Given data:
* The initial velocity of the arrow is 90 m/s.
* The time taken by the arrow is 6 seconds.
* The acceleration of the arrow while moving in the upward direction is,
[tex]g=-9.8ms^{-2}[/tex]Solution:
By the kinematics equation, the final velocity of the arrow in terms of the initial velocity, acceleration, and time is,
[tex]v-u=gt[/tex]where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity and t is the time taken,
Substituting the known values,
[tex]\begin{gathered} v-90=-9.8\times6 \\ v-90=-58.8 \\ v=-58.8+90 \\ v=31.2\text{ m/s} \end{gathered}[/tex]Thus, the final velocity of the arrow is 31.2 m/s.
Hence, option A is the correct answer.
A train departs from its station at a constant acceleration of 5 m/s2. What is the speed of the train at the end of 20s?
Variables we're given:
a: acceleration; a = 5 m/s^2
vi: initial velocity; vi = 0 m/s
t: time; t = 20s
Variable we need to solve for:
vf: final velocity
Given these values, we can find the right equation to use.
vf = vi + at
vf = 0 + 5*20
vf = 5*20 = 100
vf = 100 m/s
Forces of 3.0 N and 4.0 N act at right angles on a block of mass 2.0 kg. What is the acceleration of the block?
Using Newton's Second Law:
[tex]F_{net}=ma[/tex]But first, we need to find the net force, We can do this as follows:
Now, using Pythagorean theorem:
[tex]\begin{gathered} F_{net}=\sqrt{3^2+4^2} \\ F_{net}=\sqrt{9+16} \\ F_{net}=\sqrt{25} \\ F_{net}=5N \end{gathered}[/tex]Now, solving for a from Newton's Second Law:
[tex]\begin{gathered} a=\frac{F_{net}}{m} \\ a=\frac{5}{2} \\ a=2.5m/s^2 \end{gathered}[/tex]Answer:
2.5 m/s²
A 1800 kg car is parked at the top of a hill 4.7 m high.a. What is the gravitational potential energy of the car?b. The car's brakes fail and it rolls down the hill. At the bottom of the hill, what isthe potential energy of the car?c. At the bottom of the hill, what is the kinetic energy of the car?d. What is the velocity of the car when it reaches the bottom of the hill?
mass = m = 1800 kg
height = h = 4.7 m (top of the hill)
a) gravitational potential energy = m x g x h
PE = 1800 x 9.8m/s^2 x 4.7m = 82,908 J
b) bottom of the hill height = 0
PE = 1800 x 9.8 x 0 = 0 J
c) Kinetic energy = 1/2 x m x v^2
At the bottom of the hill PE converts into KE
82,908 J
d) Kinetic energy = 1/2 x m x v^2
82,908 =1/2 m v^2
82,908= 1/2 * 1800* v^2
82,908/ (1/2*1800) = v^2
√[82,908/ (1/2*1800)] = v
v = 9.6 m/s
True or False: There are exceptions to Newton's 1st and 3rd Laws.
The given statement is a false statement.
No, there are no exceptions to Newton's first and third laws.
Two speakers are against a wall emitting a sound at 647 Hz, 5 m apart, and facing each other. You are standing against the wall in between the speakers. Use 344 m/s for the speed of sound in air.A) What is the closest distance that you can be from the speaker on the left at which you experience total destructive interference, and the sound from the speakers cancel out? Round your answer to 2 decimal places.B) What is the closest distance that you can be from the speaker on the left at which you experience total constructive interference, and the sound intensity is doubled? Round your answer to 2 decimal places.
Given Data:
*The frequency of the sound is:
[tex]\begin{gathered} f=647\text{ Hz} \\ =647\text{ s}^{-1} \end{gathered}[/tex]*The speed of sound in air is:
[tex]v=344\text{ m/s}[/tex]*The distance between the speakers is:
[tex]d=5\text{ m}[/tex]Explanation:
The wavelength of the sound is given by:
[tex]\begin{gathered} \lambda=\frac{v}{f} \\ =\frac{344\text{ m/s}}{647\text{ s}^{-1}} \\ =0.532\text{ m} \end{gathered}[/tex]Let us consider that we are standing at point P in the figure below.
