What would the shape of the graph be if a speed of 50.0 km/hr is maintained from
10 s to 20 s? I

Therefore, the missile's individual speed would be 14c + 14c = 28c.

The missile is traveling in the opposite direction from the Earth, which has an individual speed of zero.

Both the rocket and the missile are traveling in the same direction relative to one another. The moving item whose speed is being measured and the frame of reference for the point of observation must both be in relative motion.

The difference between their respective speeds is hence the relative speed.

Therefore, the missile's individual speed would be 14c + 14c = 28c.

The missile is traveling in the opposite direction from the Earth, which has an individual speed of zero.

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The tension in the cord is equal to centripetal force which is 32 newton.

We need to know about centripetal force to solve this problem. When an object moves in a circular motion, the object is experiencing centripetal force. The magnitude of centripetal force is

Fc = m . v²/R

where Fc is the centripetal force, m is mass, v is velocity and R is the radius.

From the question, we know that

m = 2 kg

v = 4 m/s

R = 1 m

The tension in the cord is equal to centripetal force

T = Fc

T = m . v²/R

T = 2 . 4² / 1

T = 32 N

Hence, the tension in the cord is 32 newton.

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Answer:

alright

Explanation:

that's so cool I'll join

Complete Question

Suppose you have a pocket calculator through which 1.25 C of charge passes in 3.5 h? What is the average current in milliamperes produced by the solar cells .

It is 0.126mA .

An item has a fine fee if it incorporates more protons than electrons. when a thing loses a few electrons, the difference between the number of protons and electrons in the item rises. As a result, the item acquires a positive rate.

tremendous and negative fees cancel each other out when there are an equal range of high-quality and poor charges, leaving the item neutral.

general rate that passes via the calculator may be denoted by q = 1.25c

total time t = 2.75=9900s

current required is = 1.25/9900

=1.26×10^-4 A

=1.26×10^1mA

=0.126mA

Thus the solar cell must produce an average of 0.126mA of current.

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The volume of mercury is in the vial 28.71 cm³

Once you are aware of the mass of the mercury, you may calculate the volume of the vial by using the density of the mercury to determine its mass.

You are aware that the vial including the mercury weighs 439.31 g.

The vial weighs vial=50.90 g.

This indicates that the mercury will have a mass of mercury=439.31 g

50.90 g=388.41 g.

You now are aware that the density of mercury is 13.53 g/cm3.

The mass of 1 cm³ of mercury is 13.53 g.

Your sample will have a volume of 388.41g/1 cm³/13.53g = 28.71 cm.

Four sig figs, the number of sig figs you have for the mass of the empty vial, is the answer, which is rounded to four sig figs.

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The calculated greatest speed is v=0.725 vs.

The ratio of a moving object's speed to the fundamental speed of sound is known as the Mach number. Therefore, an object travelling at twice the speed of sound has a Mach number of 2. speed of sound, the rate at which sound waves move through various substances. Modern estimates of the speed of sound, in particular, for dry air at a temperature of 0 °C (32 °F), are 331.29 meters (1,086.9 feet) per second.

The sound speed is vs=331 m/s.

The duration is t=4.9s.

A=5g is the acceleration.

u=0 is the beginning speed.

Let v be the fastest possible speed without a blackout.

The initial equation of motion is given by v=u+at

v=0+5gt

v=5(9.81m/s2)(4.9s)

v=240m/s

v=240m/svs/331m/s

v=0.725 vs.

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Because of the cosmological principle, we can predict gravity works similarly on Earth as on Mars.

The cosmological principle says that an observer's view of the universe is independent of both the direction he looks and his location. This principle only applies to the universe's large-scale properties, but it does suggest that the world has no edge, implying that the big-bang origin took place not at a specific point in space, but instead throughout space at the exact time.

These two assumptions allow us to determine the history of the universe after a specific epoch known as the Planck time. Scientists are yet to discover what existed prior to Planck time.

