Answer: the child's centripetal acceleration=24.50 m/s²
Explanation:
Given that mass of child= 50 kg
radius of merry go round= 2.25m
angular speed = 3.30 rad/s
Centripetal Acceleration = v²/ r
But V= ωr
So Centripetal Acceleration = v²/ r = (ωr)²/ r
=(3.30)² x (2.25)²/ 2.25 = (3.30)² x 2.25
=24.5025m/s²
=24.50 m/s²
A truck covers 40.0 m in 9.50 s while uniformly slowing down to a final velocity of 2.75 m/s.
a. Find its original speed.
b. Find its acceleration.
Explanation:
Given that,
Distance covered, d = 40 m
Time, t = 9.5 s
Final velocity, v = 2.75 m/s
(a) Let u be the original speed of the truck. We can find it using first equation of motion.
[tex]v=u+at\\\\2.75=u+2.75\times 9.5\\\\2.75-26.125=u\\\\u=-23.375\ m/s[/tex]
(b) Acceleration = rate of change of velocity
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{2.75-(-23.375)}{9.5}\\\\=2.75\ m/s^2[/tex]
So, the original speed is -23.375 and acceleration is 2.75 m/s².
An object has a mass of 19 kg. Attached to the end of a spring, if the spring constant is 486 N/m, what is the maximum frequency?
Answer:
The frequency [tex]f = 0.8048 \ Hz[/tex]
Explanation:
From the question we are told that
The mass of the object is [tex]m = 19 \ kg[/tex]
The spring constant is [tex]k = 486 \ N/ m[/tex]
Generally the maximum frequency is mathematically represented as
[tex]f = \frac{1}{2 \pi } * \sqrt{ \frac{k}{m } }[/tex]
=> [tex]f = \frac{1}{2 * 3.142 } * \sqrt{ \frac{486}{ 19 } }[/tex]
=> [tex]f = 0.8048 \ Hz[/tex]
An object of mass 6 kg. is resting on a horizontal surface. A horizontal force
of 15 N is constantly applied on the object. The object moves a distance of
100m in 10 seconds.
(a) How much work does the applied force do?
(b) What is the kinetic energy of the block after 10 seconds?
(c) What is the magnitude and direction of the frictional force (if there is
any)?
(d) How much energy is lost during motion?
Answer:
a) The work done by the applied force is 1500 joules.
b) The kinetic energy of the block after 10 seconds is 1200 joules.
c) The magnitude of the force of friction is 3 newtons and its direction is against motion.
d) 300 joules of energy are lost during motion.
Explanation:
a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ([tex]W[/tex]), measured in joules, is defined by the following expression:
[tex]W = F\cdot \Delta x[/tex] (1)
Where:
[tex]F[/tex] - Force, measured in newtons.
[tex]\Delta x[/tex] - Distance, measured in meters.
If we know that [tex]F = 15\,N[/tex] and [tex]\Delta x = 100\,m[/tex], then the work done by the force exerted on the object is:
[tex]W = (15\,N)\cdot (100\,m)[/tex]
[tex]W = 1500\,J[/tex]
The work done by the applied force is 1500 joules.
b) At first we need to calculate the net acceleration of the object ([tex]a[/tex]), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:
[tex]\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}[/tex] (2)
Where [tex]v_{o}[/tex] is the initial velocity of the object, measured in meters per second.
We clear the acceleration within the equation above:
[tex]\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t[/tex]
[tex]a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}[/tex]
If we know that [tex]\Delta x = 100\,m[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex] and [tex]t = 10\,s[/tex], then the net acceleration experimented by the object is:
[tex]a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}[/tex]
[tex]a = 2\,\frac{m}{s^{2}}[/tex]
By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:
[tex]\Sigma F = F - f = m\cdot a[/tex] (3)
Where:
[tex]F[/tex] - External force exerted on the object, measured in newtons.
[tex]f[/tex] - Kinetic friction force, measured in newtons.