The path difference will be:
[tex]\begin{gathered} \Delta x=(5-x)-x \\ =5-2x \end{gathered}[/tex]A)
For destructive interference:
[tex]\begin{gathered} \Delta x=\frac{\lambda}{2} \\ 5-2x=\frac{0.532\text{ m}}{2} \\ 5-2x=0.266\text{ m} \\ 5-0.266\text{ m = }2x \\ \frac{4.734m}{2}=x \\ x=2.37\text{ m} \end{gathered}[/tex]B)
When the intensity doubles, the frequency also doubles. So, the wavelength will be:
[tex]\begin{gathered} \lambda^{\prime}=\frac{v}{2f} \\ =\frac{\lambda}{2} \end{gathered}[/tex]For constructive interference, path difference is:
[tex]\begin{gathered} \Delta x=\frac{\lambda^{\prime}}{2} \\ 5-2x=\frac{\lambda}{4} \\ 5-2x=\frac{0.532\text{ m}}{4} \\ 5-2x=0.133\text{ m} \\ 5-0.133\text{ m}=2x \\ \frac{4.867m}{2}=x \\ x=2.43\text{ m} \end{gathered}[/tex]Final Answer:
A) The closest distance will be:
[tex]2.37\text{ m}[/tex]B) The closest distance will be:
[tex]2.43\text{ m}[/tex]If the electric field intensity between two plates is initially E₁. What would the finalelectric field intensity between parallel plates be if we double the distancebetween the plates, and half charges on the plates, and triple the area of theplatesO a. 3E₁O b. 12 E₁OC. E₁/3O d. E₁/6
Given:
The electric field intensity between the parallel plates is,
[tex]E_1[/tex]The distance between the plates is doubled and the charges are halved.
The area of the plates is tripled.
To find:
the new electric field
Explanation:
The electric field intensity between the parallel plates is,
[tex]E_1=\frac{Q}{A\epsilon}[/tex]Q is the charge on each plate and A is the area of each plate.
Now the field intensity is,
[tex]\begin{gathered} E_2=\frac{\frac{Q}{2}}{3A\times\epsilon} \\ =\frac{Q}{6A\epsilon} \\ =\frac{E_1}{6} \end{gathered}[/tex]Hence, the new electric field intensity is d.
[tex][/tex]What is the kinetic energy of a 2.4 kg brick travelling at a speed of 12 m/s in Joules?
We are asked to determine the kinetic energy of an object. To do that we will use the following formula:
[tex]k=\frac{1}{2}mv^2[/tex]Where "m" is the mass and "v" the speed of the object. Replacing the known values we get:
[tex]k=\frac{1}{2}(2.4\operatorname{kg})(12\frac{m}{s})^2[/tex]Solving the operations we get:
[tex]k=172.8J[/tex]Therefore, the kinetic energy is 172.8 Joules.
3. Law of Acceleration: A 1400 kg car has an acceleration of 3.6 m/s^2. What is the net force applied by the engine?
ANSWER:
5040 N
STEP-BY-STEP EXPLANATION:
We have that the force can be calculated as follows:
[tex]F=m\cdot a[/tex]We know the mass and the acceleration, so we plug in and calculate the force, just like this:
[tex]\begin{gathered} m=1400\text{ kg; a = 3.6}\frac{m}{s^2} \\ F=1400\cdot3.6 \\ F=5040\text{ N} \end{gathered}[/tex]The photo shows a sea urchin embryo. A sea urchin is a multicellular organism.
The process that happens during the growth of sea urchin embryo are
The cells take in nutrients and water (b)The cells of the embryo get smaller in size (c)When the sea urchin embryo grows, increase in size can be seen for a certain period of time. For the growth of the embryo, it requires water and nutrients. Thus, embryonic cell would take in nutrients and water.
After reaching a certain size, the embryo of the sea urchin would start to divide. As the division of embryo progresses, the cells of the embryo would get smaller in size as the same cell would divide itself into many small cells.
During the process, no new cells would be added and amount of DNA in a cell remains the same.
To know more about Embryo
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which of the following represents the electron configuration for iron? please see see the image for the choices!
Electronic configuration:
In chemistry, the electronic configuration of a chemical element indicates the distribution of all available electrons of the element in the different shells or blocks (s, p, d, etc). For filling the shell or block, first of all, fill s block, then p, d, and so on.
The electronic configuration for iron is
[tex]1s^22s^22p^63s^23p^64s^23d^6[/tex]Hence, the correct answer is (b)
consider the vectors A = (1.2m)x and B = (-3.4m)×. which of these vectors has (a) the greater x component and (b) the greater magnitude
We are given the following two vectors
[tex]\begin{gathered} \vec{A}=(1.2\; m)\hat{x}\; \\ \vec{B}=(-3.4\; m)\hat{x}\; \end{gathered}[/tex](a) Greater x component
Vector A (1.2 m) has a greater x component as compared to vector B (-3.4 m) because vector A is positive and vector B is negative.