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The following are the values:-

(a)Water content (W%)  = 13.99%

(b) Total unit weight (Υ[tex]_{t}[/tex]) = 1778 kg/m³

(c) Dry unit weight (Y[tex]_{d}[/tex]) = 1559.75 kg/m³

(d) Void ratio (e)  = 0.718

(e) Porosity (n) = 0.417

Volume of mold (V[tex]_{t}[/tex]) = 0.4 m³

Dry mass of sand (W[tex]_{s}[/tex]) = 623.9 kg

Wet mass of sand (W[tex]_{t}[/tex]) = 711.2 kg

Specific Gravity (G[tex]_{s}[/tex]) = 2.68

Water content (W%) = [tex]\frac{W_{w} }{W_{s} }[/tex] × [tex]100[/tex]

Water content (W%) = [tex]\frac{711.2 - 623.9}{623.9}[/tex] × [tex]100[/tex]

Water content (W%) = 13.99%

Total unit weight (Υ[tex]_{t}[/tex]) = [tex]\frac{W_{t} }{V_{t} }=\frac{711.2}{0.4}[/tex]

Total unit weight (Υ[tex]_{t}[/tex]) = 1778 kg/m³

Dry unit weight (Y[tex]_{d}[/tex]) = [tex]\frac{W_{s} }{V_{t} }=\frac{623.9}{0.4}[/tex]

Dry unit weight (Y[tex]_{d}[/tex]) = 1559.75 kg/m³

Solid unit weight (Y[tex]_{s}[/tex]) = [tex]\frac{W_{s} }{V_{s} }[/tex]

Or

G[tex]_{s}[/tex] = [tex]\frac{Y_{s} }{Y_{w} }[/tex] ⇒ Y[tex]_{s}[/tex] = G[tex]_{s}[/tex]Y[tex]_{w}[/tex]

Y[tex]_{s}[/tex] = 2.68 × [tex]\frac{10^{-3} }{(10^{-2})^{3} }\frac{kg}{m}[/tex] = 2680 kg/m³

Now,

Void Ratio (e) = [tex]\frac{Y_{s} }{Yd}-1[/tex]

Void Ratio (e) = [tex]\frac{2680}{1559.72}-1[/tex]

Void Ratio (e) = 0.718

Porosity (n) = [tex]\frac{e}{1+e}[/tex]

Porosity (n) = [tex]\frac{0.718}{1+0.718}[/tex]

Porosity (n) = 0.417

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A laboratory sample of sand is formed inside a mold of 0.4 cubic meters. It took 711.2 kg of wet sand (dry mass of sand = 623.9 kg) to fill the mold. Assuming the specific gravity of the solid is 2.68, compute the:

(a)Water content (W%)  

(b) Total unit weight (Υ[tex]_{t}[/tex])

(c) Dry unit weight (Y[tex]_{d}[/tex])

(d) Void ratio (e)

(e) Porosity (n)

The mass of the box which the fisherman applies a horizontal force is 13.18 Kg

To solve this exercise, the formula and procedure to be applied are:

F = m * a

Where:

Information about the problem:

Applying the force formula and isolating the mass, we get:

F = m * a

m = F/a

m = 43.5 Kg * m/s² / 3.30 m/s2

m = 13.18 Kg

It is the physical magnitude that expresses the effort necessary to move the mass of one kilogram with an acceleration of one meter per second squared, this is expressed in units of the international system in Newton.

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The horizontal pulling force on the block is 22.5 N.

When a block is pulled by applying a force, friction also acts on it.

The net force is calculated by taking the net value of the two forces.

The free-body diagram(FBD) of the block,  is shown in the adjoining diagram.

Thus,

F - f = ma

Here F is the applied horizontal force and f is the frictional force.

Putting the given values in the above equation

F - 10 N = 5 x 2.5

F = 22.5 N

Thus, the horizontal pulling force on the block is 22.5 N.

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The horizontal pulling force on the block is 22.5 N.

When a block is pulled by applying a force, friction also acts on it.