If we know that [tex]F = 15\,N[/tex], [tex]m = 6\,kg[/tex] and [tex]a = 2\,\frac{m}{s^{2}}[/tex], the kinetic friction force is:
[tex]f = F-m\cdot a[/tex]
[tex]f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)[/tex]
[tex]f = 3\,N[/tex]
The work done by friction ([tex]W'[/tex]), measured in joules, is:
[tex]W' = f\cdot \Delta x[/tex] (4)
[tex]W' = (3\,N) \cdot (100\,m)[/tex]
[tex]W' = 300\,J[/tex]
And the net work experimented by the object is:
[tex]\Delta W = 1500\,J - 300\,J[/tex]
[tex]\Delta W = 1200\,J[/tex]
By the Work-Energy Theorem we understand that change in translational kinetic energy ([tex]\Delta K[/tex]), measured in joules, is equal to the change in net work. That is:
[tex]\Delta K = \Delta W[/tex] (5)
If we know that [tex]\Delta W = 1200\,J[/tex], then the change in translational kinetic energy is:
[tex]\Delta K = 1200\,J[/tex]
The kinetic energy of the block after 10 seconds is 1200 joules.
c) The magnitude of the force of friction is 3 newtons and its direction is against motion.
d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.
A ball is hit with a paddle, causing it to travel straight upward. It takes 2.90 s for the ball to reach its maximum height after being hit. Treat upward as the positive direction.
(a) What is the ball's initial velocity (in m/s), just after it is hit? _____ m/s upward
(b) What height (in m) does it reach above the level where it was hit? _____ m
Answer:
A. 28.42 m/s
B. 41.21 m
Explanation:
From the question given above, the following data were obtained:
Time (t) to reach the maximum height = 2.90 s
Initial velocity (u) =?
Maximum height (h) =?
A. Determination of the initial velocity of the ball.
Time (t) to reach the maximum height = 2.90 s
Final velocity (v) = 0 m/s (at maximum height)
Acceleration due to gravity (g) = 9.8 m/s²
Initial velocity (u) =?
v = u – gt (since the ball is going against gravity)
0 = u – (9.8 × 2.9)
0 = u – 28.42
Collect like terms
0 + 28.42 = u
u = 28.42 m/s
Thus, the initial velocity of the ball is 28.42 m/s
B. Determination of the maximum height reached by the ball.
Final velocity (v) = 0 m/s (at maximum height)
Acceleration due to gravity (g) = 9.8 m/s²
Initial velocity (u) = 28.42 m/s
Maximum height (h) =?
v² = u² – 2gh (since the ball is going against gravity)
0² = 28.42² – (2 × 9.8 × h)
0 = 807.6964 – 19.6h
Collect like terms
0 – 807.6964 = – 19.6h
– 807.6964 = – 19.6h
Divide both side by – 19.6
h = – 807.6964 / – 19.6
h = 41.21 m
Thus, the maximum height reached by the ball is 41.21 m
According to the Law of Conservation of Energy, why does the first hill on a roller coaster always have to be the tallest of all the other hills?
What does a sound wave’s frequency tell you?
a - the number of vibrations per second an object is making
b - the loudness or softness of the sound from an object
c - the amplitude of the sound wave
d - the type of medium in which the sound is traveling
Answer:
The number of vibrations per second an object is making
Explanation:
We need to find the correct option for a sound wave’s frequency.
Frequency is the number of vibrations per second an object is making. It can be calculated as total no. of vibrations divided by time taken.
Its SI unit is Hertz and it is denoted by Hz.
Hence, the correct option is (a).
which one of the following is a correct statement......
1)increasing temperature will decrease a pressure of gas
2)increase in amount of gas will decrease the pressure of gas
3)increasing the volume of gas will decrease the pressure of gas
4)all of the above
Answer:
Just 3
Explanation:
I believe the other two are incorrect
In general, how did the water pressure in the tank change when mass was added to the fluid?
Answer:
As the height increases the pressure must increase.