Vector A has a greater x component
(b) Greater magnitude
Vector B (3.4 m) has a greater magnitude as compared to vector A (1.2 m) because the absolute value of vector B is greater than the absolute value of vector A.
[tex]\begin{gathered} |\vec{A}|=|1.2|=1.2\; m \\ |\vec{B}|=|-3.4|=3.4\; m \end{gathered}[/tex]Vector B has a greater magnitude.
A golf ball with a mass of 0.08 kg falls to the ground with a velocity of 8.2 m/s. It bounces upwards with a velocity of 5.9 m/s. If the total time of impact was 0.015 s, what was the average force the golfball was under during the impact?
Given data:
* The mass of the golf ball is m = 0.08 kg.
* The initial velocity of the golf ball is u = 8.2 m/s.
* The final velocity of the golf ball is v = 5.9 m/s.
* The time of impact is t = 0.015 s.
Solution:
According to Newton's second law, the average force on the golf ball in terms of the change in momentum with time is,
[tex]F=\frac{m(v-u)}{t}[/tex]Substituting the known values,
[tex]\begin{gathered} F=\frac{0.08(5.9-8.2)}{0.015} \\ F=\frac{-0.08\times2.3}{0.015} \\ F=12.27\text{ N} \end{gathered}[/tex]Hencem
40. When a spring is stretched 0.200 meter from its equilibrium position, it possesses a potential energy of 10.0 joules. What is the spring constant for this spring?
Answer:
500 N/m
Explanation:
The potential energy of a spring is equal to
[tex]PE=\frac{1}{2}kx^2[/tex]Where k is the spring constant and x is the distance stretched.
Solving the equation for k, we get
[tex]\begin{gathered} 2PE=kx^2 \\ k=\frac{2PE}{x^2} \end{gathered}[/tex]Now, we can replace PE = 10.0 J and x = 0.200 m, so
[tex]k=\frac{2(10.0J)}{(0.2\text{ m)}^2}=\frac{20.0J}{0.04m^2}=500N/m[/tex]Therefore, the spring constant is 500 N/m
I need the answered for #4 & #5 I need help so bad on these two please lord Jesus
(4)
Converting 4.5 fm to m,
[tex]\begin{gathered} 4.5\text{ fm}\times(\frac{10^{-15}\text{ m}}{1\text{ fm}})=4.5\times10^{-15}\text{ m} \\ \end{gathered}[/tex]Converting m to nm;
[tex](4.5\times10^{-15}\text{ m})\times(\frac{10^9\text{ nm}}{1\text{ m}})=4.5\times10^{-6}\text{ nm}[/tex]Therefore, 4.5 fm equals to 4.5×10^(-6) nm.
(5)
Converting 450 miles/hr to feet/s;
[tex]\begin{gathered} 450\text{ miles/hr}=450\text{ miles/hr}\times(\frac{5280\text{ feet}}{1\text{ mile}})\times(\frac{1\text{ hr}}{60\text{ min}})\times(\frac{1\text{ min}}{60\text{ sec}}) \\ =660\text{ feet/sec} \end{gathered}[/tex]Therefore, 450 miles/hr equals to 660 feet/sec.
A net force of 70 N accélérâtes a 3-kg mass over a distance of 12 m A. What is the work done by the net force.B. What is the increase in kinetic energy of the mass?
ANSWER:
A. 840 J
B. 840 J
STEP-BY-STEP EXPLANATION:
Given:
Force (F) = 70 N
Mass (m) = 3 kg
Distance (d) = 12 m
A.
The work is given by the following equation:
[tex]\begin{gathered} W=F\cdot d \\ \\ W=70\cdot12 \\ \\ W=840\text{ J} \end{gathered}[/tex]Work is equal to 840 joules
B.
Initial speed = 0
Final speed = v
We calculate the acceleration just like this:
[tex]\begin{gathered} F=m\cdot a \\ \\ a=\frac{F}{m} \\ \\ a=\frac{70}{3}=23.33\text{ m/s}^2 \end{gathered}[/tex]Knowing this we can calculate the speed:
[tex]\begin{gathered} v^2=u^2+2ad \\ \\ v^2=0^2+2\cdot23.33\cdot12 \\ \\ v^2=559.92 \\ \\ v=\sqrt{559.92} \\ \\ v=23.66\text{ m/s} \end{gathered}[/tex]Now we calculate the increase in kinetic energy:
[tex]\begin{gathered} KE=\frac{1}{2}m(v^2-u^2) \\ \\ KE=\frac{1}{2}\cdot3\cdot(23.66^2-0^2) \\ \\ KE=839.69\cong840\text{ J} \end{gathered}[/tex]Therefore, the increase in kinetic energy is 840 joules.
find examples of levers and determine the location of the fulcrums, effort force, load force, effort arm, and load arm for each. For each lever, you will: A. Fake a picture or draw a sketch of the lever.B. State the lever’s purpose.C. Determine which of the five main functions of simple machines the lever fulfills.D. Label the load arm, effort arm, load force, effort force, and fulcrum.E. Determine if it is a 1st class lever, 2nd class lever, or 3rd class lever.
