The air is partially removed at the upper part of the tube and the valve is closed. The height of the water above its pool surface is 0.7 m.The height of the liquid above its pool surface is 0.33 m.The difference in the heights of the pool surfaces is 0.22 m.

Required:
Find the density of the unknown liquid.

Explanation:

The collision forces are equal and opposite.  Therefore, the magnitudes are equal.

Explanation:

(a) Find the magnitude of the angular acceleration of the wheel.

(b) Find the angle in radians through which it rotates in this time interval.

Now we convert rad to angle

The angle in radians = 27.9°

Answer:

The answer to this question can be defined as follows:

Explanation:

In the given question, an equation is missing which can be defined as follows:

[tex]I \omega =\sqrt{J(J+1)}h[/tex]

solution:

Angular momentum:

[tex]L=I \omega =\sqrt{J(J+1)}h[/tex]

Convert Angular momentum in terms of kinetic energy:

[tex]K = \frac{L^2}{2I}[/tex]

    [tex]= \frac{h^2(J(J+1))}{2I}[/tex]

Answer:

-0.63 rad/s²

Explanation:

Given that

Initial angular velocity of the turntable, w(i) = 33 1/3 rpm

Final angular velocity of the turntable, w(f) = 0 rpm

Time taken to slow down, t = 5.5 s

The calculation is attached in the photo below

Answer:

The width is  [tex]Z = 0.0424 \ m[/tex]

Explanation:

From the question we are told that

    The width of the slit is [tex]d = 77.7 \mu m = 77.7 *10^{-6} \ m[/tex]

    The wavelength of the light is  [tex]\lambda = 721 \ nm[/tex]

      The position of the screen is  [tex]D = 2.83 \ m[/tex]

Generally angle at which the first minimum  of the interference pattern the  light occurs  is mathematically  represented as

        [tex]\theta = sin ^{-1}[\frac{m \lambda}{d} ][/tex]

Where m which is the order of the interference is 1

substituting values

       [tex]\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ][/tex]

      [tex]\theta = 0.5317 ^o[/tex]

 Now the width of first minimum  of the interference pattern is mathematically evaluated as

       [tex]Y = D sin \theta[/tex]

substituting values

       [tex]Y = 2.283 * sin (0.5317)[/tex]

       [tex]Y = 0.02 12 \ m[/tex]

 Now the width of  the  pattern's central maximum is mathematically evaluated as

        [tex]Z = 2 * Y[/tex]

substituting values

      [tex]Z = 2 * 0.0212[/tex]

     [tex]Z = 0.0424 \ m[/tex]

Answer:

producers make its own energy frominorganic substances.

Answer:

nvbnncbmkghbbbvvvvvvbvbhgggghhhhb

Answer:

(a) V = 0.75 m/s

(b) V = 0.125 m/s

Explanation:

The speed of the flow of the river can be given by following formula:

V = Q/A

V = Q/w d

where,

V = Speed of Flow of River

Q = Volume Flow Rate of River

w = width of river

d = depth of river

A = Area of Cross-Section of River = w d

(a)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 20 m

d = 20 m

Therefore,

V = (300 m³/s)/(20 m)(20 m)

V = 0.75 m/s

(b)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 60 m

d = 40 m

Therefore,

V = (300 m³/s)/(60 m)(40 m)

V = 0.125 m/s

Answer:

h = 8.48*10^-3m

Explanation:

In order to calculate the height reached by the pendulum with the marble, you first take into account the momentum conservation law, to calculate the speed of both pendulum and marble just after the collision.

The total momentum of the system before the collision is equal to the total momentum after:

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]        (1)

Here you used the fact that the pendulum has its total mass concentrated at the end of the pendulum.

m1: mass of the marble = 0.0215kg

m2: mass of the pendulum concentrated at its end = 0.250kg

v1: horizontal speed of the arble before the collision = 5.15m/s

v2: horizontal speed of the pendulum before the collision = 0m/s

v: horizontal speed of both marble and pendulum after the collision = ?