The net force is calculated by taking the net value of the two forces.

The free-body diagram(FBD) of the block,  is shown in the adjoining diagram.

Thus,

F - f = ma

Here F is the applied horizontal force and f is the frictional force.

Putting the given values in the above equation

F - 10 N = 5 x 2.5

F = 22.5 N

Thus, the horizontal pulling force on the block is 22.5 N.

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The billiard ball involves relaxation 80 cm

Displacement refers to the space among the very last and the preliminary role. Hence the displacement of the ball can be the distinction among the preliminary and the very last displacement.

Let the preliminary role be 0

Final role = 8 cm

So the distinction among preliminary role and very last role = 0 - 88 cm

So the billiard ball involves relaxation 80 cm at the back of its orbital role.

The displacement thickness for the boundary layer is described as the gap the floor could need to pass withinside the y-course to lessen the glide passing via way of means of a extent equal to the actual impact of the boundary layer.

The distance from the boundary floor where the rate reaches 99% of the mainstream speed is used as an arbitrarily defined measure of the thickness of the boundary layer indicated by the symbol.

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The potential difference between the plates will be 1552 Volts.

Voltage, or the difference in electric potential between two points, is defined as the amount of labor per unit of charge needed to move a test charge between the two points.

Given that a 3.7-f capacitor that stores sufficient energy to operate a 75.0-w

The potential difference will be calculated by the formula below:-

Q = I t            

Where I = charge / time

Q = V * C    

V C = I t

V = I t / C

V = 75 C/s x 60 sec / 2.9 faraday

V =  1552 Volts

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The body of mass is resting on a horizontal plane, then the normal force exerted onto it by stattic and kinetic frictional forces both have an equal magnitude. In addition, the net resultant force acting upon the body of mass will be zero.

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The lines there’s nothing like the freedom of the open road are taken from the poem Song of the open road. The road in the poem signifies the symbol of freedom.

These lines are from the poem "Song of the Open Road”  written by Walt Whitman. In the poem, a journey characterized by freedom, and independence is described. The road in the poem is a symbol of self-awareness and freedom as the author travels along the road.

Whitman describes that the journey includes a test of the wisdom of the soul, revealed through provoking questions and experiences.

The poem also conveys the narrator's joy and happiness in being on the open road, which converts into a love of freedom and the open spaces that are the outdoors.

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A ¹²C atom's mass is exactly 1u, or 1/12 of its mass. As a result, the mass of an atom of hydrogen is larger than one-twelfth the mass of an atom of ¹²C.

What is atomic mass  ?

Atomic mass is the amount of matter that makes up an element's atom. It is written as a multiple of 1.992646547 x  10^-23 gram, or one-twelfth of of the mass of the carbon-12 atom, which is given an atomic mass of 12 units.

A ¹²C atom's mass is equal to one-twelfth of the unified atomic mass unit.

1u = 1/12 m ¹²C  

Additionally, since 1 mole is equal to 6.022x10^23 atoms, the unified atomic mass unit is equal to the mass (in grams) of 1 mole of a substance. Accordingly, the unified atomic mass unit in grams is:

1u = 1 g/ mol x 1mol / 6.022 x 10^23atoms  = 1.66 x 10^-27 kg

From this definition, we can determine the mass of a 1H atom in relation to a ¹²C atom:

m¹H = 1.67 x 10 ^ -27 kg

As a result, a 1H atom has a mass that is larger than 1/12 that of a ¹²C atom.

