Physics, 09.03.2020 17:02 RDS468

You get a running start, pushing a box of mass 21 kg up to a ramp of grade 0.7 and friction coefficient 0.48. The box has a speed 10.9 m/s when it starts up the ramp, the same time that you let it go. The ramp ends at a platform of height 3.14 m.(a) What is the work done by friction as the box goes up the ramp?(b) What is the speed of the box just before it hits the ground?

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Answer from: kyrajewell2016

a) -442.6 J

b) 8.8 m/s

Explanation:

The work done by the friction force is equal to:

W_f=-F_f d

where

$F_f = \mu_k mg cos \theta$ is the frictional force on the box, where

$\mu_k = 0.48$ is the coefficient of friction

m = 21 kg is the mass of the box

g=9.8 m/s^2 is the acceleration due to gravity

$\theta$ is the angle of the ramp

The negative sign is due to the fact that the frictional force is opposite to the motion of the box

d is the length of the ramp

So the work can be rewritten as

$W_f=-\mu_k mg cos \theta d$

Here we know that the grade of the ramp, which is the ratio between height and horizontal length, is 0.7:

$grade=\frac{h}{L}=0.7$

This means that

$tan \theta=0.7$

So we find the angle:

$\theta=tan^{-1}(0.7)=35.0^{\circ}$

We also know that the height of the ramp is

h = 3.14 m

So we can find the length of the ramp:

$d=\frac{h}{sin \theta}=\frac{3.14}{sin 35^{\circ}}=5.47 m$

Therefore, the work done by friction is

$W_f=-(0.48)(21)(9.8)(cos 35^{\circ})(5.47)=-442.6 J$

The final speed of the box can be found by using the law of conservation of energy. In fact, the initial kinetic energy of the box (at the bottom) + the work done by friction must be equal to the final kinetic energy of the box, as it reaches the ground after leaving the platform.

So we can write:

$KE_i+W_f=KE_f\\\frac{1}{2}mu^2 + W_f = \frac{1}{2}mv^2$

where:

u = 10.9 m/s is the initial speed of the box

m = 21 kg is the mass of the box

W_f=-442.6 J is the work done by friction

is the final speed of the box as it reaches the ground

Solving for v, we find:

$v=\sqrt{u^2+\frac{2W_f}{m}}=\sqrt{10.9^2+\frac{2(-442.6)}{21}}=8.8 m/s$

Answer from: Quest

the grams of oxygen gas that can be produced is 20.6 grams of o₂

calculation

2liclo₃ → 2 licl + 3 o₂

step 1: find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5 +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281 moles

step 2: use the mole ratio to calculate the moles of o2

from equation above, liclo₃: o₂ is 2: 3

therefore the moles of o₂=0.4281 moles x3/2 =0.6422 moles

step 3: find mass of o₂

mass= moles x molar mass

from periodic table the molar mass of o₂ = 16 x2= 32 g/mol

mass=0.6422 moles x 32 g/mol = 20.6 grams of o₂

Answer from: Quest

An atom (ours, not carroll's) consists of a positively charged nucleus, surrounded by one or more negatively charged particles called electrons. the positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral. most of an atom's mass is in its nucleus; the mass of an electron is only 1/1836 the mass of the lightest nucleus, that of hydrogen. although the nucleus is heavy, it is quite small compared with the overall size of an atom. the radius of a typical atom is around 0.1 to 0.25 nanometres (nm), whereas the radius of a nucleus is about 10-6 nm. * if an atom were enlarged to the size of the earth, its nucleus would be only 200 feet in diameter and could easily rest inside a small football stadium.read more on -

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