You solve the equation (1) for v, and replace the values of the other parameters:

[tex]v=\frac{m_1v_1+m_2v_2}{m_1+m_2}\\\\v=\frac{(0.0215kg)(5.15m/s)+(0.250kg)(0m/s)}{0.0215kg+0.250kg}=0.40\frac{m}{s}[/tex]

Next, you use the energy conservation law. In this case the kinetic energy of both marble and pendulum (just after the collision) is equal to the potential energy of the system when both marble and pendulum reache a height h:

[tex]U=K\\\\(m_1+m_2)gh=\frac{1}{2}(m_1+m_2)v^2\\\\h=\frac{v^2}{2g}[/tex]

v = 0.40m/s

g: gravitational acceleration = 9,8m/s^2

[tex]h=\frac{(0.40m/s)^2}{2(9.8m/s^2)}=8.48*10^{-3}m[/tex]

Then, the height reached by marble and pendulum is 8.48*10^-3m

Answer:

Explanation:

Average acceleration

= (final velocity - initial velocity) /time

= (50-0)km/h /30 s

= 50 * 1000 / 3600 m/s /s

= 13.89 m/s^2

Answer:

Weight of object on moon is 5N ,as we know. Weight of object on moon is 1/3 the of object on earth,so

let weight of object on earth = X

5N= X/3

X = 3×5 = 15N

Hence the weight of the object on earth will

be 15N

Answer:

E = 326.17 N/C

Explanation:

(a) In order to calculate the magnitude of the electric field between the parallel plates you first calculate the acceleration of the proton. You use the following formula:

[tex]x=v_ot+\frac{1}{2}at^2[/tex]         (1)

vo: initial speed of the proton = 0m/s

t: time that the proton takes to cross the space between the plates = 3.20*10^-6 s

a: acceleration of the proton = ?

x: distance traveled by the proton = 1.60cm = 0.016m

You solve the equation (1) for a, and replace the values of all parameters:

[tex]a=\frac{2x}{t^2}=\frac{2(0.016m)}{(3.20*10^{-6}s)^2}=3.125*10^{10}\frac{m}{s^2}[/tex]

Next, you use the Newton second law for the electric force, to find the magnitude of the electric field:

[tex]F_e=qE=ma[/tex]           (2)

q: charge of the proton = 1.6*10^-19C

m: mass of the proton = 1.77*10^-27kg

You solve the equation (2) for E:

[tex]E=\frac{ma}{q}=\frac{(1.67*10^{-27}kg)(3.125*10^{10}m/s^2)}{1.6*10^{-19}C}\\\\E=326.17\frac{N}{C}[/tex]

The magnitude of the electric field in between the parallel plates is 326.17N/C

Complete question:

A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.

Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain

Answer:

The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.

Explanation:

Given;

magnitude of applied force, F = 1.5 N

Apply Newton's second law of motion;

F = ma

[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]

The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;

[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]

Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.

Answer:

sphere, disk, hoop

Explanation:

See attached file

Answer:

Explanation:

electron rest mass = 0.511MeV/C²

postion rest mass = 0.511MeV/C²

boson rest mass = 91.188GeV/C²

= 91188 MeV/C²

Answer:

fggdfddvdghyhhhhggghh

Answer:

spring constant = 31.25N/m

Explanation:

spring constant = force/extension

mass = 250g = 0.25kg

extension = 8cm = 0.08m

force = mg = 0.25 x 10 = 2.5N

spring constant = 2.5/0.08 = 31.25N/m

Answer:

Explanation:

If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;

[tex]S = ut + \frac{1}{2}at^{2}[/tex]

S is the altitude of the ballest

u is the initial velocity

a is the acceleration of the body

t is the time taken to strike the ground

Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s

Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g

Substituting this values into the equation of the motion;

[tex]S = 0 + \frac{1}{2}gt^2\\ S = \frac{1}{2}gt^2\\[/tex]

a) An expression for the altitude of the ballast after t seconds is therefore

[tex]S = \frac{1}{2}gt^2\\[/tex]

b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);

