What happens to the mechanical advantage of a machine if the output force is less than the input force? What must happen to output distance? Give an example of a machine that does this?

Answer:

(a) V₂ = 1.86 m/s

(b) Q = 5.1 x 10⁻⁴ m³/s

Explanation:

(a)

The formula derived for Venturi tube is as follows:

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

where,

P₁ - P₂ = Difference in Pressure of Inlet and Outlet = 1.2 KPa = 1200 Pa

ρ = Density of Gasoline = 7.4 x 10² kg/m³

V₂ = Exit Velocity = ?

V₁ = Inlet Velocity

Therefore,

1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)

V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)

V₂² - V₁² = 3.24 m²/s²   ------------------- equation (1)

Now, we will use continuity equation:

A₁V₁ = A₂V₂

where,

A₁ = Inlet Area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²

A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²

Therefore,

(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂

V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)

V₁ = 0.25 V₂  

using this value in equation (1):

V₂² - (0.25 V₂)² = 3.24 m²/s²

0.9375 V₂² = 3.24 m²/s²

V₂² = (3.24 m²/s²)/0.9375

V₂ = √(3.456 m²/s²)

V₂ = 1.86 m/s

(b)

For fluid flow rate we use the following equation:

Flow Rate = Q = A₂V₂ = (2.746 x 10⁻⁴ m²)(1.86 m/s)

Q = 5.1 x 10⁻⁴ m³/s

The formula for finding variables in a  Venturi tube is shown below:

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

where, P₁ - P₂ is difference in pressure of Inlet and outlet, ρ = density, V₂ = exit velocity and V₁ is inlet velocity

P₁ - P₂ = 1.2 KPa = 1200 Pa

ρ = 7.4 x 10² kg/m³

V₂ = Exit Velocity = ?

V₁ = Inlet Velocity

We then substitute the variables into this equation.

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)

V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)

V₂² - V₁² = 3.24 m²/s²  ------ equation (1)

The  continuity equation A₁V₁ = A₂V₂ is then used

where,A₁ = Inlet area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²

A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²

(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂

V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)

V₁ = 0.25 V₂  

We then substitute the value into equation 1

V₂² - (0.25 V₂)² = 3.24 m²/s²

0.9375 V₂² = 3.24 m²/s²

V₂² = (3.24 m²/s²)/0.9375

V₂ = √(3.456 m²/s²)

V₂ = 1.86 m/s

This can be calculated using the formula

Flow Rate = Q = A₂V₂

                  = (2.746 x 10⁻⁴ m²)(1.86 m/s)

                 = 5.1 x 10⁻⁴ m³/s

Answer:

F =  3.32 x 10⁻⁶ N

Explanation:

The force of attraction between two masses is given by Newton's Law of Gravitation, as follows:

F = Gm₁m₂/r²

where,

F = Force between Romeo and Juliet = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of Romeo = 68 kg

m₂ = mass of space station = 4.58 x 10⁵ kg

r = distance = 25 m

Therefore,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(68 kg)(4.58 x 10⁵ kg)/(25 m)²

F =  3.32 x 10⁻⁶ N

Answer:

The fish displacement was -3.4 m

Explanation:

Distance and Displacement

A moving object constantly travels some distance at defined periods of time. The total moved distance is the sum of each individual distance the object traveled. It can be written as:

dtotal=d1+d2+d3+...+dn

This sum is obtained independently of the direction the object moves.

The displacement only takes into consideration the initial and final points of the path defined by the object in its moving. The displacement, unlike distance, is a vectorial magnitude and can be even zero if the object starts and ends the movement at the same point.

The displacement, when the object moves in one axis only is given by:

d = final position - initial position

We know the fist started at 7.8 m from a given reference along the x-axis.

After some undisclosed movements, it ends up at the position 4.4 m. Thus, the displacement is:

d = 4.4 m - 7.8 m = -3.4 m

This means the fish ended up to the left of the position it started from.