Part A. We will analyze the following lever:
Part B. This type of lever is seen in construction carts and is used to transport heavy materials.
Part C. The main function of this lever is to magnify the force and therefore, reduce the effort of loading and transporting heavyweights.
Part D. The lever has the following parts:
Part E. In this lever, the load is between the effort and the fulcrum therefore, this is a class two lever.
Can someone help me draw a sketch of what the notice
Let's sketch all the elements shown in the image:
Can you explain to me how to do MCQ number 62?
So we have two opposing forces all things considered. We have the force of gravity pulling the box down and the force of friction trying to resist the force of gravity.
The force of gravity is mg
but because you are trying to find acceleration in the x component, we use cos (theta) to get the horizontal component
so the force of gravity on the box is mg cos theta
Now on friction
The force of friction is (mu) * normal force
Normal force is mg sin theta because you are trying to get the vertical component of gravity for friction
so mg (mu) sin theta is the downward force
combining what we have
Net Force = mg cos theta - mg (mu) sin theta
Net force = ma, and by dividing out mass,
we get
a = g cos theta - g (mu) sin theta, which simplifies down to a = g (cos theta - mu sin theta)
I need help with this practice problems from my ACT prep guide online
Let's select the simplest way to in
What must be the current in a resistor if the drop in potential across it is 0.750V and it uses 2.00W of power?1.25A1.50A0.375A2.67A
Given data:
The voltage drop across the resistor is V=0.750 V.
The power used by resistor is P=2.00 W.
The expression for the power in terms of voltage and current,
[tex]P=VI[/tex]Substitute the given values in above equation,
[tex]\begin{gathered} 2=(0.75)I \\ I=2.67\text{ A} \end{gathered}[/tex]Thus, the current in the resistor is 2.67 A.
Final Answer: 2.67 A.
An object of mass 17.0 kg subjected to a non-zero net force moves with an acceleration of 2.2 m/s^2.(a) Determine the net force (in N) acting on it. (Enter the magnitude only.)N(b) What acceleration (in m/s2) would a 34.0-kg object have if the same net force is applied to it?m/s?
m= 17 kg
a = 2.2 m/s^2
F = m x a
F= 17 kg x 2.2 m/s^2 = 37.4 N (a)
(b)
m= 34 kg
F= 37.4 N
F= m x a
a = F/ m
a = 37.4 N / 34kg = 1.1 m/s^2 (b)
Solve 1.1. A 1C charge is originally a distance of 1m from a 0.2C charge, but is moved to a distance of 0.1 m. What is the change in electric potential energy?1. 0 J2. -9.0x10^9 J3. 1.6x10^10 J4. 9.0x10^9 J2. A 1C charge is a distance of 1m from a 0.2C charge. What is the electric potential at the position of the 0.2C charge?1. 9.0x10^9 V2. 0 V3. 4.5x10^10 V4. 1.8x10^9 V
Answer:
3. 1.6 x 10^10 J
Explanation:
The electric potential energy can be calculated as
[tex]\begin{gathered} U=k\frac{q_1q_2}{r} \\ \\ Where\text{ k = 9 }\times10^9\text{ N m}^2\text{ /C}^2 \\ q_1\text{ : Charge 1} \\ q_2\text{ : Charge 2} \\ r:\text{ distance} \end{gathered}[/tex]So, we can calculate the change in electric potential energy as
[tex]\begin{gathered} \Delta U=U_2-U_1 \\ \\ \Delta U=(9\times10^9\text{ N m}^2\text{ /C}^2)\frac{(1\text{ C\rparen\lparen0.2 C\rparen}}{(0.1\text{ m\rparen}}-(9\times10^9\text{ N m}^2\text{ /C}^2)\frac{(1\text{ C\rparen\lparen0.2 C\rparen}}{1\text{ m}} \\ \\ \Delta U=1.8\times10^{10}J-0.18\times10^{10}J \\ \Delta U=1.62\times10^{10}J \end{gathered}[/tex]Therefore, the answer is
3. 1.6 x 10^10 J
Show one side can be simplified so it is identical to the other side. Is this problem correct? sec x cot x = csc x truesec (x) cot (x)= 1/sin(x)1/sin(x) = csc (x)= csc (x) true
We are given the following expression:
[tex]\sec x\cot x[/tex]In order to simplify this expression we will use the following relationship:
[tex]\sec x=\frac{1}{\cos x}[/tex]Replacing we get:
[tex]\sec x\cot x=(\frac{1}{\cos x})(\cot x)[/tex]Now we will use the following relationship:
[tex]\cot x=\frac{\cos x}{\sin x}[/tex]Replacing we get:
[tex](\frac{1}{\cos x})(\cot x)=(\frac{1}{\cos x})(\frac{\cos x}{\sin x})[/tex]Now we cancel out the cosines of x:
[tex](\frac{1}{\cos x})(\frac{\cos x}{\sin x})=\frac{1}{\sin x}[/tex]And this expression is equivalent to cosecant of "x", therefore:
[tex]\sec x\cot x=\csc x[/tex]A rock is thrown off of a 150 foot diff with an upward velocity of 40 fus. As a result its height after t seconds is given by the formulah(t) = 150 + 40t - 5t^2a) What is its height after 4 seconds?b) What is its velocity after 4 seconds?(Positive velocity means it is on the way up, negative velocity means it is on the way down.)
