(4) A stainless steel storage tank contains 5 kg of carbon dioxide gas and 7 kg of argon gas. How many kmoles are in the tank?

Answer:

a) 121 km

b) 74°

Explanation:

To start with, we assume that there exist two components, the East and the North. We'd be representing the East, by "e" and the North, by "n"

Now we start with the

First vector:

east1 = 232 cos 30 = 201

north1 = 232 sin 30 = 116

Now, that of the second vector will be

east 2 = - 168

north2 = 0

Next, we add the two together and get

East components

201 - 168 = 33 east

North components

116 + 0 = 116

Therefore, the magnitude has to be

magnitude = √(33² + 116²)

Magnitude = √14545

Magnitude = 121

tanθ = 116/33

Tanθ = 3.51

θ = tan^-1 3.51

θ = 74° North East

(a) For the first 6.2 s, the car has velocity at time t given by

[tex]v(t)=13.5\dfrac{\rm m}{\rm s}+\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

so that after 6.2 s, it attains a velocity of

[tex]v(6.2\,\mathrm s)=25.28\dfrac{\rm m}{\rm s}[/tex]

For any time t after 6.2 s, its velocity is given by

[tex]v(t)=25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

which tells us the velocity only falls from this point onward. This means the maximum speed is 25.28 m/s, or about 25.3 m/s.

(b) Solve for t (after 6.2 s) that makes v(t) = 0 :

[tex]25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t=0[/tex]

[tex]\implies t\approx21.067\,\mathrm s[/tex]

It takes the car about 21.2 s to come to a rest, so the car travels a total of about 6.2 s + 21.2 s = 27.4 s.

(c) For the first 6.2 s, the car undergoes a displacement at time t of

[tex]x(t)=\left(13.5\dfrac{\rm m}{\rm s}\right)(6.2\,\mathrm s)+\dfrac12\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)(6.2\,\mathrm s)^2[/tex]

[tex]\implies x\approx120.218\,\mathrm m[/tex]

For time t beyond 6.2 s, its displacement is

[tex]x(t)=120.218\,\mathrm m+\left(25.28\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

The car comes to a rest after 21.2 s (accelerating at a rate of -1.2 m/s^2), so that its total displacement is

[tex]x(21.2\,\mathrm s)\approx386.49\,\mathrm m[/tex]

so the car travels a total distance of about 387 m.

Answer:

Explanation:

Displacement vector along x axes = 4.5 - 2.5 = 2 m

Displacement vector along y axes = 3 - 2 = 1 m

Displacement vector along z axis = 3.5- 4 = - 0.5 m

Displacement vector = 2 i + j - 0.5 k m

The question is incomplete. Here is the complete question.

The map (in the attachment) shows Olivia's trip to the coffee shop. She gets on her bike at Loomis and then rides south 0.9mi to Broadway. She turns east onto Broadway, rides 0.8mi to where Broadway turns, and then continues another 1.4mi to the shop.

What is the magnitude of the total displacement of her trip?

Whta is the direction of the total displacement of her trip?

Answer: Magnitude = 2.6mi

              Direction: 54.65° east

Explanation: Displacement is the change in postition of a moving object.

There are a few ways to determine total displacement. For this case, the Perpendicular Components of a Vector method will be used.

For this method, total displacement is given by:

[tex]\Delta d_{t}=\sqrt{(\Delta d_{x})^{2}+(\Delta d_{y})^{2}}[/tex]

[tex]\theta=tan^{-1}(\frac{\Delta d_{y}}{\Delta d_{x}})[/tex]

[tex]\Delta d_{x}[/tex] is the x-component of total displacement and it is the sum of each individual x-components;

[tex]\Delta d_{y}[/tex] is the y-component of total displacement and it is the sum of each individual y-components;

θ is the angle the resulting displacement;

For Olivia's trip, there are no x-component of the first part and for the third part, the path she bikes is a hypotenuse of a right triangle. So, that right triangle's x-component is:

[tex]sin30=\frac{x}{1.4}[/tex]

