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Radda [10]
3 years ago
10

A science teacher who was teaching a unit on minerals modeled mineral formation for the class. She heated granular sugar and a v

ery small amount of water in a pan until the sugar melted and boiled. After a few minutes, she poured the mixture out into a pan, and the mixture took the shape of the pan and hardened. As it hardened, the sugar molecules in the mixture formed crystals.
Minerals that might form in the igneous environment the teacher modeled include
A.
magnetite and quartz.
B.
calcite and epidote.
C.
gypsum and feldspar.
D.
halite and feldspar.
Physics
2 answers:
Olenka [21]3 years ago
5 0

Answer:

Magnetite and quartz

Explanation:

Elza [17]3 years ago
3 0

Answer:

It is A

Explanation:

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What is a successful result of science
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I think the answer is discovery.
7 0
2 years ago
A car is parked on a steep incline, making an angle of 37.0° below the horizontal and overlooking the ocean, when its brakes fai
patriot [66]

Answer:

a) The speed of the car when it reaches the edge of the cliff is 19.4 m/s

b) The time it takes the car to reach the edge is 4.79 s

c) The velocity of the car when it lands in the ocean is 31.0 m/s at 60.2º below the horizontal

d) The total time interval the car is in motion is 6.34 s

e) The car lands 24 m from the base of the cliff.

Explanation:

Please, see the figure for a description of the situation.

a) The equation for the position of an accelerated object moving in a straight line is as follows:

x =x0 + v0 * t + 1/2 a * t²

where:

x = position of the car at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the car starts from rest and the origin of the reference system is located where the car starts moving, v0 and x0 = 0. Then, the position of the car will be:

x = 1/2 a * t²

With the data we have, we can calculate the time it takes the car to reach the edge and with that time we can calculate the velocity at that point.

46.5 m = 1/2 * 4.05 m/s² * t²

2* 46.5 m / 4.05 m/s² = t²

<u>t = 4.79 s </u>

The equation for velocity is as follows:

v = v0  + a* t

Where:

v = velocity

v0 =  initial velocity

a = acceleration

t = time

For the car, the velocity will be

v = a * t

at the edge, the velocity will be:

v = 4.05 m/s² * 4.79 s = <u>19.4 m/s</u>

b) The time interval was calculated above, using the equation of  the position:

x = 1/2 a * t²

46.5 m = 1/2 * 4.05 m/s² * t²

2* 46.5 m / 4.05 m/s² = t²

t = 4.79 s

c) When the car falls, the position and velocity of the car are given by the following vectors:

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

v =(v0x, v0y + g * t)

Where:

r = position vector

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity

v = velocity vector

First, let´s calculate the initial vertical and horizontal velocities (v0x and v0y). For this part of the problem let´s place the center of the reference system where the car starts falling.

Seeing the figure, notice that the vectors v0x and v0y form a right triangle with the vector v0. Then, using trigonometry, we can calculate the magnitude of each velocity:

cos -37.0º = v0x / v0

(the angle is negative because it was measured clockwise and is below the horizontal)

(Note that now v0 is the velocity the car has when it reaches the edge. it was calculated in a) and is 19,4 m/s)

v0x = v0 * cos -37.0 = 19.4 m/s * cos -37.0º = 15.5 m/s

sin 37.0º = v0y/v0

v0y = v0 * sin -37.0 = 19.4 m/s * sin -37.0 = - 11. 7 m/s

Now that we have v0y, we can calculate the time it takes the car to land in the ocean, using the y-component of the vector "r final" (see figure):

y = y0 + v0y * t + 1/2 * g * t²

Notice in the figure that the y-component of the vector "r final" is -30 m, then:

-30 m = y0 + v0y * t + 1/2 * g * t²

According to our reference system, y0 = 0:

-30 m = v0y * t + 1/2 g * t²

-30 m = -11.7 m/s * t - 1/2 * 9.8 m/s² * t²

0 = 30 m - 11.7 m/s * t - 4.9 m/s² * t²

Solving this quadratic equation:

<u>t = 1.55 s</u> ( the other value was discarded because it was negative).

Now that we have the time, we can calculate the value of the y-component of the velocity vector when the car lands:

vy = v0y + g * t

vy = - 11. 7 m/s - 9.8 m/s² * 1.55s = -26.9 m/s

The x-component of the velocity vector is constant, then, vx = v0x = 15.5 m/s (calculated above).

