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3 years ago

On heating from low temperature at 1 atm of pressure which substance is most likely to sublime rather than melt?

1 answer:

5 0

Since the choices are not given, I can't give an exact answer. Sublimation is the physical change occurring from solid to gas. This is very rare because the molecular arrangement of a solid and gas are opposite to each other. Thus, it depends on the properties of the substance. For example, napthalene could sublime at room temperature.

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A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa

Nataly_w [17]

Answer:

$V_p = 267.258 m/s$

$V_n = 38.375 m/s$

Explanation:

using the law of the conservation of the linear momentum:

$P_i = P_f$

where $P_i$ is the inicial momemtum and $P_f$ is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

$P_i = m(270 m/s)$

$P_f = mV_P + M_nV_n$

where m is the mass of the proton and $V_p$ is the velocity of the proton after the collision, $M_n$ is the mass of the nucleus and $V_n$ is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m $V_p$ + 14m $V_n$

then, m is cancelated and we have:

270 = $V_p$ + $14V_n$

This is a elastic collision, so the kinetic energy K is conservated. Then:

$K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2$

and

Kf = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

then,

$\frac{1}{2}m(270)^2$ = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

here we can cancel the m and get:

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

now, we have two equations and two incognites:

270 = $V_p$ + $14V_n$ (eq. 1)

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

in the second equation, we have:

36450 = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$ (eq. 2)

from this last equation we solve for $V_n$ as:

$V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

and replace in the other equation as:

270 = $V_p +$ 14 $\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

so,

$V_p = -267.258 m/s$

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

$V_n = \frac{270+267.258}{14}$

$V_n = 38.375 m/s$

3 0

3 years ago

How is a revolution related to a year

nikdorinn [45]

Revolution means orbiting around another body.
<span>A year is the time for a planet to complete one orbit around the Sun. 100% sure

</span>

4 0

2 years ago

Read 2 more answers

A chocolate chip cookie is an example of a (2 points) a homogeneous mixture b heterogeneous mixture c suspension d colloid

kumpel [21]

Answer:

I think it is heterogeneous mixture. have a good day

8 0

2 years ago

Read 2 more answers

Which of the following must an engineer consider when designing a car's tires?

Vsevolod [243]

When designing tires for a car, an engineer must consider the materials that will be used to fabricate the tires ... both the suitability of the materials for the purpose, and their cost.

For example, the engineer should not design car tires to be made of glass or of silk, since these would not hold up well when driving on concrete roads, and they would need to be replaced too often.

For another example, he should not design car tires to be made of gold or polished diamond, since these would be beyond the price range of most car owners, and also, these tires would be very susceptible to being stolen by nefarious and dishonest individuals in the middle of the night.

I hope this answer is helpful. The question you posted is a very difficult one, because there are no answer choices listed, and also because the question has the mysterious unknown term "the following" in it.

8 0

2 years ago

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid

nataly862011 [7]

Answer:

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load $m_{LS}$ = 10000 kg

mass of flat bed $m_{FB}$ = 20000 kg

initial speed of truck $v_{0}$ = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs $F_{N}$ -------------let this be equation 1

where $F_{N}$ = normal force = mg

Fs,max = μs mg

ma $_{max}$ = μs mg

divide through by mass

a $_{max}$ = μs g ---------- let this be equation 2

in equation 2, we substitute in our values

a $_{max}$ = 0.5 × 9.8 m/s²

a $_{max}$ = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

$v_{f}$ ² = $v_{0}$ ² + 2aΔx

where $v_{f}$ is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² / 9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

3 0

3 years ago