Physics, 30.01.2020 09:53 macp98463oz47dl

The system shown to the right is a standard atwood machine. if the mass of b is 6.25 kg and the mass of a is 1.38 kg, how fast is block b accelerating?

The system shown to the right is a standard atwood machine. if the mass of b is 6.25 kg and the mass

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Answer from: aprilkenedy12

6.3 m/s^2

Explanation:

Let's write the equations of motion for the two blocks:

$m_B g - T = m_B a\\T - m_A g = m_A a$

where

T is the tension in cable

mA = 1.38 kg is the mass of block A

mB = 6.25 kg is the mass of block B

g = 9.8 m/s^2 is the gravitational acceleration

a = ? is the acceleration of the system

Solving the first equation for T we find

T= m_B g - m_B a

And substituting into the second one:

$m_B g - m_B a - m_A g = m_A a\\a=\frac{m_B - m_A}{m_A + m_B}g=\frac{6.25 kg - 1.38 kg}{6.25 kg + 1.38 kg}(9.8 m/s^2)=6.3 m/s^2$

Answer from: akerner

The acceleration 'a' of Block B is 6.26 $ms^{-2}$ .

Step-by-Step Explanation:

The mass of Block B is more than Block A so under the influence of gravitational acceleration 'g' = 9.8 $ms^{-2}$ The system will move such that Block B moves downward and Block A moves upward with an acceleration 'a'.

The net force 'F' that produces this acceleration 'a' is the difference in the weights of the two blocks.

F = ma

F = (mB) g - (mA) g where mA = mass of block A , mB = mass of block B

⇒ ma = (mB) g - (mA) g where m is the total mass of the system

(mA + mB)a = (mB) g - (mA) g

Hence; acceleration 'a' is given as:

a = (mB - mA)g/(mA + mB)

a = (6.25 - 1.38)9.8/(6.25+1.38)

$a = 4.87(9.8)/7.63\\a = 47.73/7.63\\a = 6.26 ms^{-2}$

Answer from: Quest

Amethod of procedure that has characterized natural science since the 17th century, consisting in systematic observation, measurement, and experiment, and the formulation, testing, and modification of hypotheses.

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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