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4 years ago

How much heat energy (in megajoules) is needed to convert 7 kilograms of ice at –9°C to water at 0°C?

1 answer:

8 0

The heat required to change the phase of a substance can be calculated by using the formula,

q = mCΔT where q is the heat needed, m is the mass of the substance, C is the specific heat capacity and ΔT is the change in temperature.

q = 7000 g (4.18 J/ g °C) (0°C - (-9°C))
q = 263340 J or .26334 MJ
Hope it's correct! ( If so rank as brainliest answer) :)

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Technology can best be defined as what?

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3 years ago

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A) One Strategy in a snowball fight the snowball at a hangover level ground. While your opponent is watching this first snowfall

Alexandra [31]

Answers:

a) $\theta_{2}=38\°$

b) $t=0.495 s$

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{1} t_{1}$ (1)

Where:

$V_{o}=14.1 m/s$ is the initial speed of snowball 1 (and snowball 2, as well)

$\theta_{1}=52\°$ is the angle for snowball 1

$t_{1}$ is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

$y=0$ is the final height of the snowball 1

$g=-9.8m/s^{2}$ is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{2} t_{2}$ (3)

Where:

$\theta_{2}$ is the angle for snowball 2

$t_{2}$ is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}$ (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate $t_{1}$ from (2):

$0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (5)

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$ (6)

Substituting (6) in (1):

$x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})$ (7)

Rewritting (7) and knowing $sin(2\theta)=sen\theta cos\theta$ :

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})$ (8)

$x=-\frac{(14.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(52\°))$ (9)

$x=19.684 m$ (10) This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate $t_{2}$ from (4) and substitute it on (3):

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$ (11)

$x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})$ (12)

Rewritting (12):

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})$ (13)

Finding $\theta_{2}$ :

$2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})$ (14)

$2\theta_{2}=75.99\°$

$\theta_{2}=37.99\° \approx 38\°$ (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$ .

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of $t_{1}$ and $t_{2}$ from (6) and (11), respectively:

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$

$t_{1}=-\frac{2(14.1 m/s)sin(52\°)}{-9.8m/s^{2}}$ (16)

$t_{1}=2.267 s$ (17)

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$

$t_{2}=-\frac{2(14.1 m/s)sin(38\°)}{-9.8m/s^{2}}$ (18)

$t_{2}=1.771 s$ (19)

Since snowball 1 was thrown before snowball 2, we have:

$t_{1}-t=t_{2}$ (20)

Finding the time difference $t$ between both:

$t=t_{1}-t_{2}$ (21)

$t=2.267 s - 1.771 s$

Finally:

$t=0.495 s$

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4 years ago

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ICE Princess25 [194]

Answer:

You can make a horizontale transverse by moving a slinky vertically up and down!

Explanation:

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4 years ago

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The speed of light in air is 3.0 x 10^8 m/s. The speed of light in particular glass is 2.3 x 10^8 m/s. Use the information to de

olga_2 [115]

Answer:

33.61°

Explanation:

Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.

Or ,

n = c/v.

v is the velocity in the medium (2.3 × 10⁸ m/s)

c is the speed of light in air = 3.0 × 10⁸ m/s

So,

n = 3.0 × 10⁸ / 2.3 × 10⁸

n = 1.31

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence ( 25.0° )

${\theta_r}$ is the angle of refraction ( ? )

${n_r}$ is the refractive index of the refraction medium (air, n=1)

${n_i}$ is the refractive index of the incidence medium (glass, n=1.31)

Hence,

$1.31\times {sin25.0^0}={1}\times{sin\theta_r}$

Angle of refraction = $sin^{-1}0.5536$ = 33.61°

5 0

3 years ago

A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the lad

pantera1 [17]

Answer: 45.3°

Explanation:

Given,

Length of ladder = l

Weight of ladder = w

Coefficient of friction = μs = 0.495

Smallest angle the ladder makes = θ

If we assume the forces in the vertical direction to be N1, and the forces in the horizontal direction to be N2, then,

N1 = mg and

N2 = μmg

Moment at a point A in the clockwise direction is

N2 Lsinθ - mg.(L/2).cosθ = 0

μmgLsinθ - mg.(L/2).cosθ = 0

μmgLsinθ = mg.(L/2).cosθ

μsinθ = cosθ/2

sin θ / cos θ = 1 / 2μ

Tan θ = 1 / 2μ

Substituting the value of μ = 0.495, we have

Tan θ = 1 / 2 * 0.495

Tan θ = 1 / 0.99

Tan θ = 1.01

θ = tan^-1(1.01)

θ = 45.3°

3 0

3 years ago