If a fisherman applies a horizontal force with magnitude 42.5 nn to the box and produces an acceleration of magnitude 3.60 m/s2m/s2, what is the mass of the box?

According to Special Relativity

two objects in relative motion always see the other moving through time slower.

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Answer:

0.3km/min

Explanation:

2.4/8=24/80=3/10=0.3

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 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 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 correct answer is a) r = 30.7, b) r = 50 c) 0.822 cm

(a)

According to Snell's laws :

The incident ray, the refracted ray, and their normal boundaries between the two media all lie on the same plane.

i = 50o

Using Snell's Law

ni × sin(i) = nr × sin(r)

1.0 × sin(50) = 1.5 × sin(r)

r = 30.7

(b)

The Angle of refraction will be same as Angle of incidence, i = 50

So,

Final Angle of refraction, r = 50

(c)

Thickness, t = 5.0 cm

Difference = t  ×  [sin(50) - sin(30.7)]

x = 5.0  ×  [sin(50) - sin(30.7)]

x = 1.28 cm

The ray is displaced by , = 1.28 × sin(90 - 50) cm

= 0.822 cm

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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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A to find the total field you have to sum over all d's corresponding to each increment of length dx'. Summing of infinitesimals is called integration.

Combination is the computation of a vital. Integrals in math's are utilized to find numerous helpful amounts like regions, volumes, relocation, and so on. Researchers who predicted the idea of a basic, was the old Greek researcher Endous of Cnidus, who lived around 408-355 years BC. At the point when we talk about integrals, it is connected with typically unmistakable integrals. The endless integrals are utilized for antiderivatives. Vital conditions are arranged by three distinct divisions, making eight various types: Essential conditions are significant in numerous applications. The positive indispensable of a capability f (x) with lower cutoff and maximum breaking point b might be characterized as the distinction where F (x) is an antiderivative of the capability f (x).

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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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Correct question is -

A student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim. Which of the following measuring tools can the student use to test the validity of the claim?

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.

b. A meterstick to measure the distance of the track that the car travels on.

c. A motion detector that is oriented perpendicular to the direction that the car travels.

d. A mass balance to determine the mass of the car

a. Photogates positioned at the beginning, give up, and at numerous locations along the music that the car travels on.

b. A meterstick to degree the distance of the track that the automobile travels on.

on this scenario, a student is provided with a battery-powered toy vehicle that the manufacturer claims will continually function at a consistent speed. The student have to layout an test so as to check the validity of the declare.

consequently, to check the validity of the declare, the student must use the following measuring tools;

subsequently, with those two gadgets the student can successfully measure or determine the validity of the declare.

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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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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.

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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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:

alright

Explanation:

that's so cool I'll join

The momentum of the car, given the data d from the question is 0.0462 Kg.m/s

Momentum is defined as the product of mass and velocity. It is expressed as

Momentum = mass × velocity

With the above formula, we can determine the momentum of the car. Details below

The momentum of the car can be on rained as follow:

Momentum = mass × velocity

Momentum of car = 0.03 × 1.54

Momentum of car = 0.0462 Kg.m/s

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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 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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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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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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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 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 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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Answer: Carbon and oxygen combine together to give two compounds carbon dioxide (CO 2) and carbon monoxide (CO) Thus the ratio of different weights of oxygen (32 and 16) combining with a fixed weight of carbon (12) is 32 : 16 i.e. 2 :1, which is simple whole number ratio. Illustration 2:

Explanation:

Use the dot product identity,

[tex]\vec x \cdot \vec y = \|\vec x\| \|\vec y\| \cos(\theta)[/tex]

where [tex]\theta[/tex] is the measure of the angle between the two vectors [tex]\vec x,\vec y[/tex]. Note that

[tex]\vec x \cdot \vec x = \|\vec\| \|\vec x\| \cos(0) = \|\vec x\|^2[/tex]

Applying the identity to [tex]\vec x = \vec y = \vec A + \vec B[/tex] and using the distributive and commutative properties of the dot product, we get

[tex](\vec A + \vec B) \cdot (\vec A + \vec B) = (\vec A \cdot \vec A) + 2 (\vec A \cdot \vec B) + (\vec B \cdot \vec B)[/tex]

which is equivalent to

[tex]\|\vec A + \vec B\|^2 = \|\vec A\|^2 + 2\|\vec A\|\|\vec B\| \cos(\theta) + \|\vec B\|^2[/tex]

Now, we're given [tex]\|\vec A\| = 13.9[/tex], [tex]\|\vec B\| = 5.9[/tex], and [tex]\theta=53^\circ[/tex], so that

[tex]\|\vec A + \vec B\| = \sqrt{13.9^2 + 2\cdot13.9\cdot5.9\cos(53^\circ) + 5.9^2} \approx \boxed{18.1}[/tex]

When a body moves in a simple harmonic motion, its acceleration at the end of its path is more than that of the acceleration at equilibrium.

The acceleration of the body moving in Simple Harmonic Motion(SHM) is given by,

Acceleration = ω^2 A

Here w is the angular velocity of the body and A is the amplitude.

A is the maximum distance of the body from the equilibrium position.

When the body is at an equilibrium position, its amplitude is equal to zero.

Hence, acceleration = ω^2 x 0 = 0

When the body is at the end of its path, its amplitude is equal to A.

Hence, acceleration = ω^2 x A

Therefore,

when a body moves in a simple harmonic motion, its acceleration at the end of its path is more than that of the acceleration at equilibrium.

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Answer: yes

Explanation:

so basically i need points