Explanation:
When we add masses to the fluid, the amount of fluid in the tank increases, therefore its height increases and the pressure is described by the expression
P = ρ g h
where rho is constant for a given fluid and h is the height measured from the surface of the fluid.
As the height increases the pressure must increase.
A 0.55-kg particle has a speed of 5.0 m/s at point A and kinetic energy of 7.6 J at point B.
a. What is its kinetic energy at A?
b. What is its speed at point B?
c. What is the total work done on the particle as it moves from A to B?
Answer:
(a) The kinetic energy at A is 6.875 J.
(b) The speed at point B is 5.26 m/s.
(c) The total work done on the particle as it moves from A to B is 0.725 J.
Explanation:
Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.
Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and at rest, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless a change in speed occurs or the body returns to its state of rest by applying a force.
Kinetic energy is represented by the following formula:
Ec = ½ *m*v²
Where Ec is kinetic energy, which is measured in Joules (J), m is mass measured in kilograms (kg), and v is velocity measured in meters over seconds (m/s).
(a) In this case, you know:
m= 0.55 kgv= 5 m/sReplacing:
Ec = ½ *0.55 kg*(5 m/s)²
and solving you get:
Ec= 6.875 J
The kinetic energy at A is 6.875 J.
(b) In this case, you know:
Ec= 7.6 Jm= 0.55 kgReplacing:
7.6 J = ½ *0.55 kg*v²
and solving you get:
[tex]v=\sqrt{\frac{7.6 J}{\frac{1}{2}*0.55 kg } }[/tex]
v= 5.26 m/s
The speed at point B is 5.26 m/s.
(c) The total work done on the particle as it moves from A to B is the difference between the work done at the end point (point B) and the work done at the start point (point A). Considering that work is equal to kinetic energy, then:
Work= Ec at point B - Ec al point A= 7.6 J - 6.875 J
Work= 0.725 J
The total work done on the particle as it moves from A to B is 0.725 J.
a body is pulled through a distance of 500m by a force of 20N. if the power developed is 0.4KW, calculate the time for which the force act?
Answer:
25000N
Explanation:
distance travelled(d)=500m
force(f)=20N
Power=0.4KW=0.4×1000Watt=400watt
Then,Time(T)=?
Now,
Power(P)=W/T
400=F×D/T.(w=f×d)
or, 400=20×500/T
or, 400=10000/T
or, T=10000/400
or,T=25 sec
Therefore,time taken=25 sec
A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied in the negative Z- direction at point E, and a force R of 75 ibf is applied in the negative Z-direction at point F. Rod AC has a diameter of 1 inch from A to ca diameter of 15 inches from 3 tic, and all fillet radii are 0.25 in. The rod is made of 1020 CD steel, and has a Sy of 135 ksi, a Sur of 150 ksi, and an ef = 0.15. Determine the effective stress at point B
Answer:
The answer is "effective stress at point B is 7382 ksi "
Explanation:
Calculating the value of Compressive Axial Stress:
[tex]\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\[/tex]
Calculating Shear Transverse:
[tex]\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}[/tex]
[tex]= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi[/tex]
[tex]= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in[/tex]
[tex]\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\[/tex]
[tex]= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi[/tex]
Suppose that you take a 10 kg mass on the surface of the earth and then place it on the moon. What will
the mass of the object on the moon be?
Answer:
10 kg
Explanation:
The mass of an object does not change even if the amount of gravtiy changes.
Balboa Park in San Diego has an outdoor organ. When the air temperature increases, how does the fundamental frequency of the one of the organ pipes change?
a. increases
b. decreases
c. stays the same
d. is impossible to determine. (The thermal expansion of the pipe is negligible.)
Answer:
a. increases
Explanation:
Outdoor pipe organ is a musical instrument that is used to play music. It produces some soothing notes and is very peaceful to the ears. The organ pipe is the source of sound for the outdoor organ. And the length of the pipe determines the wavelength of the sound. When the air temperature of the surrounding is increased, the fundamental frequency of one of the organ pipes will be increased as the speed of the sound will increased.
