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VladimirAG [237]
3 years ago
9

In an inelastic collision, which of the following statements is always true?

Physics
1 answer:
xz_007 [3.2K]3 years ago
8 0
Well C is definitely one of the correct answers.
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A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the
sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg

Since the frictional force f_k is equal to f_k=\mu_k N=\mu_k mg, then we have that the acceleration is:

a=\frac{\mu_k mg}{m}=\mu_k g

Now, from the definition of acceleration we get:

a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}

And, as the final velocity is zero because the box gets to a stop, we have:

t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}

Finally, we calculate the displacement:

x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm

This means that the box moves 7.29cm backwards relative to the belt.

4 0
3 years ago
A force of 2N will stretch a rubber band 0.02m. Assuming that Hooke's Law applies, answer the following: How far will a 1600N fo
denis-greek [22]
Hooke's Law states that the extension is directly proportional to the force applied so:
F/x = constant

F₁/x₁ = F₂/x₂
2 / 0.02 = 1600 / x₂
x₂ = 16 m

Elastic work = 1/2 Fx
= 1/2 * 1600 * 16
= 12.8 kJ
7 0
3 years ago
Elemental analysis of the unknown gas from part a revealed that it is 30.45% n and 69.55% o by mass. What is the molecular formu
Anna71 [15]

Answer:

the molecular formula for the gas is NO₂

Explanation:

since it contains

Nitrogen = n → 30.45%

Oxygen = o → 69.55%

and 30.45%+69.55% = 100% , then the gas only contains nitrogen and oxygen

Also we know that the proportion of oxygen over nitrogen  is

proportion of oxygen over nitrogen  = moles of oxygen / moles of nitrogen

since

moles = mass / molecular weight

then for a sample of 100 gr of the unknown gas

mass of oxygen = 69.55%*100 gr = 69.55 gr

mass of Nitrogen = 30.45%*100 gr = 30.45 gr

proportion of oxygen over nitrogen = (mass of oxygen/ molecular weight)/(mass of nitrogen / molecular weight of nitrogen ) =  (69.55 gr/ 16 gr/mol) /( 30.45 gr /14 gr/mol) = 1.998 mol of O/ mol of N≈ 2 mol of O/ mol of N

therefore there are 2 atoms of oxygen per atom of nitrogen

thus the molecular formula for the gas is:

NO₂

6 0
3 years ago
With certain exceptions, Class E airspace extends upward from either 700 feet or 1,200 feet AGL to, but does not include,A) 14,5
DedPeter [7]

Answer:

B) 18,000 feet MSL

Explanation:

There are three-dimensional parts in the navigation airspace in the world. The class E airspace is mostly used in the regions with coastal areas that are relatively populated. If we consider certain forms of exceptions, the class E airspace can move in the upward direction to few feet (i.e. 1200 ft). However, this doesn't include 18,000 feet MSL.

6 0
3 years ago
What is the<br> cancer treatments
ad-work [718]

One is chemo. Chemo is a special magnetic field like to treat cancer


3 0
3 years ago
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