Physics, 05.07.2019 01:30 1230bering

Ablock is at rest on the incline shown in the gure. the coefficients of static and kinetic friction are 0.6 and 0.51, respec- tively the acceleration of gravity is 9.8 m/s 29° what is the frictional force acting on the 46 kg mass?

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Answer from: sarrivera579

The frictional force acting on the block is 236.67 N.

The given parameters;

coefficient of static friction, $\mu_s$ = 0.6coefficient of static friction, $\mu_k$ = 0.51inclination angle plane, θ = 29⁰mass of the block, m = 46 kg

The normal force on the block is calculated as follows;

$F_n = W \times cos(\theta)\\\\F_n = mg \times cos(\theta)\\\\F_n = 46 \times 9.8 \times cos(29)\\\\F_n = 394.45 \ N$

Since the block is at rest, the static frictional force acting on the block is calculated as follows;

$F_s = \mu_s F_n\\\\F_s = 0.6\times 394.45\\\\F_s = 236.67 \ N$

Thus, the frictional force acting on the block is 236.67 N.

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Answer from: tpsavage9755

Normal reaction force on the block while it is at rest on the inclined plane is given as

$F_n = mgcos\theta$

here we know that

m = 46 kg

$\theta = 29^o$

now we will have

F_n = 46*9.8*cos29 = 394.3 N

now the limiting friction or maximum value of static friction on the block will be given as

$F_s = \mu_s * F_n$

F_s = 0.6 * 394.3 = 236.56 N

Above value is the maximum value of force at which block will not slide

Now the weight of the block which is parallel to inclined plane is given as

$F_{||} = mg sin\theta$

here we know that

$F_{||} = 46*9.8 sin29 = 218.55 N$

Now since the weight of the block here is less than the value of limiting friction force and also the block is at rest then the frictional force on the block is static friction and it will just counter balance the weight of the block along the inclined plane.

So here friction force on the given block will be same as its component on weight which is 218.55 N

Answer from: Quest

1.the balanced chemical equation is as below

cu + 2 ag(no)₃ → cu(no₃)₂ +2 ag

2.the much silver produced is 50.9 g

step 1: find the moles of cu

moles = mass÷ molar mass

from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236 moles

step 2: use the mole ratio to determine the moles of ag

from equation above cu: ag is 1: 2

therefore the moles of ag =0.236 moles x 2/1 =0.472 moles

step 3: find the mass of ag

mass = moles x molar mass

from periodic table the molar mass of ag =107.87 g/mol

mass = 0.472 moles x 107.87 g/mol = 50.9 g

3. the much silver collected worth is 1.797 oz

convert 50.9 g to oz

that is 1 gram = 0.0353 oz

50.9 grams =? oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1 gram} =1.797 oz

Answer from: Quest

option number c is the right answer.

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Ablock is at rest on the incline shown in the gure. the coefficients of static and kinetic friction...