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aleksandrvk [35]
2 years ago
13

Block A with a mass of 10 kg rests on a 30 degree incline. the coefficient of kinetic friction is 0.20. theattatched string is p

arallel to the incline and passes over amassless frictionless pulley at the top. block B with a massof 8.0kg is attached to the dangling end of the string. theacceleration of B is:
a. 0.69 up
b. 0.69 down
c. 2.6 up
d. 2.6 down
e. 0

Physics
1 answer:
Tomtit [17]2 years ago
6 0

Answer:

Please find attached

Explanation:

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The temperature, T, of a gas is jointly proportional to the pressure P of the gas and the volume V occupied by the gas. Use C as
AnnZ [28]

Answer:

T=C*P*V

Explanation:

It is said that a variable - let's call 'y' -, is proportional to another - let's call it 'x' - if x and y are multiplicatively connected to a constant 'C'. It means that their product (x*y) can be always equaled to the constant 'C' or their division (\frac{x}{y}) can be always equaled to 'C'. The first case is the case of the inverse proportionality: It is said that x and y are inversely proportional if

x*y=C

The second case is the case of the direct proportionality: It is said that x and y are directly proportional if

\frac{x}{y} =C : x is directly proportional to y.

or

\frac{y}{x} =C : y is directly proportional to x.

Always that any text does not specify about directly or inversely proportionality, it is assumed to mean directly automatically.

For our case, we are said that the temperature T is proportional to the pressure P and the volume V (we assume that it means directly); it is a double proportionality but follows the same rules:

If T were just proportional to P, we would have:

\frac{T}{P} =C

If T were just proportional to V, we would have:

\frac{T}{V} =C

As T is proportional to both P and V, the right equation is:

\frac{T}{P*V}=C

In order to isolate the temperature, let's multiply (P*V) at each side of the equation:

\frac{T}{P*V}*(P*V)=C*(P*V)\\T=C*P*V

3 0
2 years ago
A rubber ball with a mass 0.20 kg is dropped vertically from a height of 1.5 m above the floor. The ball bounces off of the floo
Digiron [165]
Potential Energy = mass * Hight * acceleration of gravity
PE=hmg
PE = 1.5 * .2 * 9.81
PE = 2.943
it lost .6 so 2.943 - .6 = 2.343
now your new energy is 2.343 so solve for height
2.343 = mhg
2.334 = .2 * h * 9.81
h = 1.194
the ball after the bounce only went up 1.194m
8 0
2 years ago
A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi
RSB [31]

Answer:

r = 4.44 m

Explanation:

 

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

         B = ρ g V

Now let's use Newton's equilibrium relationship

         B - W = 0

         B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

         σ = W / A

         W = σ A

The area of ​​a sphere is

           A = 4π r²

       W = W₁ + σ 4π r²

The volume of a sphere is

           V = 4/3 π r³

Let's replace

     ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

     P V = n RT

    P = ρ RT

    ρ = P / RT

 

    P / RT g 4/3 π r³ - σ 4 π r² = W₁

    r² 4π (P/3RT  r - σ) = W₁

Let's replace the values

     r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

     r² (11.81 r -0.060) = 13000 / 4pi

     r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

       r = 4.44 m

3 0
2 years ago
If it were possible to remove gravity and friction, think about what would happen to a football if it were tossed into the air.
elena-14-01-66 [18.8K]
Ignoring fluid resistance, football will <span>maintain a constant speed until other forces accelerate the football.</span>
6 0
2 years ago
Read 2 more answers
A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl
tatuchka [14]

By Newton's second law, assuming <em>F</em> is horizontal,

• the net <u>horizontal</u> force on the <u>larger</u> block is

<em>F</em> - <em>µmg</em> = 3<em>mA</em>

where <em>µmg</em> is the magnitude of friction felt by the larger block due to rubbing with the smaller one, <em>µ</em> is the coefficient of static friction between the two blocks, and <em>A</em> is the block's acceleration;

• the net <u>vertical</u> force on the <u>larger</u> block is

4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0

where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;

• the net <u>horizontal</u> force on the <u>smaller</u> block is

<em>µmg</em> = <em>ma</em>

where <em>µmg</em> is again the friction between the two blocks, but notice that this points in the same direction as <em>F</em>. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net <u>vertical</u> force on the <u>smaller</u> block is

<em>mg</em> - <em>mg</em> = 0

where <em>mg</em> is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for <em>A</em> and <em>a</em> :

<em>A</em> = (<em>F</em> - <em>µmg</em>) / (3<em>m</em>)

<em>a</em> = <em>µg</em>

5 0
1 year ago
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