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Question:

Given that mH = 1.007825 u, is the mass of a hydrogen atom 1H greater than, less than, or equal to 1/12 the mass of a 12C atom? Select the correct answer and explanation. Given that = 1.007825 , is the mass of a hydrogen atom greater than, less than, or equal to 1/12 the mass of a atom? Select the correct answer and explanation. 1/12 of the mass of a 12C atom is greater than 1 u. Therefore, the mass of the hydrogen atom is greater than 1/12 of the mass of a 12C atom. 1/12 of the mass of a 12C atom is less than 1 u. Therefore, the mass of the hydrogen atom is less than 1/12 of the mass of a 12C atom. 1/12 of the mass of a 12C atom is exactly 1 u. Therefore, the mass of the hydrogen atom is less than 1/12 of the mass of a 12C atom. 1/12 of the mass of a 12C atom is exactly 1 u. Therefore, the mass of the hydrogen atom is greater than 1/12 of the mass of a 12C atom. 1/12 of the mass of a 12C atom is exactly 1 u. Therefore, the mass of the hydrogen atom is equal to 1/12 of the mass of a 12C atom.

The angle between the two pipes is 88.88°

The objective is to find the angle between the vectors u and v.

First, construct the vectors u and v,

The water main with 20% grade in the north direction is,

20% = rise/ run

rise/run = 20/100 = 1/5

That is rise = 1 and run = 5

Hence, rise corresponds to the z- component, and run corresponds to the y-component to the vector u. There is no x-component in this direction since the vector does not go to the east at all.

Therefore, the vector

                v = xi + yj + zk

Here x = 0, y = 5, z = 1

               v = 0i +5j +k is parallel to the pipe in the north direction.

The water main with 10% grade in the east direction is,

10% = rise/run

rise/run = 1/10

Thus, rise =1 and run = 10

Hence, rise corresponds to the z-components, and run corresponds to the x-components to the vector v. There is no y-component in this direction since the vector does not go to the north at all.

Therefore, the vector

         v = xi + yj + zk

Here x = 10, y = 0, z = 1

         v =10i +0j + k is parallel to the pipe in the east direction.

The angle between two vectors is,

Ф= [tex]cos^{-1} \frac{u-v}{|u| |v|}[/tex]

The dot product of the two vector u and v is,

u.v = (0.5,1).(10,0,1)

    = 0.10+5.0+1.1

    =1

The magnitude of these two vectors is,

|u| =[tex]\sqrt{x^{2} + y^{2}+z^{2} }[/tex]

|u| = [tex]\sqrt{0^{2} + 5^{2} +1^{2} }[/tex]

    =[tex]\sqrt{0+25+1}[/tex]

    = [tex]\sqrt{26}[/tex]

|v| = [tex]\sqrt{x^{2} + y^{2} +z^{2} }[/tex]

|v| = [tex]\sqrt{0^{2}+5^{2} +1^{2} }[/tex]

   = [tex]\sqrt{100+0+1}[/tex]

   =[tex]\sqrt{101}[/tex]

Therefore, the angle between the two pipes is,

Ф = [tex]cos^{-1} \frac{u-v}{|u||v|}[/tex]

   = [tex]cos^{-1}[/tex][tex]\frac{1}{\sqrt{26}\sqrt{101} }[/tex]

   ≈ 1.55rad

  = 88.88°

Hence, the angle between the two pipes is 88.88°

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The complete question is:

A water main is to be constructed with a ​20% grade in the north direction and a 10​% grade in the east direction. Determine the angle required in the water main for the turn from north to east.

The acceleration of a car moving along straight road that increases its speed from 0 to 100 km/h in 10 s is 10 km/h-s.

The rate at which an object's velocity with the minimum time changes is called an accelerator in mechanics. They were the vector quantities and accelerations.

Acceleration has dimensions of velocity (L/T) divided by time, i.e. L T−2. SI unit of acceleration is the meter per second squared (m s−2); or "meter per second per second", as the velocity in meters per second changes by the acceleration value, every second

The direction of net load acting on an item affects the direction of its acceleration.

Average acceleration = (change in velocity)/(time it takes)

The acceleration of a car moving along straight road that increases its speed from 0 to 100 km/h is:

car's acceleration = (100 km/h)/(10 s) = 10 km/h/s

Thus, acceleration of a car moving along a straight road that increases its speed from 0 to 100 km/h in 10 s is 10 km/h-s.

