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3 years ago

64 g of sulfur dioxide (SO2) contains 32 g of oxygen. Calculate how much sulfur it contains.

2 answers:

5 0

In sulfur dioxide, there are 2 oxygen atoms and 1 sulfur atom. As there are 32g of sulfur and 32g of oxygen, that would mean that each oxygen atom would weigh about 16g. Given that, the mass of a single sulfur atom is twice that of a single oxygen atom.

vfiekz [6]3 years ago

3 0

Yeah

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A vehicle reaches a speed of 7.5 m/s over 15 seconds. What is its acceleration if it

denpristay [2]

$acceleration = \frac{velocity}{time} = \frac{7.5}{15} = 0.5ms^-^2$

So the answer is option b.

8 0

3 years ago

An infinitely long cylindrical insulating shell of inner radius a and outer radius b has a uniform volume charge density p. Dete

xenn [34]

Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;

c) r>b E=ρ b (b-a)/r*εo

Explanation: In order to solve this problem we have to use the Gaussian law in diffrengios regions.

As we know,

∫E.dr= Qinside/εo

For r<a --->Qinside=0 then E=0

for a<r<b er have

E*2π*r*L= Q inside/εo in this case Qinside= ρ.Vol=ρ*2*π*r*(r-a)*L

E*2π*r*L =ρ*2*π*r* (r-a)*L/εo

E=ρ*(r-a)/εo

Finally for r>b

E*2π*r*L =ρ*2*π*b* (b-a)*L/εo

E=ρ*b* (b-a)*/r*εo

3 0

3 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

3 years ago

I need help with this someone help me?

Brilliant_brown [7]

Graph A so answer B also why isn’t answer a with graph A and B with graph B etc like that’s just confusing lol

3 0

3 years ago

Two persons manage to push a motorcar of mass 1200 kg at uniform velocity along a level road, The same motorcar can be pushed by

densk [106]

Answer:

40N by each person.

Explanation:

8 0

3 years ago