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2 years ago

How many moles of MgCl2 are there in 329 g of the compound? Your Answer:

Physics

2 answers:

Nookie1986 [14]2 years ago

6 0

Answer:3.46

Explanation:

mass=329

molar mass=71

no. of moles=329/71

=3.46

Debora [2.8K]2 years ago

5 0

Answer:3.46

Explanation:

mass=329

molar mass=71

no. of moles=329/71

=3.46

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A balance is best used to measure a sample’s

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This force involves the attraction between objects with mass.

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B.) Gravitational Mass

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A person drives a car around a circular road with a constant speed of 20 m/s. The

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Answer:

16 m/s^2

Explanation:

acceleration tangential = (v^2)/r

a=400/25

a=16 m/s^2

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As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

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2 years ago

Find an expression for the minimum frictional coefficient needed to keep a car with speed v on a banked turn of radius R designe

solniwko [45]

"v0" means that there are no friction forces at that speed
mgsinΘ = (mv0²/r)cosΘ → the variable m cancels 
sinΘ/cosΘ = tanΘ = v0² / gr
Θ = arctan(v0² / gr) 

When v > v0, friction points downslope: 
mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: 
gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ 
µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) 
where Θ is defined above. 

When v > v0, friction points upslope: 
mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: 
gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ 
µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) 
where Θ is defined above.

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2 years ago