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2 years ago

On a hypothetical scale X The ice point is 40° and steam point is 120°.

Physics

1 answer:

arlik [135]2 years ago

7 0

Answer:

The reading of Y is -10°.

Explanation:

For scale X, the ice point is 40° and steam point is 120°.

Difference between the two extremes for scales X = 120 - 40 = 80

For scale X, the ice point and steam points are -30° and 130° respectively.

Difference between the two extremes for scales X = 130 - (-30) = 160

Comparing both scales:

One unit of scale X = x

One unit of scale Y = y

Scale X has 80 divisions while scale Y has 160

80x = 160y

x = 2y

50° in scale X = 10x + ice point in X scale

10 divisions in Y scale = 20y

Reading of Y scale = ice point of Y + 20y

= -30° + 20°

= -10°

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The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviatio

enot [183]

Answer:

2.25 %

Explanation:

65-95-99.7 is a rule to remember the precentages that lies around the mean.

at the range of mean ( $\mu$ ) plus or minus one standard deviation ( $\sigma$ ), $P([\mu-\sigma \leq X \leq \mu+\sigma])\approx 68.3%$

at the range of mean plus or minus two standard deviation, $P([\mu -2\sigma \leq X \leq \mu+2\sigma])\approx 95.5%$

at the range of mean plus or minus three standard deviation, $P([\mu - 3\sigma\leq X \leq \mu+3\sigma])\approx 99.7%$

So, note that they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) ( $P(X \leq \mu+2\sigma)$ )

So we know that the 95.5% is between $\mu - 2\sigma = 0.3 -2*0.1 = 0.28$ and $\mu + 2\sigma = 0.3 +2*0.1 = 0.32$ , hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than 0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.

So $P(X \leq \mu+2\sigma) \approx 2.25%$

8 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

A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial

Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

$t=\frac{2usin\theta}{g}$

Here θ = 30° and t = 20 s

a) $t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s$

Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

$H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m$

Maximum altitude = 490.5 m

c) Range of projectile is given by

$R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m$

Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

We have equation of motion s = ut + 0.5 at²

Horizontal motion

u = 169.91 m/s

a = 0 m/s²

t = 15 s

Substituting

s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

Vertical motion

u = 98.1 m/s

a = -9.81 m/s²

t = 15 s

Substituting

s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

$\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m$

Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

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2 years ago

To persuade my audience that the use of mobile communication devices by drivers—even when they are hands-free—is contributing to

pickupchik [31]

Answer:

specific purpose statement

Explanation:

It is a specific purpose statement made for a persuasive speech on the question of fact.

A specific purpose statement helps to build on the general purpose (that is to inform) and to make it more specific to the audience. So if the first speech is an informative speech, our general purpose is to inform our audience about a very specific realm of knowledge.

A specific purpose statement is given to audience to persuade on specific information.

8 0

2 years ago

A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text

antoniya [11.8K]

Answer:

The energy of the system is 15 J.

Explanation:

Given that,

Energy E = 2.5 J

Amplitude = 10 cm

We need to calculate the spring constant