The velocity vector when the car lands is:

v = (15.5 m/s, -26.9 m/s)

We have to express it in magnitude and direction, so let´s find the magnitude:

|v| = \sqrt{(15.5 m/s)^{2} + (-26.9 m/s)^{2}} = 31.0m/s

To find the direction, let´s use trigonometry again:

sin α = vy / v

sin α = 26.9 m/s / 31.0 m/s

α = 60.2º

(notice that the angle is measured below the horizontal, then it has to be negative).

Then, the vector velocity expressed in terms of its magnitude and direction is:

vy = v * sin -60.2º

vx = v * cos -60.2º

v = (31.0 m/s cos -60.2º, 31.0 m/s sin -60.2º)

<u>The velocity is 31.0 m/s at 60.2º below the horizontal</u>

d) The total time the car is in motion is the sum of the falling and rolling time. This times where calculated above.

total time = falling time + rolling time

total time = 1,55 s + 4.79 s = <u>6.34 s</u>

e) Using the equation for the position vector, we have to find "r final 1" (see figure):

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

Notice that the y-component is 0 ( figure)

we have already calculated the falling time and the v0x. The initial position x0 is 0. Then.

r final 1 = ( v0x * t, 0)

r final 1 = (15.5 m/s * 1.55 s, 0)

r final 1 = (24.0 m, 0)

<u>The car lands 24 m from the base of the cliff.</u>

PHEW!, it was a very complete problem :)

5 0
2 years ago
An ancient club is found that contains 100 g of pure carbon and has an activity of 6.5 decays per second. Determine its age assu
never [62]

Answer:

The age of living tree is 11104 years.

Explanation:

Given that,

Mass of pure carbon = 100 g

Activity of this carbon is = 6.5 decays per second = 6.5 x60 decays/min =390 decays/m

We need to calculate the decay rate

R=\dfrac{-dN}{dt}=\lambda N=\dfrac{0.693}{t_{\frac{1}{2}}}N....(I)

Where, N = number of radio active atoms

t_{\frac{1}{2}}=half life

We need to calculate the number of radio active atoms

For N_{12_{c}}

N_{12_{c}}=\dfrac{N_{A}}{M}

Where, N_{A} =Avogadro number

N_{12_{c}}=\dfrac{6.02\times10^{23}}{12}

N_{12_{c}}=5.02\times10^{22}\ nuclie/g

For N_{c_{14}}

N_{c_{14}}=1.30\times10^{-12}N_{12_{c}}

N_{c_{14}}=1.30\times10^{-12}\times5.02\times10^{22}

N_{c_{14}}=6.526\times10^{10}\ nuclei/g

Put the value in the equation (I)

R=\dfrac{0.693\times6.526\times10^{10}\times60}{5700\times3.16\times10^{7}}

R=15.0650\ decay/min g

100 g carbon will decay with rate

R=100\times15.0650=1507\ decay/min

We need to calculate the total half lives

(\dfrac{1}{2})^{n}=\dfrac{390}{1507}

2^n=\dfrac{1507}{390}

2^n=3.86

n ln 2=ln 3.86

n=\dfrac{ln 3.86}{ln 2}

n =1.948

We need to calculate the age of living tree

Using formula of age

t=n\times t_{\frac{1}{2}}

t=1.948\times5700

t=11103.6 =11104\ years

Hence, The age of living tree is 11104 years.

5 0
3 years ago
Assume patmos=1.00atm. what is the gas pressure pgas? express your answer in pascals to three significant figures.
hodyreva [135]
<span>Answer: Well, let's start by finding the pressure due to the "extra" height of the mercury. p = 1.36e4 kg/m³ · (0.105m - 0.05m) · 9.8m/s² = 7330 N/m² = 7330 Pa The pressure at B is clearly p_b = p_atmos = p_gas + 7330 Pa The pressure at A is p_a = p_gas = p_atmos - 7330 Pa c) 1 atm = 101 325 Pa Then p_gas = 101325 Pa - 7330 Pa = 93 995 Pa</span>
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3 years ago
Read 2 more answers
Which has an effect on acceleration (speeding up, slowing down, or changing direction)?
7nadin3 [17]

Explanation:

If acceleration points in the same direction as the velocity, the object will be speeding up. The acceleration points in the same direction as the velocity if the car is speeding up, and in the opposite direction if the car is slowing down.

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