[tex]\frac{1}{2} =\frac{x}{1.4}[/tex]

x = 0.7

Then,

[tex]\Delta d_{x}[/tex] = 0 + 0.8 + 0.7

[tex]\Delta d_{x}[/tex] = 1.5

Related to y, there are no y-component in the second part of Olivia's trip and for the third part:

[tex]cos30=\frac{y}{1.4}[/tex]

[tex]\frac{\sqrt{3} }{2} =\frac{y}{1.4}[/tex]

y = 1.21

Then,

[tex]\Delta d_{y}[/tex] = 0.9 + 0 + 1.21

[tex]\Delta d_{y}[/tex] = 2.11

Total displacement is

[tex]\Delta d_{t}=\sqrt{(1.5)^{2}+(2.11)^{2}}[/tex]

[tex]\Delta d_{t}=\sqrt{6.7021}[/tex]

[tex]\Delta d_{t}=[/tex] 2.6

Magnitude of Olivia's total displacement is 2.6mi

On the map, joining the initial and final points gives a vector pointing towards east at angle:

[tex]\theta=tan^{-1}(\frac{2.11}{1.5})[/tex]

[tex]\theta=tan^{-1}(1.41)[/tex]

θ = 54.65°

Direction of total displacement is 54.65° East.

Answer:

1.9 mi, 350.7°

Explanation:

If continued another 1.2 mi*

solving for x

0 + 0.8 + 1.2(cos30) = 1.83923048 --> 1.84

solving for y

-0.9 + 0 + 1.2(sin30) = -0.3

^negative because it is going downward

a) solving for magnitude

[tex]\sqrt{(1.84)^2+(-0.3)^2} = \sqrt{3.4756} = 1.86429611... = 1.9 mi[/tex]

b) solving for direction of total displacement

[tex]tan^-1 = (\frac{x}{y} ) \\tan^-1 = (\frac{-0.3}{1.84}) = -0.16 \\[/tex]

∘, measured counterclockwise from the eastward direction

360 - 0.16 = 359.84°

*replace any of the needed values in the equation, such as 1.2 mi to 1.4 mi

Answer:

The  value in Newton is [tex]W =  631.92 \  N[/tex]

The  value in pounds is    [tex]W  = 142 \ lb[/tex]

Explanation:

From the question we are told that

  The  mass of the spacecraft is  [tex]m =  70 \  kg[/tex]

   The distance above  the earth is  [tex]d =  275 \  km  =  275000 \  m[/tex]

Generally the gravitational force with respect to the earth is mathematically represented as

       [tex]W =  \frac{G * m *  m_e}{ (d + r_e)^2}[/tex]

Here [tex]m_e[/tex] is the mass of earth with value [tex]m_e =  5.978 *10^{24} \  kg[/tex]

       [tex]r_e[/tex] is the radius of the earth with value  [tex]r_e  =  6371  \ km  =  6371000 \ m[/tex]

   G is the gravitational constant with value [tex]G  =  6.67 *10^{-11}  \  m^3/ kg\cdot s^2[/tex]

So

     [tex]W =  \frac{ 6.67 *10^{-11} *  70 *  5.978 *10^{24}}{ (275000 + 6371000)^2}[/tex]

     [tex]W =  631.92 \  N[/tex]

Converting to  pounds

    [tex]W =  \frac{631.92  }{4.45}[/tex]

        [tex]W  = 142 \ lb[/tex]

Answer:

the statements, the correct one is A

a downward force of gravity and an upward force exerted by the surface

Explanation:

When the disc is hit, a thrust force is exerted in the direction of movement, at the moment the disc moves this force loses contact and becomes zero.

When the movement is already established there are two main forces: gravity that acts downwards and the reaction force to the support of the disk called normal that acts upwards.

As it is not mentioned that there is friction, this force that opposes the movement is zero.

Analyzing the statements, the correct one is A

Answer:

model is viable if the assumptions that answer it are in accordance with the fundamental principles or laws of physics and if it gives conclusions that can be tested with experiments.