The fish displacement was -3.4 m

Answer:

yeah

Explanation:

yeah yeah yeah yeah

Answer:

The electrostatic force will remain the same

Explanation:

From the question we are told that  

     The charge on the each conducting sphere is [tex]q = 2 \ C[/tex]

      The radius each sphere is [tex]r_1 = r_2 = r = 0. 1\ m[/tex]

Generally electrostatic force between the sphere is mathematically represented as

             [tex]F = \frac{k * q_1 q_2 }{d^2}[/tex]

Here  k is the coulomb constant ,

         d  is the distance between the two sphere which is measured from one center of the sphere to the other center of the sphere  

Now from the question we are told that the radius of the spheres is doubled (i.e  from 0.10 m  to 0.2 m ) , this will not affect the distance between the sphere because their center are still in the same position

and given there is no change in the distance between the spheres , the electrostatic force will remain the same

Answer:

Explanation:

According to law of conservation of momentum

m1u1+ m2u2 = (m1+m2)v

m1 and m2 are the masses of the object

u1 and u2 are their initial velocities

v is their common velocity

Given

m1 = 3kg

m2 = M

u1 = 8m/s

u2 = 0m/s (object at rest)

v = 6.0m/s

Substitute and get M

3(8) + M(0) = (3+M)6

24 = 18+6M

24-18 = 6M

6 = 6M

M = 1kg

Hence the mass M is 1kg

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The force of friction is equal to the product of the vertical force applied by the surface to the object in the coefficient of friction.

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In this question ,

surface vertical force = Weight of the object

Thus ;

svf = ( mass ) × ( gravity acceleration )

_________________________________

If gravity acceleration is 10 :

svf = 10 × 10 = 100 N

So ;

frictional force = 100 × 0.20

frictional force = 20 N

##############################

If gravity acceleration is 9.8 :

svf = 10 × 9.8 = 98 N

So ;

frictional force = 98 × 0.20

frictional force = 19.6 N

_________________________________

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We are given:

Homer the Astronaut:

Mass of Homer the astronaut(m1) = 200 kg

initial velocity of Homer the astronaut(u1) = 5 m/s

Larry the Pal:

Mass of Larry the Pal (m2) = 150 kg

initial velocity of Larry the Pal (u2) = 0 m/s

Since they will move together after the collision, they will have the same velocity:

v1 = v2 = V

Solving for the Final velocity:

from the law of conservation of momentum:

m1u1 + m2u2 = m1v1 + m2v2

since v1 = v2 = V:

m1u1 + m2u2 = V(m1 + m2)

replacing the variables with the given values

200 * 5 + 150 * 0 = V(200 + 150)

1000 = 350V

V  = 1000 / 350

V = 2.86 m/s

Answer:

B

Explanation:

it could be because they have a difference in genes

Answer:

b

Explanation:

Answer:

big bang theory

Explanation:

Edwin Hubble is credited for the initial development of the Big Bang theory, an idea which helps to explain the formation of the universe over 15 billion years ago.

Answer:

The total energy is 53 Joule

Explanation:

Mechanical Energy

The mechanical energy of an object of mass m, speed v, and at a height h is:

[tex]\displaystyle E = m.g.h+\frac{mv^2}{2}[/tex]

The ball has a mass of m=0.5 Kg, a speed v = 4 m/s, and at a height of h=10 m. Thus the total energy is:

[tex]\displaystyle E = 0.5\cdot 9.8\cdot 10+\frac{0.5\cdot 4^2}{2}[/tex]

E = 49 J + 4 J = 53 J

The total energy is 53 Joule

Answer:

The average speed is 22.2 km/h

Explanation:

Average Speed

Given an object travels a total distance d and took a total time t, then the average speed is:

[tex]\displaystyle \bar v=\frac{d}{t}[/tex]

The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:

[tex]\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h[/tex]

Then he drives d2=7 km at v2=43 km/h taking a time of:

[tex]\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h[/tex]

The total time is

t=0.467 h + 0.163 h = 0.63 h

The total distance is

d = 7 km + 7 km = 14 km

The average speed is:

[tex]\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h[/tex]

The average speed is 22.2 km/h

Answer:

[tex]\mid \vec C\mid=31.9[/tex]

