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Zolol [24]
4 years ago
8

The double inclined plane supports two blocks A and B, each having a weight of 10 lb. If the coefficient of kinetic friction bet

ween the blocks and the plane is 0.1, determine the acceleration of each block. (The blocks are connected to each other by a string and pulley system so their accelerations are equal, block on the left is on an incline of 60 degrees and the block on the right is on an incline of 30 degrees.)
Physics
1 answer:
Vladimir79 [104]4 years ago
8 0

Answer:

Explanation:

Left block is on surface with higher inclination so it will go down . If T be tension

For motion of block A ,

net force = mgsin60 - (T + mg cos 60 x μ ) , μ is coefficient of friction .

ma =  mgsin60 - T -  mg cos 60 x .1

10a = 277.13  - T - 16

= 261.13 - T

T = 261.13 - 10a

For motion of block B

T - mg sin30 - mgcos30 x μ = ma

T- 160 - 27.71 = 10 a

261.13 - 10a - 160 - 27.71 = 10a

73.42 = 20a

a = 3.67 ft / s²

common acceleration = 3.67 ft / s²

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Answer: .4 m/s^2= acceleration

Explanation:

f = m*a

We can rearrange this equation to solve for acceleration. Therefore,

a=f/m

a= 28N/70kg

a= 0.4 m/s^2

8 0
3 years ago
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3 years ago
easy bio A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 438 N and have a center of
goblinko [34]

Answer:

1.034 m above the floor

Explanation:

The location of center of body for a compound body, when the weights are given is calculated as:

\bar x = \frac{W_1x_1+W_2x_2+W_3x_3+W_4x_4+W_5x_5+.......+Wnx_n}{W_1+W_2W_3W_4W_5+.....+W_n}

where,

\bar x is the center of gravity of the entire body

W = weight of the individual body

x = center of gravity of the individual body

Thus on substituting the values we get,

\bar x = \frac{438\times 1.28+144\times 0.760+87\times 0.250}{438+144+87}

or

\bar x = \frac{691.83}{669}

or

\bar x =1.034m

Hence, <u>the center of gravity of the entire body lies </u><u>1.034 m</u><u> </u><u>above the floor</u>

6 0
3 years ago
Blocks A (mass 3.50 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block
Vedmedyk [2.9K]

Answer:

(a) V (A) =  0.7 m/s,

(b) V (A) =  0.7 m/s,

(c) V (B) =  0.7 m/s

(d) u= - 0.60 m/s

(e) v = 0.75 m/s

Explanation:

Given:

M(A) =3.50 Kg, M(B)=6.50 Kg, V(A) = 2.00 m/s, V(B) = 0 m/s

Sol:

a)  law of conservation of momentum

M(a) x V(A) + M(B) x V(B) = ( M(a) + M(B) ) V      (let V is Common Velocity of Both block)

so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V

after solving V =  0.7 m/s

After the collision the velocities of the both block will be as the the spring is compressed maximum.

V (A) =  0.7 m/s

b)  V(A) =  0.7 m/s ( Part (a) and Part (a) are repeated )

c) as stated above the in the Part (a)

V(B) =  0.7 m/s

d) When the both blocks moved apart after the collision:

Let u=velocity of block A after the collision.

and v = velocity of block B after the collision.

then conservation of momentum

M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u

⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s =  3.50 Kg x u + 6.50 Kg x v

⇒ 2.00 m/s = u + 1.86 v -----eqn (1)  ( dividing both side by 3.50 Kg)

For elastic collision  

the velocity relative approach = velocity relative separation

so 2.00 m/s = v-u  ----- eqn (2)

⇒v = u + 2.00 m/s

putting this value in eqn (1) we get

2.00 m/s = u + 1.86 (v + 2.00 m/s)

u= - 0.60 m/s

e) putting v= 2.00 m/s in eqn (1)

2.00 m/s = - 2.32 m/s + 1.86 v

v = 0.75 m/s

5 0
4 years ago
Liquid water can be separated into hydrogen gas and oxygen gas through electrolysis. 1 mole of hydrogen gas and 0.5 moles of
den301095 [7]

The temperature of the oxygen gas is  243.75 K.

Using ideal gas law to explain the answer, the absolute temperature of the gas will decrease if the number of moles of the gas increases and it will increase if the volume and/or pressure of the gas increases.

The reaction of the given elements;

H_2 \ + \ \frac{1}{2} O_2 \ --->\ \ H_2O

volume of the collected oxygen gas, V = 10 L

pressure of the gas, P = 1 atm

number of moles of the gas, n = 0.5

Using ideal law the temperature of the oxygen gas is calculated as follows;

PV = nRT\\\\T = \frac{PV}{nR} \\\\where;\\R \ is \ the \ ideal \ gas \ constant = 0.08205 \ L.atm/K.mol\\\\T = \frac{1 \times 10 }{0.5 \times 0.08205} \\\\T = 243.75 \ K

Thus, the temperature of the gas is 243.75 K.

Using ideal gas law to explain the answer. The absolute temperature of the oxygen gas is directly proportional to the product of its pressure and volume and inversely proportional to its number of moles. That is the absolute temperature of the gas will decrease if the number of moles of the gas increases and it will increase if the volume and/or pressure increases.

Learn more here: brainly.com/question/16617695

8 0
3 years ago
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