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3 years ago

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‏Let's say that the teacher raised the lawn bowling ball to a height of 1.6 meters before they released it . THE LAWN BOWLING BA

LL HAS A MASS OF 4KG . Remember that " g " ALWAYS equals 9.8 m / s² . Calculate the gravitational potential energy of the lawn bowling ball at this position

1 answer:

WARRIOR [948]3 years ago

3 0

ya ok do ok bro al ya al ya la ya al

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Which uses direct current?

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Answer:

Direct current is used in any electronic device with a battery for a power source. It is also used to charge batteries, so rechargeable devices like laptops and cell phones

Explanation:

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3 years ago

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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

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

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

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3 years ago

A ball traveling at 15 m/s hits a bat with a force of 200N. How much force does the bat (moving at 20m/s)

just olya [345]

Answer:

200 N

Explanation:

Given that,

A ball traveling at 15 m/s hits a bat with a force of 200 N.

We need to find the force that the bat moving at 20 m/s hit the ball with.

We know that, this probelm is based on Newton's third law of motion. The force that the ball exerting on bat should be equal to the force that the bat exerting in the ball but in opposite direction.

It would mean that the ball hits the ball with a force of 200 N. Hence, the correct option is (a).

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4 years ago

Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc

Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

$W= F*d= m*g*d$

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

$W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]$

Converting from Joules to Kcals:

$\frac{10620.75}{4186} =2.537 Kcal$

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

x ---> 100%

$x=\frac{2.537*100}{20} =12.685$

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

$\frac{9000Kcals}{12.685Kcals} =709.5 times$

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

$Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\$

$P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]$

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3 years ago

The gymnast has a mass of 45 kg gravitational field strength = 9.8 N/kg - Calculate the weight of the gymnast. Use the equation:

tia_tia [17]

Answer: B. The gravitational field strength of Planet X is Wx/m.

Explanation:

Weight is a force, and as we know by the second Newton's law:

F = m*a

Force equals mass times acceleration.

Then if the weight is:

Wx, and the mass is m, we have the equation:

Wx = m*a

Where in this case, a is the gravitational field strength.

Then, isolating a in that equation we get:

Wx/m = a

Then the correct option is:

B. The gravitational field strength of Planet X is Wx/m.

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3 years ago