\mu\left(2.0\,\mathrm{kg}\right)\left(9.8\dfrac{\rm m}{\mathrm s^2}\right)(0.20\,\mathrm m)[/tex]

[tex]\implies \boxed{\mu\approx0.45}[/tex]

The coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]

What is coefficient of friction?

Coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.

Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.

There are two forces performing work on the block: the restoring force of the spring and kinetic friction.

By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:

[tex]W_{total}=\Delta K[/tex]

or

[tex]W_{friction}+W_{spring}=0-K=-K[/tex]

where K is the block's kinetic energy at the equilibrium point,

[tex]K=\dfrac{1}{2}(2)(2.6)^2=6.76 \ J[/tex]

Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is

[tex]W_{spring}=-\dfrac{1}{2}(250)(0.20)=-5\ J[/tex]

Compute the work performed by friction:

[tex]W_{friction}-5 =-6.76\ J=-1.76\ J[/tex]

By Newton's second law, the net vertical force on the block is

∑ F = n - mg = 0   ==>   n = mg

where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.

So we have

[tex]W_{friction}=-f(0.20)[/tex]

[tex]-1.76=\mu (2)(9.8)(0.2)[/tex]

[tex]\mu =0.45[/tex]

Thus the coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]

To know more about Coefficient of friction follow

https://brainly.com/question/136431

b. The stream of water flowing through a hole at depth h = 10 cm in a tank holding water to height H = 40 cm. . 3 At what distance x does the stream strike the floor?​

Answers

Answer:

34.64 cm

Explanation:

Given that:

The depth of the hole h = 10 cm

height of the water holding in the tank H = 40 cm

For a stream of flowing water, the distance (x) at which the stream strikes the floor can be  computed by using the formula;

[tex]x = 2 \sqrt{h(H-h)}[/tex]

[tex]x = 2 \sqrt{10(40-10)}[/tex]

[tex]x = 2 \sqrt{10(30)}[/tex]

[tex]x = 2 \sqrt{300}[/tex]

[tex]x = 2 \times 17.32[/tex]

x = 34.64 cm

A pulse traveled the length of a stretched spring the pulse transferred...A)energy only B)mass only C)both energy and mass D) neither energy nor mass

Answers

Answer:

A

Explanation:

So a pulse is a part of a mechanical wave, and mechanical waves are energy transfer trough some medium, in this case a stretched spring. So the correct answer is (A) energy only. The pulse cant be transferred into mass.

Explain how solar radiation is converted into thermal energy

Answers

Answer:

See the explanation below

Explanation:

Explain how solar radiation is converted into thermal energy

From the concept of the particulate nature of matter, when the gases(air) receives a sufficient amount of radiation from the sun they tend to move in a constant rapid random motion as they move they hit one another and hence heat in form of thermal energy

write physical quantities and its unit​

Answers

length= metre

mass= kg

time= second

temperature = kelvin

current= ampere

luminous intensity= candela

Amount of substance = mole

etc

I hope this will help you

stay safe

The human eye can readily detect wavelengths from about 400 nm to 700 nm. Part A If white light illuminates a diffraction grating having 910 lines/mm , over what range of angles does the visible m

Answers

Answer:

The correct answer is "[tex]21.344^{\circ}[/tex]" and "[tex]39.56^{\circ}[/tex]".

Explanation:

According to the question,

Slit width,

[tex]d=\frac{1}{910 \ lines/mm}[/tex]

  [tex]=\frac{1}{910\times 10^3}[/tex]

  [tex]=1.099\times 10^{-6} \ m[/tex]

The condition far first order maxima will be:

⇒ [tex]d Sin \theta = 1 \lambda[/tex]

Now,

⇒ [tex]\Theta_{min} = Sin^{-1} (\frac{\lambda}{d} )[/tex]

             [tex]=Sin^{-1} (\frac{400\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]

             [tex]=21.344^{\circ}[/tex]

⇒ [tex]\Theta_{max} = Sin^{-1} (\frac{\lambda}{d} )[/tex]

             [tex]=Sin^{-1} (\frac{700\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]

             [tex]=39.56^{\circ}[/tex]

If the average time it takes for the cart from point 1 to point 2 is 0.2 s, calculate the angle θ from the horizontal of the track. Assume the track is frictionless. Hint: use the definitions of acceleration and Newton’s second law.

Answers

Answer:

hehe

Explanation:

I dont know because I am a noob ant study

The mass is released from the top of the incline and slides down the incline. The maximum velocity (taken the instant before the mass reaches the bottom of the incline) is 1.06 m/s. What is the kinetic energy at that time

Answers

Answer:

0.28 J

Explanation:

Let the mass of the object is 0.5 kg

The maximum velocity of the object is 1.06 m/s.

We need to find the kinetic energy at that time. It is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times 0.5\times (1.06)^2\\\\K=0.28\ J[/tex]

So, the required kinetic energy is equal to 0.28 J.

5. A big wheel has a diameter of 5 m and a mass of 1500 kg when fully laden with people. a) Work out the moment of inertia of the big wheel. (Hint: which shape from the ones given on p114 would be most suitable? b) When the wheel is rotating at full speed, a person has a linear velocity of 3 m/s. What is the angular velocity of this person? c) What is the rotational kinetic energy at this speed? d) A motor takes 10 seconds to accelerate the wheel from rest to a linear velocity on the circumference of 3 m/s. What is the power of the motor?​

Answers

Answer:

a)  I = 3.75 10⁴  kg  m², b)  w = 0.6 rad / s, c)  K = 6.75 10³ J, d) P = 6.75 10² W

Explanation:

This is a rotations exercise

a) the proper shape for a wheel is that of a rim where most of the weight is in the circumference plus the point weights of the people sitting on its periphery.