Explanation:

A model in physics must be verified by experiments that are carried out to measure the consequences derived from it.

A model is viable if the assumptions that answer it are in accordance with the fundamental principles or laws of physics and if it gives conclusions that can be tested with experiments. Models that meet these conditions are said to be viable

The area of ANY circle is (π) · (radius²).

So ...

(π) · (radius²) = 154 cm²

radius² = (154 cm²) / (π)

radius² = 49.02 cm²

radius = √(49.02 cm²)

radius = 7 cm

Answer:

[tex] \boxed{\sf Radius \ of \ circle \ (r) = 7 \ cm} [/tex]

Given:

Area of circle = 154 cm²

To Find:

Radius of circle (r)

Explanation:

[tex] \bold{Area \: of \: circle = \pi r^2}[/tex]

[tex] \sf \implies \pi {r}^{2} = 154 \\ \\ \sf \implies {r}^{2} = \frac{154}{\pi} \\ \\ \sf \implies {r}^{2} = \frac{154}{ \frac{22}{7} } \\ \\ \sf \implies {r}^{2} = \frac{154 \times 7}{22} \\ \\ \sf \implies {r}^{2} = \frac{1078}{22} \\ \\ \sf \implies {r}^{2} = 49 \\ \\ \sf \implies {r}^{2} = {7}^{2} \\ \\ \sf \implies r = \sqrt{ {7}^{2} } \\ \\ \sf \implies r = 7 \: cm[/tex]

Answer:

Explanation: From 13:30 to 15:00, it past: 1 h 30 mins = 1.5

Then, the distance covered by Peter: 40x1.5= 60 miles

From 13:45 to 15:00, it pasts; 1 h 15min =1.25

Then, the distance covered by Philip. 30 x 1.25 = 37.5 miles

Lastly, the distance between them: 60-37.5= 22.5 miles

So the answer is 22.5

Answer:

A) Total Distance = 2000 m and Total displacement = 0 m

B) Average Speed = 44.44 m/min and Average Velocity = 0 m/min

C) Average Speed = 0.7407 m/s and Average velocity = 0 m/s

Explanation:

A) Distance to reach ice cream shop from home = 1000 meters

Therefore distance to get back home would also be 1000 meters.

Total distance traveled = 1000 + 1000 = 2000 metres

Since journey started at home and ended at home, then total displacement = 0 metres.

B) Average speed = Total distance/total time.

Total time = 10 + 15 + 20 = 45 minutes

Since total distance = 2000 m

Then;

Average speed = 2000/45

Average speed = 44.44 m/min

Average velocity = Total displacement/total time

Average velocity = 0/45 = 0 m/min

C) We now want answers in B to be in m/s.

Total time = 45 minutes.

From conversion, 60 seconds make 1 minute. Thus, 45 minutes = 45 × 60 = 2700 seconds

Thus;

Average Speed = 2000/2700

Average Speed = 0.7407 m/s

Average displacement = 0/2700 = 0 m/s

Answer:

V= 14.2m/s

Explanation:

We know that acceleration= dv/dt

So 16m/s²=dv/ dt = v dv/ds

So this wil be

Integral of 16m/s² ds [at 2,2]= integral of v dv at[ 0, v]

So 16[s (3/2)/3/2] at ( s,3) = v²/2

At s= 6m

So v² = 64/3( 6^1.5-3^1.5)

= 14.2m/s

Answer:

10m

Explanation:

if it was at the 2.0 line it would be 10 m

If a typical atom has a diameter of about 1.0×10⁻¹⁰ m, then there are approximately atoms are there along a 2.0-centimeter line.

In positional notation, significant figures refer to the digits in a number that is trustworthy and required to denote the amount of something, also known as the significant digits, precision, or resolution.

As given in the problem If a typical atom has a diameter of about 1.0×10⁻¹⁰ m, then we have to find out approximately how many atoms are there along a 2.0-centimeter line,

diameter of the one atom =  1.0×10⁻¹⁰

approximate number of atoms in 2 cm line = 2 ×10⁻² /( 1.0×10⁻¹⁰ )

                                                                         =2 ×10⁸ atoms

Thus, there are approximately 2 ×10⁸ atoms are there along a 2.0-centimeter line.