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Force necessary to overcome static friction= (μs mg ) / (cos Ф + μs sinФ)

We have to find Force(F) in terms of m, g,  Ф ,and μs

To get the value of F first we have to find the value of N by using the formula of friction,

F sin Ф + N - mg =0

N = mg - F sinФ

As the body does not move so the net force is zero, therefore we have,

FcosФ -μs N = 0

Now we have to put the value of N in the equation. To know the force necessary to overcome the static friction we have,

F cosФ - μs ( mg - FsinФ) = 0

F cosФ - μs mg + μs FsinФ = 0

F( cos Ф + μs sinФ ) = μs mg

F = μs mg/ (cosФ +μs sin Ф)

Therefore we get the value of F i.e., force in terms of m, g, Ф, and μs.

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The complete question is:

If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as Fcosθ−μsN=0 Fsinθ+N−mg=0 To find the magnitude of force F, we have to solve a system of two equations with both F and the normal force N unknown. Use the methods we have learned to find an expression for F in terms of m, g, θ, and μs (no N).

The larger magnification will be based on Focal length

The narrower the focal length, the smaller the magnification, and the larger the focal length, the larger the magnification.

it is obvious that the angle of the photograph from the objective lens or reflect to the eyepiece could be narrower the higher the focal duration. therefore, the photo may be considered via the eyepiece deeper with out dropping attention the narrower the angle (higher focal period).

The intensity of consciousness is shorter than the focal period. It approach that in comparison to large focal lengths, decrease focal lengths provide you with much less focuser play.

although you could nevertheless acquire the equal magnification with either focal period, the shorter focal duration will lose awareness a great deal earlier than the bigger focal period.

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the calculated displacement is 1000 m.

The equation a = v/t denotes acceleration (a), which is the change in velocity (v) over the change in time (t). This enables you to calculate the change in velocity in meters per second squared (m/s2). Since acceleration is a vector quantity, both its magnitude and its direction are included. The following equation can be used to determine distance from acceleration: D = v*t + 1/2*a*t2. where and is the acceleration, t is the passage of time, and v is the velocity.

Velocity= 25 m/s

time= 40 seconds

displacement= velocity x time

=25 x 40

=1000 m

the calculated displacement is 1000 m.

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The relative velocity of the first stone concerning the second stone is 20 m/s.

The time interval at which two stones are dropped from the tower is t.

t = 2 seconds

Let the initial velocity of the first stone be u1.

Let the initial velocity of the second stone be u2.

Let the final velocity of the first stone be v1.

Let the final velocity of the second stone be v2.

The velocity of the first stone is,

[tex]v _{1} = u _{1} + at[/tex]

[tex]v _{1} = 0 + gt[/tex]

[tex]v _{1} = gt[/tex]

The velocity of the second stone is,

[tex]v _{2} = g(t - 2)[/tex]

The relative velocity of the first stone concerning the second stone is,

[tex]v = v _{1} - v _{2}[/tex]

= g - g ( t - 2)

= gt -gt - 2g

= 2 × g

= 2 × 10

= 20 m/s

Therefore, the relative velocity of the first stone concerning the second stone is 20 m/s.

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Its acceleration is parallel to the x-axis

Using Newton Second Law of motion

Where M is mass of the object, [tex]\overrightarrow{F}[/tex] = Net force,[tex]\overrightarrow{a}[/tex] = acceleration.

Thus net force and acceleration has always the same direction.

Now net force on the box is in the positive x-direction then acceleration on the box is in the positive x-direction.

Acceleration is given by

[tex]\overrightarrow{a}[/tex] = [tex]\frac{d\overrightarrow{v}}{dt}[/tex]

Where [tex]\overrightarrow{v}[/tex] = velocity of the box and t = time

thus acceleration is parallel to change in velocity but not velocity. Thus it is not necessary that velocity of the box is parallel to the x-axis. Because if initially box is moving in arbitary direction then only its x-component of velocity changes only but another component of velocity remain same.

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Answer: yes

Explanation:

so basically i need points

The basic harmonic motion of the mass-spring system lasts for 5.024 seconds.