Explanation:

Consider the vectors:

[tex]\vec A=9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}}[/tex]

[tex]\vec B=-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}}[/tex]

Calculate the magnitude of

[tex]\vec C=-2\vec B-\vec A[/tex]

Substitute the values of the vectors:

[tex]\vec C=-2(-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}})-(9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}})[/tex]

Operate and remove parentheses:

[tex]\vec C=19\mathbf{\hat{i}}+26.8\mathbf{\hat{j}}-9.4\mathbf{\hat{i}}+3.6\mathbf {\hat{j}}[/tex]

Operating both components separately:

[tex]\vec C=9.6\mathbf{\hat{i}}+30.4\mathbf{\hat{j}}[/tex]

Now find the magnitude of C:

[tex]\mid \vec C\mid=\sqrt{9.6^2+30.4^2}[/tex]

[tex]\mid \vec C\mid=\sqrt{1016.32}[/tex]

[tex]\mathbf{\mid \vec C\mid=31.9}[/tex]

The force required to hold it completely submerged under water is 0.252 N

As a result of the low density (ρ1 = 0.0839 g/cm3 = 83.9 kg/m3)of the ball compared to that of water (ρ2 =1000 kg/m3), the buoyant force that is acting on the ball is greater than its weight.

Therefore, the minimum force required to hold the ball submerged under water can be calculated using the relation

Radius = 3.77/2 cm = 0.0377/2 m = 0.01885 m

Volume of sphere = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ = 2.805 e-5 m³

Mass of sphere = 4/3 π r³ ρ1 = 4/3 * 3.142 * 0.01885³ * 83.9 = 0.0023 kg

Weight of sphere = 4/3 π r³ ρ1 g = 4/3 * 3.142 * 0.01885³ * 83.9 * 9.8 = 0.023 N

Volume of water displaced = 4/3 π r³ = 2.805 e-5

Buoyant force = weight of water displaced = 4/3 π r³ ρ2 g = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ * 1000 * 9.8 = 0.275 N

F = 0.275 - 0.023 = 0.252 N

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The force required to hold it completely submerged under water is 0.25 N

The density of the ball ([tex]\rho_b[/tex]) = 0.0839 g/cm³ = 83.9 kg/m³

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

Diameter = 3.77 cm = 0.0377 m

radius of ball = 0.0377/2 = 0.01885 m

The volume (V) = [tex]\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*0.01885^3=2.8*10^{-5}\ m^3[/tex]

Let us assume the acceleration due to gravity (g) = 9.8 m/s², Hence:

The force is required to hold it completely submerged under water (F) is:

[tex]F=\rho_w Vg-\rho_b Vg=1000*(2.8*10^{-5})*9.81-83.9*(2.8*10^{-5})*9.81\\\\[/tex]

F = 0.25 N

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Answer:

Explanation:

all biotic and abiotic factors around a living organism is its environment

Answer: An environment is the surroundings or conditions in which a person, animal, or plant lives or operates. the natural world, as a whole or in a particular geographical area, especially as affected by human activity. The best synonym for environment would be nature.

Explanation: Environment is a place where different things are such as a swampy or hot environment. They constantly interact with it and adapt themselves to conditions in their environment. In the environment there are different interactions between animals, plants, soil, water, and other living and non-living things. Environment plays an important role in healthy living and the existence of life on planet earth. Earth is a home for different living species and we all are dependent on the environment for food, air, water, and other needs. Therefore, it is important for every individual to save and protect our environment. To divide environments' sorts we can mention 3 kinds of environments Natural, industrial, and social environment. Natural environment: Include water, light, land, air and all organisms that live in nature.

Answer:

Definition: Pressure:- Force per unit area surface is called pressure. P = F/ A Pressure = Force /Area Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed. Gauge pressure is the pressure relative to the ambient pressure. 

Answer:

A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force.

Pressure is defined to be the amount of force exerted per area.

Formula of pressure = P = F÷A

Where, P = Pressure,

            F = Force applied and,

            A = Unit area.

Examples of Pressure is Air pressure.