Jack and Rob try to move the new couch. Jack pulls with 30N of force and Rob pushes with 25 N of force.
Net Force = (your answer)
Body Diagram or Explanation = (your answer)
Answer:
Fr = 55 [N]
Explanation:
We must perform a free body diagram to understand the forces acting on the couch. In the attached image we see the force that Jack exerts just like that of Rob acting in the same direction.
We must perform a summation of force on the X-axis, to determine the resulting force.
∑Fx = Fr
Fr = 30 + 25
Fr = 55 [N]
number 8 please i need help asap
Nose bleeding occurs as we move towards high altitude, why?
Answer:
Because as we climb higher the amount of oxygen in the air decreases and this makes the air thinner and dryer. And at higher altitudes the blood pressure inside of our body is more than the atmospheric pressure which forces the blood to come out from openings like the nose.
Explanation:
A weight of 1400 pounds is suspended from two cables as shown in the figure. What is the tension in the left cable? _________ pounds (Round to one decimal place as needed).
Answer:
Following are the solution to this question:
Explanation:
Law:
[tex]\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}[/tex]
[tex]= 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}[/tex]
[tex]\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}[/tex]
[tex]\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\[/tex]
Engineers are developing a process to produce phosphorus pentachloride.
Which two criteria are the most essential when developing this process?
Answer:
The two criteria that are most essential are options C and D
Explanation:
In trying to develop a new process/procedure for the commercial/industrial production of a particular substance (in this case phosphorus pentachloride), it is important that the workers that would operate the equipment required for this new procedure are able to operate it safely, i.e without fear of injury or losing one's life. If this fear is present or any adverse effect occurs, it would not only be counterproductive but would also make workers avoid using this procedure and may be forced to revert to previous safe procedure.
It is also essential for the new procedure to be able to get the most desired product out of the reactant, meaning the rate and amount of desired product (phosphorus pentachloride) that would be obtained from a given amount of reactant must be very high (a better upgrade when compared to the previous procedure). This is going to cause for a better amount of phosphorus pentachloride produced as against the previous amounts produced from previous procedure.
From the above, it can be deduced that options C and D are two criteria that are most essential when trying to develop a novel process to produce phosphorus pentachloride.
Answer: the guy above me is correct
Explanation:
Who’s has bigger cross sectional area capillaries or aorta?
Answer:
Even though the cross-sectional area of each capillary is extremely small compared to that of the large aorta, the total cross-sectional area of all the capillaries added together is about 1,300 times greater than the cross-sectional area of the aorta because there are so many capillaries
Explanation:
High above the town of Whoville there lived a Grinch and his dog Max. The Grinch being antisocial has never gone to school so needs your help with some physics concepts.
a) Start by stating Newton's Three Laws of Motion
b) While watching the Who's skate on Who lake, the Grinch notices that when Sue Who is spinning as she pulls in her arms she spins faster. Explain why this happens to the Grinch.
c) The Grinch notices that as time progresses his cave gets messier. Explain why this occurs.
Answer:
a) Newton proposed three laws to explain the motion of things.
b) bodies are in motion the kinetic moment is preserved,
c) entropy
Explanation:
In this exercise you are asked to explain various concepts of physics
a) Newton proposed three laws to explain the motion of things.
* The first law or inertia states that a body in an inertial system (without acceleration) remains stationary or with constant velocity if the sum of the forces is zero
* The second law states that the force (interaction) is proportional to the masses and the acceleration of the bodies
F = m a
* The third law or action and reaction, says that the forces always go in pairs, that is, if a body interacts with a body 2, and body 2 will also interact with body 1, with the same force but with the opposite direction, each force is applied to a body
b) when bodies are in motion the kinetic moment
L = I w
it is preserved, if it does not interact with the outside.