We are going to approximate the reda with a thin ring

           I = M r²

           I = 1500 5²

           I = 3.75 10⁴  kg  m²

b) angular and linear velocity are related

         v = w r

         w = v / r

         w = 3/5

         w = 0.6 rad / s

c) the expression for kinetic energy is

         K = ½ I w²

         K = ½ 3.75 10⁴   0.6²

         K = 6.75 10³ J

d) the power is

         P = W / t

         

to find the work we use the relationship between work and the variation of kinetic energy

         W = ΔK = K_f - K₀

the system part of rest wo = 0

         W = K_f

         W = 6.75 10³ J

we calculate

         P = 6.75 10³/10

         P = 6.75 10² W

3) A fahrenheit thermometer shaded from the sun on a hot day reads 101⁰ what is the temperature on the centigrade scale?

Answers

Answer:

38.33°C

Explanation:

Applying,

180/100 = (F-32)/C............. Equation 1

Where F = Temperature of the hot day in fehrenheit, C = Temperature of the hot day in centigrade.

make C the subject of the equation

C = 100(F-32)/180.............. Equation 2

From the question,

Given: F = 101°F

Substitute into equation 2

C = 100(101-32)/180

C = 38.33°C

What distance do I cover if I travel 10 m E, then 6 mW, then 12 m E?
A. 16 m
B. 28 m
C. 16 m E
D. 28 m E

Answers

Answer:

C. 16 m E

Explanation:

Applying,

The law of addition of vector: Vector in the same direction are added while vector in opposite direction are substracted

From the question above,

Step 1: Total distance covered towards east = 10+12 = 22 m E

Step2: Total distance covered towards west = 6 m W

Therefore, the resultant distance traveled = 22-6 = 16 m E

Hence the right option is C. 16 m E

A hockey ball is flicked of the ground with initial velocity of 2.0m/s upwards and 10m/s horizontally. Calculate the distance travelled from the point where the ball is flicked and to the point where the ball hits the ground.

Answers

Answer:

imma try and fail again and again

5. For the speaker in this circuit, the voltage across it is always proportional to the current through it. Find the maximum amount of power that the circuit can deliver to the speaker.

Answers

Answer:

speaker64

--------

34x

Explanation:

64-34

x

speaker

4

2

4

788

- circuit

voltage

100000

x.34

Sorry but you have no picture shown

uniform electric field of magnitude 365 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.00 cm. (a) What is the work done by the field on the electron? 1.753e-18 J (b) What is the change in potential energy associated with the electron? J

Answers

Answer:

a)   W = - 1.752 10⁻¹⁸ J,  b)    U = + 1.752 10⁻¹⁸ J

Explanation:

a) work is defined by

         W = F . x

the bold letters indicate vectors, in this case the force is electric

         F = q E

we substitute

         F = q E x

the charge of the electron is

         q = - e

         F = - e E x

let's calculate

         W = - 1.6 10⁻¹⁹  365  3 10⁻²

         W = - 1.752 10⁻¹⁸ J

b) the change in potential energy is

          U = q ΔV

the potential difference is

          ΔV = - E. Δs

 

we substitute

         U = - q E Δs

the charge of the electron is

           q = - e

          U = e E Δs

we calculate

           U = 1.6 10⁻¹⁹ 365  3 10⁻²

           U = + 1.752 10⁻¹⁸ J

A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50, how thick would you make the coating?

Answers

Answer:

[tex]t=0.50cm[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lamda=3c[/tex]m

Refraction Index [tex]n=1.50[/tex]

Generally the equation for Destructive interference for Normal incidence is mathematically given by

[tex]2nt=m(\frac{1}{2})\lambda[/tex]

Since  Minimum Thickness occurs at

At [tex]m=0[/tex]

Therefore

[tex]t=\frac{\lambda}{2}[/tex]

[tex]t=\frac{3}{4(1.50)}[/tex]

[tex]t=0.50cm[/tex]

What is the net force on the side of the container The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0.850 ATM. A faucet with an opening 1.0 inch diameter is located at the bottom of the container.

Answers

Answer:

26467.21 N

Explanation:

Initial height of water ( h1 )  = 5 m

diameter of container ( d1 )= 100 cm

pressure inside the container ( p1 )= 0.850 atm

Diameter of faucet ( d2 )= 1 inch

Calculate the value of the net force on the side of container

lets assume ; pressure outside the container ( p2 ) = 1 atm

Fnet = ( P1A1  + mg ) - ( P2A2 )

       = [ ( 0.85 * 101325 ) ( π(1/2)^2 ) + mg ) - [ ( 101325 )( π )(0.0127)^2 ]

       = [ 16902.2766 + pvg ] - [ 51.3161 ]

where ; pvg = pAhg = 1000 * π ( 1/2 )^2 * 5 * 9.8 = 9616.25

       = [ 16902.2766 + 9616.25 ] - [ 51.3161 ]   = 26467.21 N ( downwards )

Explain how the gravitational force between the earth and the sun changes as the earth moves from position A to B as shown in the figure. Sun Earth at position B Earth at position A​

Answers

Answer:

The distance between sun & Earth at position A is less than the earth at position B. The gravitational force of two bodies is inversely proportional to the square of the distance. So At position A gravitational force is more & it decreases as it rotate towards position B.

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