Answer:

dont die bro

Explanation:

Answer:

a = 0.62 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics. But first, we must convert speeds from kilometers per hour to meters per second.

[tex]v_{f} =v_{o} -a*t[/tex]

[tex]55[\frac{km}{hr}]*\frac{1hr}{3600s}*\frac{1000m}{1km} =15.27[\frac{m}{s} ]\\10[\frac{km}{hr} ]*\frac{1hr}{3600s}*\frac{1000m}{1km} = 2.77[\frac{m}{s} ][/tex]

where:

Vf = final velocity = 2,77 [m/s]

Vo = initial velocity = 15.27 [m/s]

t = time = 20 [s]

a = acceleration [m/s²]

Now replacing:

[tex]2.77=15.27-a*(20)\\20*a=12.49\\a = 0.62[m/s^{2}][/tex]

Answer:

It would be 10.00

Explanation:

Hope this helps its different for everyone what was it for u it was D for me

Answer:

the focal length is 4cm

Explanation:

use the formula 1+1/m = f

it is obtained from the mirror formula

m is the magnification and is the ratio of the image height to the object height

however they should have the same unit . that means you would have to do a little conversion from cm to mm or vice versa .

Steps 1 and 2)

The variables are W = work, P = power, and t = time. In this case, W = 9514 joules and P = 347 watts.

The goal is to solve for the unknown time t.

-----------------------

Step 3)

Since we want to solve for the time, and we have known W and P values, we use the equation t = W/P

-----------------------

Step 4)

t = W/P

t = 9514/347

t = 27.4178674351586

t = 27.4 seconds

-----------------------

Step 5)

The lawn mower ran for about 27.4 seconds. I rounded to three sig figs because this was the lower amount of sig figs when comparing 9514 and 347.

-----------------------

Note: we don't use the mass at all

Answer:

357 J

Explanation:

The elastic potential energy of arrow in the stretched bow is 357 J.

The kinetic energy of the arrow after it has been shot is given by half of the product of the arrow's mass and velocity of the arrow.

Here there are no other forms of energy at play here. Only potential and kinetic energy.

As we know that in any system the energy is conserved accordingly the elastic potential energy of the arrow will be equal to the kinetic energy of the bow after it is released i.e., 357 J.

Answer:

False

Explanation:

Answer:

False

Explanation:

Hope this helped, Have a Wonderful Day/Night!!

Answer:

+1

-1

Explanation:

The oxidation number of Na in NaCl is +1 and that of Cl is -1.

When NaCl is oxidized we have;

        NaCl →  Na⁺  + Cl⁻

In this process, Na will lose an electron to become Na⁺ and thus isoelectronic with Ne, Cl will gain the electrons to become Cl⁻ and thus isoelectronic with Argon.

Answer:

they are cooler than the rest if the sun

Answer:

angular speed = 67.91 rad/s

Explanation:

Given that,

The diameter of a tire, d = 72 cm

The radius of a tire, r = 36 cm = 0.36 m

Linear speed of an automobile, v = 88 km/h = 24.45 m/s

We need to find the angular speed of the tires about their axles. The relation between angular speed and angular speed is given by :

[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{24.45 }{0.36}\\\\\omega=67.91\ rad/s[/tex]

So, the angular speed of the tire is 67.91 rad/s.

Answer:

Answer is: c. It must lose two electrons and become an ion.

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

Explanation:

Answer:

the third option

Explanation:

It is the proportion predicted to be present in the early universe.

The hydrogen and helium abundance helps us to model the expansion rate of the early universe.

In the abundance of hydrogen and helium, we can say that they account for nearly all the nuclear matter in today's universe.

In big Bang model, the universe is mostly light or protons.

The Schramm's model for relative abundances indicate that helium is about 25% by mass and hydrogen about 73% with all other elements constituting less than 2%.

Several proponents of big Bang theory has proposed similar relative abundance for hydrogen and helium. In all  it is clear that hydrogen and helium constitute of more than 98% of the ordinary matter in the universe.

Finally, the hydrogen and helium abundance helps us to model the expansion rate of the early universe.

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