The quantity I called the moment of inertia is I = m r²
In this case
in the initial instant with outstretched arms
L = I₀ w₀
in the final moment. By shrinking your arms
[tex]L_{f}[/tex] = I_{f} w_{f}
as the body does not interact with another
L₀ = L_{f}
I₀ w₀ = I_{f} w_{f}
w_{f} = [tex]\frac{I_{o} }{I_{f} }[/tex] w₀
we substitute
w_{f} = \frac{r_{o} }{r_{f} }^2 w₀
as the distance of the arms extended is greater than when they are retracted (r₀> [tex]r_{f}[/tex]), the final angle speed increases
c) This is an application of the concept of entropy that establishes that disorder is the most probable state of a system, which is why the cave increases its disorder or entropy over time since by not fixing things, they are increasingly dispersed
Please HElp!!
Name five conductors and insulators
Answer:
Conductors- copper, aluminum, gold, and silver.
Insulators- glass, air, plastic, rubber, and wood.
Explanation:
Answer: 5 conductors are copper, aluminum, gold, iron, and silver and 5 insulators are glass, air, plastic, rubber, and wood.
A baseball player slides into home base and experiences coefficients of friction of μ8 = 0.63 and μk = 0.49. The normal force acting on the player is 660N. How much frictional force acts on the player?Round your answer to the nearest hundredth if necessary. Ff = ________ N
Answer:
323.4 N
Explanation:
Given that,
The coefficient of static friction, [tex]\mu_s=0.63[/tex]
The coefficient of kinetic friction, [tex]\mu_k=0.49[/tex]
We need to find the frictional force acting on the player. As the baseball player is sliding into home base, we will use the coefficient of kinetic friction to find the frictional force.
[tex]F=\mu _kN\\\\=0.49\times 660\\\\=323.4\ N[/tex]
So, the frictional force acting on the player is 323.4 N.
1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Chuck, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, Andrea's tangential speed is which of the following?a) twice Chuck's.b) the same as Chuck's.c) half of Chuck's.d) impossible to determine.2) When the merry go round is rotating at a constant angular speed, Andrea's tangential speed is:______.a) twice Chuck's.b) the same as Chuck's.c) half of Chuck's.d) impossible to determine.
Explanation:
The tangential speed of Andrea is given by :
[tex]v=r\omega[/tex]
Where
r is radius of the circular path
ω is angular speed
The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'
r' = 2r
New angular speed,
[tex]v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v[/tex]
New angular speed is twice that of the Chuck's speed.
A ball is thrown straight up in the air. For which situations are both the instantaneous velocity and the acceleration zero?
a. on the way up
b. at the top of its flight path
c. on the way down
d. halfway up and halfway down
e. none of the above
Answer: e: none of the above.
Explanation:
For any object in the air, the gravitational acceleration will be always -9.8m/s^2, where the negative sign is because gravity pulls the object down.
The instantaneous velocity, as a function of time, for the case of the ball, is
V(t) = (-9.8m/s^2)*t + v0
Where v0 is the velocity at which the ball is thrown up.
The velocity will be zero when the ball is at the top of its flight pat, in that point the sign of the velocity changes, it stops being positive (so the ball stops going up) and becomes negative (so the ball starts to fall down).
Now, while the instantaneous velocity can be zero during the flight, the acceleration does not, it only will be zero when the object hits the ground. Then the only correct option will be e: none of the above.
A proton with an initial speed of 600,000 m/s is brought to rest by an electric field.1. What was the potential difference that stopped the proton?