[tex]576 = \frac{1}{2}(32)t^2\\\\\\576*2 = 32t^2\\1152 = 32t^2\\t^2 = \frac{1152}{32} \\t^2 = 36\\t = \sqrt{36}\\ t = 6.0secs[/tex]

This means that the ballast strikes the ground after 6secs

c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.

v = 0 + 32(6)

v = 192ft

Answer:

a) virtual

Explanation:

This is a lens problem, in this case the equation that describes the process is the constructor equation

        1 / f = 1 / p + 1 / q

where f is the focal length, weights the distance to the object and qq the distance to the image

In a projection system the image that we see is q, which can have several characteristics

a) virtual It can never be projected, since this image is formed by the extensions of the light rays, therefore it is not real but a construction of the brain that interprets where the rays must come from.

b) Real. This image can be projected since light rays pass through the image

c) Inverted. Inverted images are real so they can be projected, so rays pass through the image

d) expanded. In this case the image is greater than the object, this occurs when the object the distance to the image is greater than the distance to the object, therefore the distance q is negative, therefore this image is straight and is formed by the extensions from the rays and you can't project

Answer:

option A) virtual

Explanation:

hope this helps :)

Speed = (distance) / (time)

Speed = (1,233 mile) / (2.4 hour)

Speed = 513.75 mile/hour

Speed = (513.75 mi/hr) x (1609.344 meter/mi) x (1 hr / 3600 sec)

Speed = (513.75 x 1609.344 / 3600) (mile-meter-hour/hour-mile-second)

Speed = 229.7 meter/second

Answer:

1) Φ=zero

2)  λ = 2π / k

3)   T = 2π / w

4)  v = w / k

Explanation:

The equation of a traveling wave is

     y = A sin (ka - wt + Ф)

Let's answer using this equation the different questions

1) we see that the equation given in the problem the phase is zero

2) wavelength

     k = 2π /λ

      λ = 2π / k

3) The perido

angular velocity is related to frequency

       w = 2π f

frequency and period are related

       f = 1 / T

       w = 2 π / T

        T = 2π / w

4) the wave speed is

         v = λ f

          λ = 2π / k

          f = w / 2π

          v = 2π /k  w /2π

           v = w / k

Answer:

hand saws

power saws

Circular Saw

Explanation:

that is all that i know

Plz check attachment for answer.

Hope it's helpful

Answer:

A)    θ = 28.1º , B)         v = 2.47 m / s

Explanation:

A) The angle of the ramp can be found using trigonometry

         sin θ = y / L

         Φ = sin⁻¹ y / L

         θ = sin⁻¹ (115/244)

         θ = 28.1º

B) For this pate we can use the relationship between work and kinetic energy

       W =ΔK

where the work is

       W = -fr x

the negative sign is due to the fact that the friction force closes against the movement

Lavariacion of energy cineta is

         ΔEm = ½ m v² - mgh

        -fr x = ½ m v² - m gh

the friction force has the equation

         fr = very N

at the highest part there is no speed and we take the origin from the lowest part of the ramp

To find the friction force we use Newton's second law. Where one axis is parallel to the ramp and the other is perpendicular

Axis y . perpendicular

            N- Wy = 0

            cos tea = Wy / W

            Wy = W cos treaa

             N = mg cos tea

we substitute

- (very mg cos tea) x = ½ m v²2 - mgh

            v2 = m (gh- very g cos tea x)

   let's calculate

           v = Ra (9.8 0.80 - 0.2 9.8 0.0 cos 28.1)

           v = RA (7.84 -1.729)

           v = 2.47 m / s

Answer:

the length that would produce a sound tone under the same experimental contditions must be increased by  Δl = [tex]\frac{v}{2f}[/tex]

Explanation:

Recall

V = f ×λ

where λ is ⁴/₃l₂ for second resonance

f = [tex]\frac{3v}{4l_{2} }[/tex]

l₂ = [tex]\frac{3v}{4f}[/tex]

where λ is 4l₁ for 1st resonance

f = [tex]\frac{v}{4l_{1} }[/tex]

l₁ = [tex]\frac{v}{4f}[/tex]

∴ Δl = l₂ - l₁ =  [tex]\frac{3v}{4f}[/tex] ⁻  [tex]\frac{v}{4f}[/tex]

Δl=  [tex]\frac{2v}{4f}[/tex]

Δl = [tex]\frac{v}{2f}[/tex]

Therefore, the length should increase by [tex]\frac{v}{2f}[/tex]

Answer:

Thermodynamics is a branch of physics which deals with the energy and work of a system. It was born in the 19th century as scientists were first discovering how to build and operate steam engines. Thermodynamics deals only with the large scale response of a system which we can observe and measure in experiment.

