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3 years ago

There are two unitless vectors:

Physics

1 answer:

Hunter-Best [27]3 years ago

4 0

Answer:

$200.38^0$

Explanation:

Given the forces:

F1 = 8.92 i + 17.37 j

F2 = 8.31 i - 10.97 j

If a third vector is added to them such that they add up to to the null vector as F1 + F2 + F3 = 0

then to get F3:

F3 = -F2-F1

F3 = -(8.31 i - 10.97 j)-(8.92 i + 17.37 j )

F3 = -8.31 i + 10.97 j-8.92 i - 17.37 j

F3 = -8.31i-8.92i+10.97j-17.37j

F3 = -17.23i-6.4j

from the vector:

x = -17.23 and y = -6.4

angle of the third vector with respect to the +x-axis is expressed as:

$\theta = tan^{-1}\frac{y}{x}\\ \theta = tan^{-1}\frac{-6.4}{-17.23}\\\theta = tan^{-1}0.3715\\\theta = 20.38^0$

Hence the angle the vector makes with the x axis will be $\theta = 180+20.38 = 200.38^0$

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How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

$T_{f}$ = (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

$T_{f}$ = 466.8 K

so change in temperature ΔT

ΔT = 466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x $_{i\\}$ ² - x $_{f\\}$ ² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q = 16 J

Therefore, the required heat energy is 16 J

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2 years ago

the gravitational force between the sun and earth is larger than sun and the moon. what do you think the reason for this is?

Tems11 [23]

Answer: The greater an object's mass, the more gravitational force it exerts.

Explanation: So, to begin answering your question, Earth has a greater gravitational pull than the moon simply because the Earth is more massive. Sorry if I get this wrong. I am in 5th grade! ♥

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2 years ago

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Why do elements within a group have similar chemical properties

CaHeK987 [17]

Elements in the same group in the periodic table have similar chemical properties. This is because their atoms have the same number of electrons in the highest occupied energy level. Group 1 elements are reactive metals called the alkali metals.Group 0 elements are unreactive non-metals called the noble gases.

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3 years ago

The weight of a person is 500N and his foot imprint area is 0.5m^2.Calculate the total pressure exerted by person when he stands

Ghella [55]

Answer:

Pressure on both feet will be $500N/m^2$

Explanation:

Weight of the person F = 500 N

Area of foot print $A=0.5m^2$

Area of both the foot $A=2\times 0.5=1m^2$

We have to find pressure on both the feet

Pressure is equal to ratio of force and area

So pressure $P=\frac{F}{A}$

$P=\frac{500}{1}=500N/m^2$

So the pressure on both feet will be $500N/m^2$ when person stands on both feet.

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3 years ago

PLEASE HELP LOTTA POINTS

sergiy2304 [10]

There's a nasty wrinkle here that's kind of sneaky, and makes the work harder than it should be.

Look at the first question. There's a number there that's dropped in so quietly that you're almost sure to miss it, but it changes the whole landscape of both of these problems. That's where it says

" ... 20 cm mark (30 cm from the fulcrum) ... " .

That tells us that the yellow bar resting on the pivot is actually a meter stick, but the pictures don't show the centimeter marks on the stick. The left end of the stick is "0 cm", the right end of the stick is "100 cm", and the pivot is under the "50 cm" mark.

When the question talks about hanging a weight, it tells the centimeter mark on the stick where the weight is tied. To solve the problem, we have to first figure out how far that is from the pivot, then calculate how far from the pivot to put the weight on the other side, and finally what centimeter mark that is on the stick.

How to solve the problems:

-- The "moment" of a weight is (the weight) x (its distance from the pivot) .

-- To balance the stick, (the sum of the moments on one side) = (the sum of the moments on the other side).

= = = = = = = = = =

#1). Only one moment on the left side.

(160 gm) x (30 cm from pivot) = 4,800 gm-cm

To balance, we need 4,800 gm-cm of moment on the right side.

(500 gm) x (distance from pivot) = 4,800 gm-cm

Distance from pivot = (4,800 gm-cm) / (500 gm) = 9.6 cm

The 500 gm has to hang 9.6 cm to the right of the pivot. But that's not the answer to the problem. They want to know what mark on the stick to hang it from. The pivot is at the 50cm mark. The 500gm has to hang 9.6 cm to the right of the pivot. That's the 59.6 cm mark on the stick.

= = = = =

#2). There are 2 weights hanging from the left side. We have to find the moment of each weight, add them up, then create the same amount of moment on the right side.

one weight: 120gm, hanging from the 25cm mark.

That's 25cm from the pivot. Moment = (120gm) (25cm) = 3,000 gm-cm

the other weight: 20gm, hanging from the 10cm mark;

That's 40cm from the pivot. Moment = (20gm) (40cm) = 800 gm-cm

Add up the moments on the left side:

(3,000 gm-cm) + (800 gm-cm) = 3,800 gm-cm.

To balance, we need 3,800 gm-cm of moment on the right side.

(500 gm) x (its distance from the pivot) = 3,800 gm-cm

Distance from the pivot = (3,800 gm-cm) / (500 gm) = 7.6 cm

The pivot is at the 50cm mark on the stick. You have to hang the 500gm from 7.6cm to the right of that. The mark at that spot on the stick is (50cm + 7.6cm) = 57.6 cm .

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2 years ago

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