2. What was the initial kinetic energy of the proton, in electron volts?
Answer:
(1) the potential difference that stopped the proton is 1878.75 V
(2) the initial kinetic energy of the proton is 1878.75 eV
Explanation:
Given;
initial speed of the proton, v = 600,000 m/s
mass of proton, m = 1.67 x 10⁻²⁷ kg
(1) The work done in bringing the proton to rest is given as;
[tex]W = eV[/tex]
Apply work energy theorem;
[tex]K.E =W\\\\ \frac{1}{2} mv^2 = eV\\\\V = \frac{mv^2}{2e}[/tex]
where;
V is the potential difference
[tex]V = \frac{1.67\times 10^{-27} \times\ 600,000^2}{2 \ \times \ 1.6 \times 10^{-19}} \\\\V = 1878.75 \ V[/tex]
(2) the initial kinetic energy of the proton, in electron volts;
[tex]K.E = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times \ 1.67\times 10^{-27} \times 600,000^2 = 3.006 \times 10^{-16} \ J\\\\K.E = \frac{3.006 \times 10^{-16} \ J \ \ eV}{1.6 \times 10^{-19} \ J} = 1878.75 \ eV[/tex]
A girl uses paper-clips to balance a toy bird on her finger. What effect does it have on the centre of mass?
why was bowling one of the first racially integrated sports
Answer:
bowling was one of the first racially integrated sports because bowling alleys were primarily located in urban areas. Racial integration was inevitable since it was promoted by US Armed Forces during 1940s and its image as sport for the common man made it a choice of activity for Americans!!
Explanation:
The maximum value of the electric field in an electromagnetic wave is 2.0 V/m. What is the maximum value of the magnetic field in that wave
Answer:
The maximum value of the magnetic field in the given wave is 6.67 nT.
Explanation:
Given;
maximum value of the electric field, E₀ = 2.0 V/m
The maximum value of the magnetic field in the given wave is calculated as;
[tex]B_o = \frac{E_o}{c}[/tex]
where;
c is the speed of light = 3 x 10⁸ m/s
[tex]B_o = \frac{E_o}{c} \\\\B_o = \frac{2}{3 \ \times \ 10^8} \\\\B_o = 6.67 \ \times \ 10^{-9} \ T\\\\B_o = 6.67 \ nT[/tex]
Therefore, the maximum value of the magnetic field in the given wave is 6.67 nT.
a body of mass 1kg is made to oscillate on a spring of force constant 16 n/m calculate 1 the angular frequency 2 the frequency of oscillation
Explanation:
Given that,
Mass of a body, m = 1 kg
Force constant, k = 16 N/m
We need to find the angular frequency and the frequency of oscillation.
(a) The angular frequency of a body is given by :
[tex]\omega=\sqrt{\dfrac{k}{m}} \\\\=\omega=\sqrt{\dfrac{16}{1}} \\\\=4\ rad/s[/tex]
(b) The frequency of oscillation is given by :
[tex]f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{4}{2\pi}\\\\=\dfrac{2}{\pi}\ Hz[/tex]
Hence, this is the required solution.
The 1kg rock is tied to a string and swung in a circular path as shown. The 1 meter string is tied to a post, and during the motion, the string has a 30 angle with the post. The rock makes 100 rounds in 1 minute. The centripetal force on the rock is
Answer: 54.8
Explanation:
The centripetal force on the rock moving along the string is 104.66 N.
The given parameters;
the mass of the rock, m = 1 kgangle of inclination of the string, θ = 30⁰angular velocity of the rock, ω = 100 rev/minlength of the string, r = 1 mThe centripetal acceleration of the rock is calculated as follows;
[tex]\omega _f^2 = \omega _i^2 + 2\alpha (\theta)\\\\(100 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s} )^2 = 0 + 2\alpha (30 ^0 \times \frac{\pi \ rad}{180^0} )\\\\109.69 = 1.048\alpha \\\\\alpha = \frac{109.69}{1.048} \\\\\alpha =104.66 \ rad/s^2[/tex]
[tex]a_c = \alpha \times r\\\\a_c = 104.66 \times 1 = 104.66 \ m/s^2[/tex]
The centripetal force on the rock is calculated as follows;
[tex]F_c = ma_c[/tex]
[tex]F_c = 104.66 \times 1\\\\F_c = 104.66 \ N[/tex]
Thus, the centripetal force on the rock moving along the string is 104.66 N.
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