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2 years ago

-2 5/8 is bigger than -5/12

Physics

1 answer:

Neko [114]2 years ago

8 0

Answer:

hey...................................

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A rock is dropped off a cliff and falls the first half of the distance to the ground in 2.0 seconds. how long will it take to fa

son4ous [18]

Answer:

The answer is 0.83 seconds.

Explanation:

The formula of free fall is following:

$h=1/2*g*t^2$

Where g=9.8 m/s^2 and t=2 seconds, the rock takes:

$h=1/2*9.8*2^2=19.6$

19.6 meters. This is the half distance of the cliff. The whole distance is 39.2 meters. So it takes:

$39.2=1/2*9.8*t^2\\t^2=8\\t=2.83$

2.83 second to fall down completely. The rock takes the second half of the cliff in 0.83 seconds

8 0

3 years ago

A slinky forms it’s third harmonic standing wave when the input frequency is 24 Hz. What is the fundamental frequency of the sli

jarptica [38.1K]

Answer: The fundamental frequency of the slinky = 8Hz

An input frequency of 28 Hz will not create a standing wave

Explanation:

Let Fo = fundamental frequency

At third harmonic,

F = 3Fo

If F = 24Hz

24 = 3Fo

Fo = 24/3 = 8Hz

If an input frequency = 28 Hz at 3rd harmonic

Let find the fundamental frequency

28 = 3Fo

Fo = 28/3

Fo = 9.33333Hz

Since Fo isn't a whole number, it can't create a standing wave

6 0

3 years ago

Illustrates an Atwood's machine. Let the masses of blocks A and B be 7.00 kg and 3.00 kg , respectively, the moment of inertia o

Harman [31]

Answer:

A) 1.55

B) 1.55

C) 12.92

D) 34.08

E) 57.82

Explanation:

The free body diagram attached, R is the radius of the wheel

Block B is lighter than block A so block A will move upward while A downward with the same acceleration. Since no snipping will occur, the wheel rotates in clockwise direction.

At the centre of the whee, torque due to B is given by

${\tau _2} = - {T_{\rm{B}}}R$

Similarly, torque due to A is given by

${\tau _1} = {T_{\rm{A}}}R$

The sum of torque at the pivot is given by

$\tau = {\tau _1} + {\tau _2}$

Replacing ${\tau _1}$ and ${\tau _2}$ by ${T_{\rm{A}}}R$ and $- {T_{\rm{B}}}R$ respectively yields

$\begin{array}{c}\\\tau = {T_{\rm{A}}}R - {T_{\rm{B}}}R\\\\ = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R\\\end{array}$

Substituting $I\alpha$ for $\tau$ in the equation $\tau = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R$

$I\alpha=\left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R$

$\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$

The angular acceleration of the wheel is given by $\alpha = \frac{a}{R}$

where a is the linear acceleration

Substituting $\frac{a}{R}$ for $\alpha$ into equation

$\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$ we obtain

$\frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$

Net force on block A is

${F_{\rm{A}}} = {m_{\rm{A}}}g - {T_{\rm{A}}}$

Net force on block B is

${F_{\rm{B}}} = {T_{\rm{B}}} - {m_{\rm{B}}}g$

Where g is acceleration due to gravity

Substituting ${m_{\rm{B}}}a$ and ${m_{\rm{A}}}a$ for ${F_{\rm{B}}}$ and ${F_{\rm{A}}}$ respectively into equation $\frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$ and making a the subject we obtain

$\begin{array}{c}\\{m_{\rm{A}}}g - {m_{\rm{A}}}a - \left( {{m_{\rm{B}}}g + {m_{\rm{B}}}a} \right) = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g - \left( {{m_{\rm{A}}} + {m_{\rm{B}}}} \right)a = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)a = \left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g\\\\a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}\\\end{array}$

Since ${m_{\rm{B}}}$ = 3kg and ${m_{\rm{B}}}$ = 7kg

g=9.81 and R=0.12m, I=0.22 ${\rm{ kg}} \cdot {{\rm{m}}^2}$

Substituting these we obtain

$a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}$

$\begin{array}{c}\\a = \frac{{\left( {7{\rm{ kg}} - 3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {7{\rm{ kg}} + 3{\rm{ kg}} + \frac{{0.22{\rm{ kg/}}{{\rm{m}}^2}}}{{{{\left( {0.120{\rm{ m}}} \right)}^2}}}} \right)}}\\\\ = 1.55235{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Therefore, the linear acceleration of block A is 1.55 ${\rm{ m/}}{{\rm{s}}^2}$

(B)

For block B

${a_{\rm{B}}} = {a_{\rm{A}}}$

Therefore, the acceleration of both blocks A and B are same

1.55 ${\rm{ m/}}{{\rm{s}}^2}$

(C)

The angular acceleration is $\alpha = \frac{a}{R}$

$\begin{array}{c}\\\alpha = \frac{{1.55{\rm{ m/}}{{\rm{s}}^2}}}{{0.120{\rm{ m}}}}\\\\ = 12.92{\rm{ rad/}}{{\rm{s}}^2}\\\end{array}$

(D)

Tension on left side of cord is calculated using

$\begin{array}{c}\\{T_{\rm{B}}} = {m_{\rm{B}}}g + {m_{\rm{B}}}a\\\\ = {m_{\rm{B}}}\left( {g + a} \right)\\\end{array}$

$\begin{array}{c}\\{T_{\rm{B}}} = \left( {3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} + 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 34.08{\rm{ N}}\\\end{array}$

(E)

Tension on right side of cord is calculated using

$\begin{array}{c}\\{T_{\rm{A}}} = {m_{\rm{A}}}g - {m_{\rm{A}}}a\\\\ = {m_{\rm{A}}}\left( {g - a} \right)\\\end{array}$

$\begin{array}{c}\\{T_{\rm{A}}} = \left( {7{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} – 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 57.82{\rm{ N}}\\\end{array}$

6 0

2 years ago

What is the dimensional formula of pressure ??

Igoryamba

Answer:

pressure is equal to the net amount of force acting per unit area. Dimensional Formulae of force is M1L1T-2 and of area is L2. Therefore Pressure's dimension can be obtained by calculating Force by Area. Dimensional formula of pressure difference is M1L-1T-2.

3 0

2 years ago

Determine the ratio of Earth's gravitational force exerted on an 80-kg person when at Earth's surface and when 1400 km above Ear

postnew [5]

m = mass of the person = 80 kg

M = mass of earth = 5.98 x 10²⁴ kg

R = radius of earth = 6.37 x 10⁶ m

h = height above the earth's surface = 1400 km = 1.4 x 10⁶ m

r₁ = initial distance of the person from the center of earth when on surface = R = 6.37 x 10⁶ m

r₂ = final distance of the person from the center of earth when at some height = R + h = 6.37 x 10⁶ + 1.4 x 10⁶ = 7.77 x 10⁶ m

F₁ = Gravitational force of earth on the person when at surface

Gravitational force of earth on the person when at surface is given as

F₁ = G M m/r₁² eq-1

F₂ = Gravitational force of earth on the person when at some height

Gravitational force of earth on the person when at some height is given as

F₂ = G M m/r₂² eq-2

dividing eq-1 by eq-2

F₁ /F₂ = (G M m/r₁² )/(G M m/r₂²)

F₁ /F₂ = r₂²/r₁²

inserting the values

F₁ /F₂ = (7.77 x 10⁶)²/(6.37 x 10⁶)²

F₁ /F₂ = 1.49

3 0

3 years ago