Learn more about significant figures here,

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Length = (75.00 m)

Length = (75 meter) x (3.28084 foot/meter) x (12 inch/foot)

Length = (75 x 3.28084 x 12) (meter-foot-inch / meter-foot)

Length = 2,952.76 inches

Answer:

Displacement vector represents the average velocity of the bird.

Explanation:

Given that,

A bird flutters around in a tree in a path described by the dark line.

Suppose, given vectors

(a). Time, (b). Displacement, (c). speed, (d). distance

We know that,

Vector quantity :

Vector quantity has direction and magnitude.

Average velocity :

Average velocity is equal to the displacement divided by time.

In mathematically,

[tex]v=\dfrac{D}{t}[/tex]

Where, D = displacement

t = time

We need to find the vector which is represents the average velocity of the bird

Using given data

Average velocity of the bird shows the displacement over the time.

Displacement is the vector quantity.

Hence, Displacement vector represents the average velocity of the bird.

Answer:

Average net force = 0.62 N

Explanation:

We are given;

Mass; m = 1.7 kg

Initial velocity; u = 12.5 m/s

Final velocity; v = 25 m/s

time; t = 25 seconds

Now, we are told that the final velocity was 30° west of North. So, resolving this velocity along the horizontal gives;

v = 25 cos 30°

Now, using Newton's first equation of motion gives;

v = u + at

Where a is acceleration

Plugging in the relevant values gives;

25 cos 30° = 12.5 + 25a

21.6506 - 12.5 = 25a

a = (21.6506 - 12.5)/25

a = 0.3660 m/s²

Now, magnitude of the average net force would be; F = ma

F = 1.7 × 0.366

F ≈ 0.62 N

Answer:

Its according from what it is made with if wax it takes longer but if with a craft paper it takes lesser time if paper approximately 20 to 25 minutes

Answer:

I think B tell me if it's right

Explanation:

If an engineer is designing a tire for heavy machinery then a statement that describes the clearest criterion for the solution would be that it must function safely under the load of 4500 kg, therefore the correct answer is option C.

Mechanical advantage is defined as a measure of the ratio of output force to input force in a system, It is used to analyze the forces in simple machines like levers and pulleys.

Mechanical advantage = output force(load) /input force (effort)

As given in the problem statement that an engineer is designing a tire for heavy machinery and we have to find the statement which describes the clearest criterion for the solution,

As he is designing heavy machinery that must be able to support a large amount of weight,

Thu, the best option that is satisfying the criteria is option C.

Learn more about Mechanical advantages, here

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Answer:

Hearing, Vision and Metabolism.

Answer:Isobars and isotherms

Explanation:

Answer:

It's increasing with time

In the state where I live, your driver's license is not a proof that you have liability insurance.  You don't need liability insurance to get a driver's license, but you need it in order to operate a car that you own.

It may be different in the state where YOU live.

NEXT QUESTION

Explanation:

In this problem, we are meant to slove for the resultant and the direction of the the vectors given

Given data

let the sail to a point due north be y= 116km

and the point due east be x= 121 km

(a) How far (in km)

The resultant between the two points is the distance between them

[tex]r=\sqrt{x^2+y^2} \\\\r=\sqrt{121^2+116^2} \\\\r=\sqrt{14641+13456} \\\\r=\sqrt{28097} \\\\r=167.62[/tex]

The distance between the points is 167.62

(b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination

the direction can be gotten using

tan∅= y/x

∅= tan-1 (y/x)

∅= tan-1(116/121)

∅= tan-1(0.958)

∅= 43.77°

The direction is 43.77°

Answer:

The magnetic field inside the cylindrical resistor is [tex]\dfrac{\mu_{0}ir}{2\pir_{0}^2}[/tex]

Explanation:

Given that,

Distance from the axis of the cylinder = r

We need to calculate the magnetic field inside the cylindrical resistor

Using formula of magnetic field

[tex]\oint{\vec{B}\cdot\vec{dl}}=\mu_{0}i_{encl}[/tex]

[tex]B\cdot(2\pi r)=\mu_{0}\dfrac{i\pir^2}{\pi r_{0}^2}[/tex]

Where, r₀ = radius

r = distance

i = current

[tex]|\vec{B}(r)|=\dfrac{\mu_{0}ir}{2\pir_{0}^2}[/tex]

Hence, The magnetic field inside the cylindrical resistor is [tex]\dfrac{\mu_{0}ir}{2\pir_{0}^2}[/tex]

Answer: Hydrogen Peroxide

Explanation: Hydrogen peroxide breaks down into oxygen and water. As a small amount of hydrogen peroxide generates a large volume of oxygen, the oxygen quickly pushes out of the container. The soapy water traps the oxygen, creating bubbles, and turns into foam.

Answer: Heart rate provides an objective measurement of how hard your body is working. The higher the exercise intensity, the higher your heart rate will be.

Explanation:

Answer:

You're getting exercise and training your heart!

Explanation:

Answer:

Eat more healthy foods.  Workout and build your immune system.

Explanation:

Eat  Healthy foods like Carrots, Apples, Bannas, Pears, and anything that deals with not much of any sugar. An example of unhealthy foods is Cakes, Chocolates, Candy, and more. Drink a lot of water.

Answer:

We’re imagining imagining imagining an imagination...

Answer:

the intercept is the orgin (0,0)

Answer:

The magnitude of the acceleration of the car after the brakes failed is 4 m/s²

Explanation:

The car was originally traveling at 30.0 m/s, that is

The initial velocity, [tex]u[/tex] = 30.0 m/s

The time spent while the car manages to brake is 5.0 seconds, that is

time, [tex]t[/tex] = 5.0 secs

and the distance traveled during this time is

distance, [tex]s[/tex] = 125 m

From one of the equations of kinematics for linear motion,

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

Where [tex]a[/tex] is the acceleration

We can determine the deceleration of the car during the first 5.0 seconds

Hence,

From,

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

[tex]125 = 30.0(5.0) + \frac{1}{2}(a)(5.0)^{2}[/tex]

[tex]125 =150.0 + 12.5a[/tex]

[tex]12.5a = 125 - 150.0[/tex]

[tex]12.5a = -25\\a = \frac{-25}{12.5}\[/tex]

[tex]a = - 2.0 m/s^{2}[/tex] (Negative sign indicates deceleration)

Now we will calculate the final velocity reached at this time

From,

[tex]v^{2} = u^{2} + 2as[/tex]

Where [tex]v[/tex] is the final velocity

[tex]v^{2} = 30.0^{2} + 2(-2.0)(125)\\v^{2} = 400\\v = \sqrt{400} \\v = 20 m/s \\[/tex]

This is the final velocity reached by the car during the first 5.0 seconds

Now, for the magnitude of the acceleration of the car after the brakes failed,

After the brakes failed,

it travels an additional 150 m further down the road, that is

s = 150m

an additional 5.0 seconds, that is

t = 5.0 seconds

Also, from

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

The initial velocity here will be the final velocity for the first 5.0 seconds, that is,

u = 20 m/s

Hence,

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

[tex]150 = 20(5.0) + \frac{1}{2}(a)(5.0)^{2}[/tex]

[tex]150 = 100 + 12.5a\\12.5a = 150 - 100\\12.5a = 50\\a = \frac{50}{12.5} \\a = 4m/s^{2}[/tex]

Hence, the magnitude of the acceleration of the car after the brakes failed is 4 m/s²

Answer:

The correct option is a

Explanation:

This question seeks to test a general rule in physics (on charges) which states that like charges repel but unlike charges attract. This means that, a negatively charged substance will repel or not attract another negatively charged material and the same applies to a positively charged substance also. However, a negatively charged substance will attract a positively charged material and vice versa, hence only a negatively charged rod will attract a positively charged hanging ball.