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3 years ago

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A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m

eter diameter barrel spins rapidly in a circle, making a revolution every 1.65 seconds. Determine the net force (in N) acting on her. Use g = 9.8 m/s/s.

1 answer:

elena-14-01-66 [18.8K]3 years ago

3 0

Answer:

<u><em>1,230N</em></u>

Explanation:

<u>1. Name of the variables:</u>

$f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration$

<u>2. Formulae:</u>

$f=\dfrac{number\text{ }of\text{ }revolutions}{time}$

$\omega=2\pi f$

$a_c=\omega^2 r$

$F_c=m\times a_c$

<u>3. Solution (calculations)</u>

$f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}$

$\omega=2\pi\times0.\overline{60}\approx 3.808rad/s$

$a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2$

$F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N$

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A skateboarder shoots off a ramp with a velocity of 5.1 m/s, directed at an angle of 55° above the horizontal. The end of the ra

ExtremeBDS [4]

Answer

given,

initial velocity of skateboard = 5.1 m/s

angle above the horizontal = 55°

height of the ramp = 1 m

a) maximum height of projectile

$H = \dfrac{u^2sin^2 \theta}{2g}$

$H = \dfrac{5.1^2\times sin^2 55^0}{2\times 9.81}$

H = 0.889 m

the maximum height of the skateboard above the ground

= 1 + 0.889

= 1.889 m

b) time to reach the height

$t = \dfrac{u\ sin\theta}{g}$

$t = \dfrac{5.1\ sin55^0}{9.8}$

t = 0.426 s

horizontal distance = u cos θ × t

= 5.1 × cos 55° × 0.426

horizontal distance = 1.25 m

7 0

3 years ago

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$

5 0

3 years ago

Two people stand across from one another at the top edges of identical buildings, 50 meters above the ground. One person throws

cricket20 [7]

Answer:

C

Explanation:

Hope this helps!!!!

6 0

3 years ago

How many protons, electrons, and neutrons are present in an atom of cr-24?

mote1985 [20]

The number of protons, electrons, and neutrons in the atom of cr-24 is 28, 24, and 24 respectively.

<h3>What is atom?</h3>

An atom is the smallest unit of matter, consisting of the positively charged nucleus and the electrons which move around it. The atom can not be divided further.

The atom of a matter is made by three elements-

1) Neutron-Neutron is the element of an atom, which has zero charge.
2) Proton-Proton is the element of an atom, which has a positive charge.
3) Electron-Electron is the element of an atom, which has a negative charge.

The atom of cr-24 has the atomic number 24. The atomic number of atom is equal to the number of proton and electron in an atom. Thus,

$\rm Proton=24\\\rm Electron=24$

Now, if this atom has the mass number of 52 (most common). Then the number of neutron is,

$\rm Neutrons=Mass\; Number-Atomic\; Number\\\rm Neutrons=52-24\\\rm Neutrons=28$

Thus, the number of protons, electrons, and neutrons in the atom of cr-24 is 28, 24, and 24 respectively.

Learn more about the atom here;

brainly.com/question/17545314

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2 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

3 years ago