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sertanlavr [38]

2 years ago

6

Consider the vector b⃗ with magnitude 4.00 m at an angle 23.5∘ north of east. what is the x component bx of this vector? express

your answer in meters to three significant figures.

1 answer:

BlackZzzverrR [31]2 years ago

6 0

Decomposing the vector b on the x-axis and the y-axis, we get a rectangle triangle where the two sides are bx (x-axis) and by (y-axis), and b is the hypothenuse.
The component in x, bx, is equal to the product between the hypothenuse and the cosine of the angle between b and the x-axis, which is $23.5 ^{\circ}$ :
$b_x = b \cos (23^{\circ})=(4.00 m)(\cos (23^{\circ}))=3.68 m$

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In a house the temperature at the surface of a window is 28.9 °C. The temperature outside at the window surface is 7.89 °C. Heat

Alenkasestr [34]

Answer:

-13.18°C

Explanation:

To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.

Its definition is given by the function

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

Where,

Q = The amount of heat transferred

t = time

k = Thermal conductivity constant

A = Cross-sectional area

$\Delta T =$ The difference in temperature between one side of the material and the other

d= thickness of the material

The problem says that there is a loss of heat twice that of the initial state, that is

$Q_2 = 2*Q_1$

Replacing,

$kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}$

$\frac{\Delta T}{x}=2*\frac{\Delta T}{x}$

$\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}$

$\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}$

Solvinf for $T_o$ ,

$T_o = -13.18$

Therefore the temprature at the outside windows furface when the heat lost per second doubles is -13.18°C

3 0

2 years ago

An object thrown in the air has a velocity after t seconds that can be described by v(t) = -9.8t + 24 (in meters/second) and a h

marin [14]

Answer and Explanation: Kinetic energy is related to movement: it is the energy an object possesses during the movement. it is calculated as:

$K=\frac{1}{2}mv^{2}$

For the object thrown in the air:

$K=\frac{1}{2}.2.[v(t)]^{2}$

$K=(-9.8t+24)^{2}$

$K=96.04t^{2}-470.4t+576$

Kinetic energy of the object as a function of time: $K=96.04t^{2}-470.4t+576$

Potential energy is the energy an object possesses due to its position in relation to other objects. It is calculated as:

$U=mgh$

For the object thrown in the air:

$U=9.8.2.h(t)$

$U=9.8.2.(-4.9t^{2}+24t+60)$

$U=-96.04t^{2}+470.4t+1176$

Potential energy as function of time: $U=-96.04t^{2}+470.4t+1176$

Total kinetic and potential energy, also known as mechanical energy is

TME = $96.04t^{2}-470.4t+576$ + ( $-96.04t^{2}+470.4t+1176$ )

TME = 1752

The expression shows that total energy of an object thrown in the air is constant and independent of time.

6 0

2 years ago

Dao makes a table to identify the variables used in the equations for centripetal acceleration. A 2 column 5 rows. The first col

Zanzabum

Answer:

Column X. Tangential Speed

Column Y. radius

Explanation:

The equation for centripetal acceleration is

$a_{c}$ = v² / r

Where v is the tangential velocity of the body and the radius of curvature.

To analyze this equation you must place the tangential velocity in one column and in the other the turning radius

Let's check the answers

Column X. Tangential Speed

Column Y. radius

This is the correct answer.

5 0

2 years ago

Read 2 more answers

2. Harry is pushing a car down a level road at 2.0 m/s with a force of 243 N. The total force

Arte-miy333 [17]

b) Equal to 243 N.

Explanation:

The total force acting on the car in the opposite direction including the road friction and air resistance is equal to 243 N.

This is in conformity with newton's third law of motion.

Newton's third law of motion states that "action and reaction are equal and opposite in direction. "

The action force is that of the pull by Harry acting on the car.
The reaction force is in the opposite direction.
Both action and reaction force equal and opposite and magnitude and direction

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0

2 years ago

The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 600°C is 1 × 10-6. Calculate the number of vacancies

algol [13]

Answer :

The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 × 10^6cm^3/m^3

= 10.35 × 10^6 g/m^3

Avogadro's number = 6.022 × 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 × 10^-6

Nag = (Na * Da)/Aag

Where,

Nag = Total number of lattice sites in Ag

Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6

= 5.778 × 10^22/m^3.

6 0

2 years ago