Answer:

thermodynamics is the branch of physics which deals with the study of heat and other forms energy and their mutual relationship(relation ship between them)

Explanation:

i hope this will help you :)

Given that,

radius = 1.24 m

According to question,

The rope cannot push outwards. It must always have some slight tension or the bucket will fall.

We need to calculate the tension in the rope

At the top the force of gravity is

[tex]F=mg[/tex]

The force needed to move the bucket in a circle is centripetal force.

So, if mg is ever greater than centripetal force then the bucket and the contents will start to fall.

The rope have a tension of less than zero.

We need to calculate the velocity of swing bucket

Using centripetal force

[tex]F=\dfrac{mv^2}{r}[/tex]

[tex]mg=\dfrac{mv^2}{r}[/tex]

[tex]g=\dfrac{v^2}{r}[/tex]

[tex]v^2=gr[/tex]

[tex]v=\sqrt{gr}[/tex]

Put the value into the formula

[tex]v=\sqrt{9.8\times1.24}[/tex]

[tex]v=3.49\ m/s[/tex]

Hence, The minimum tension in the rope is less than zero .

The bucket swings with the velocity of 3.49 m/s.

Answer:

in all cases with increasing temperature the density should decrease.

Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body,

Explanation:

Let's start for ya dream gas

        PV = nRT

Since it indicates that the pressure is constant, we see that the volume is directly proportional to the temperature.

The density of is defined by

        ρ = m / V

As we saw that volume increases with temperature, this is also true for solid materials, using linear expansion. Therefore in all cases with increasing temperature the density should decrease.

Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body, since all the radiation that falls on it is absorbed or emitted.

This type of construction has a characteristic curve where the maximum of the curve is dependent on the tempera, but independent of the material with which it is built, to explain the behavior of this curve Planck proposed that the diaconate in the cavity was not continuous but discrete whose energy is given by the relationship

             E = h f

Answer:

272.89g

Explanation:

Find the diagram to the question in the attachment below;.

Using the principle of moment to solve the question which states that the sum of clockwise moment is equal to the sum of anticlockwise moment.

Moment = Force * Perpendicular distance

Taking the moment of force about the pivot.

Anticlockwise moment:

The 85g mass will move in the anticlockwise

Moment of 85g mass = 85×36.6

= 3111gcm

Clockwise moment.

The mass of the metre stick M situated at the centre (50cm from each end) will move in the clockwise direction towards the pivot.

CW moment = 11.4×M = 11.4M

Equating CW moment to the ACW moment we will have;

11.4M = 3111

M = 3111/11.4

M = 272.89g

The mass of the metre stick is 272.89g

Answer:

Δf = 73.72Hz

Explanation:

In order to calculate the difference in frequency between the direct waves and the reflected waves, you first take into account the Doppler's effect for an observer getting closer to the source:

[tex]f'=f(\frac{v+v_o}{v-v_s})[/tex]         (1)

You can assume that the reflected waves come from a source "the whale". Then you have:

f': frequency of the reflected waves = ?

f: frequency of the source = 22.0*kHz = 22.0*10^3 Hz

v: speed of the sound in water = 1482m/s

vs: speed of the source = 4.95m/s

vo: speed of the observer = 0m/s

You replace the values of the parameters in the equation (1):

[tex]f'=(22.0*10^3Hz)(\frac{1482m/s}{1482m/s-4.95m/s})=22073.72Hz[/tex]

Then, the difference in frequency is:

[tex]\Delta f = f'-f=22000Hz-22073.72Hz=73.72Hz[/tex]