We have a 16-pound mass that is fastened to a spring with a 25-pound-per-foot spring constant. This entire spring and mass system is moving harmonically.

Simple harmonic motion (sometimes referred to as SHM) is a particular type of periodic motion in mechanics and physics in which the restoring force on the moving object is inversely proportional to the magnitude of the object's displacement and acts in the direction of the object's equilibrium position.

The particle's acceleration in a simple harmonic motion is geared toward its mean location and directly proportionate to its displacement. Simple harmonic motion results in the conservation of the particle's total energy. SHM is a cyclical motion.

Simple harmonic motion is demonstrated by bungee jumping. Due to the flexibility of the bungee cord, the jumper is experiencing SHM as it oscillates up and down.

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At least eight different genes have inherited mutations that cause the disease xeroderma pigmentosum. More than half of all cases in the US have mutations in the XPC, ERCC2, or POLH genes.

An alteration to an organism's DNA sequence is known as a mutation. Viral infection, mistakes in DNA replication during cell division, and exposure to mutagens are all ways in which mutations can develop.

Examples of inherited mutations include sickle cell disease, hemophilia, and cystic fibrosis. Additional mutations could happen naturally at any time during a person's lifetime. They're known as novel mutations. The damaged cells are small in number.

An alteration to an organism's DNA sequence is known as a mutation. Viral infection, mistakes in DNA replication during cell division, and exposure to mutagens are all ways in which mutations can develop.

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The mass of the girl is 30.6 Kg.

Inertia, a fundamental property of all matter, can be quantified as mass. In essence, it's the resistance of a mass of matter to changing its direction or speed in reaction to the application of a force. The more mass a body possesses, the smaller the change that is brought about by an applied force. The dimensionless quantity mass is used to express the amount of matter contained in a particle or object (symbolized m).

Given,

The pressure exerted by the girl = 15000 [tex]\frac{N}{m^{2}}[/tex]

Total area of contact = 0.02[tex]m^{2}[/tex]

As we know,

Pressure  = [tex]\frac{Force}{area}[/tex]

Force = Area x Pressure

Force = 0.02 x 15000

Force = 300N

For the conversion of weight into mass, use the formula "w = m x g." The gravitational pull on an object is referred to as its weight. The equation for the statement is written as w = m x g, or w = mg, by scientists.

Weight of the girl = Mass x Gravitational constant

300N = Mass of the girl x 9.8

Mass of the girl = 300 / 9.8

Mass of the girl =  30.6 Kg

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The position when time, t =35s is  87.59 m.

The acceleration represented by the graph is 0.143 m/s².

a = Δv/Δt

where;

a = (6 - 5) / (7 - 0)

a = 0.143 m/s²

The position of the object is determined from the product of velocity and time of motion.

s = vt + ¹/₂at²

where;

s = 0 + ¹/₂(0.143)(35)²

s = 87.59 m

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Their final velocity would be 0.118 m/s

Calculation

M1= 168000kg, V1= 0.291 m/s

M2= 117000kg, V2 = 0.128 m/s

According to conservation of momentum,

M1V1 + M2V2 = V(M1+M2)

⇒V = [tex]\frac{168000 * 0.291-117000*0.128}{168000+117000}[/tex]

=[tex]\frac{48888-14976}{285000}[/tex]

=0.118 m/s

Thus, the trains will finally get to the velocity of 0.118 m/s

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The correct answer is 223 days.

The relationship between the duration of revolution and the separation between the sun is shown by Kepler's third law. Using the notions of circular motion and the gravitational and centripetal forces, we may obtain this equation.

According to Kepler's third rule, the semi-major axis of an orbit is linked to the orbital period of a planet around the sun as follows:

p² = a³

where an is the semi-major axis/distance to the star and p is the orbital period in years.

It is said that a = 0.72 AU for Venus.

P= √(0.72 AU)^3 = 0.61 years.

365 days in a year = 222.9 ≈ 223 days.

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