Given the formula:
[tex]h(t)=150+40t-5t^2[/tex]Let's solve for the following:
• (a). What is its height after 4 seconds?
To find the height after 4 seconds, substitute 4 for t and solve for h(4):
[tex]\begin{gathered} h(4)=150+40(4)-5(4)^2 \\ \\ h(4)=150+160-80 \\ \\ h(4)=230 \end{gathered}[/tex]The height after 4 seconds is 230 feet.
• (b). What is its velocity after 4 seconds?
To find the velocity, we have:
[tex]v(t)=\frac{dh}{dt}=\frac{d}{dt}(150+40t-5t^2)[/tex]Now find the derivative:
[tex]v(t)=40-10t[/tex]Substitute 4 for t and solve for v(4):
[tex]\begin{gathered} v(4)=40-10(4) \\ \\ v(4)=40-40 \\ \\ v(4)=0\text{ m/s} \end{gathered}[/tex]Therefore, after 4 seconds, the velocity is 0 m/s.
ANSWER:
a) 230 ft
b) 0
Is it possible to have an average velocity of O for some motion but an average speed of 120 km/h for that motion? Provide an quantitative example.
ANSWER:
It is possible as long as the starting point is the same as the end point.
STEP-BY-STEP EXPLANATION:
Yes, it can be possible.
If displacement is zero,, i.e. initial and final point of motion is some then, average velocity will be zero, but average speed will not be zero.
Consider the following move:
Return back to A is some time interval further so:
[tex]\begin{gathered} V_{\text{ave}}=\frac{0}{2+2}=0\text{ m/s} \\ S_{\text{ave}}=\frac{240+240}{2+2}=\frac{480}{4}=120\text{ m/s} \end{gathered}[/tex]When riding a bike, the energy transfers from the rider's body or legs to the feet, to the pedals.TrueFalse
The mechanical energy is contributed through feet.
Thus, the statement is true.
A woman carries a 12 kg box holding it in both armsa. How much force must be exerted on each of her arms to support the box?b. How will this force change if she holds the box with only one arm?
a)58.8 N
b)117.6 N
Explanation
Weight is a quantity representing the force exerted on a particle or object by an acceleration field , on earth we have weigth due to acceleratioi of gravity ,its value is
[tex]g=\text{ 9.8}\frac{m}{s^2}[/tex]caused by the presence of a massive second object,the earth
so, the weight is given as
[tex]\begin{gathered} weigth=mg \\ where\text{ m is the mass} \end{gathered}[/tex]Step 1
given
[tex]mass=\text{ 12 kg}[/tex]a) if she holds the box holding in both arms
the sum of force must equal zero, hence
[tex]\begin{gathered} arm1+arm2-weigth=0 \\ arm1+amr2=weigth \end{gathered}[/tex]so, the sum of the forces on both arms must equal tehh weigth of the box,
let's find the weigth of the box
[tex]\begin{gathered} weigth=\text{ mg} \\ w=12\text{ kg*9.8}\frac{m}{s^2} \\ w=117.6\text{ Newtons} \end{gathered}[/tex]so
[tex]\begin{gathered} arm1+amr2=117.6\text{ N} \\ arm1=\text{ arm2, so} \\ 2(arm1)=117.6 \\ arm1=\frac{117.6N}{2} \\ arm=58.8\text{ N} \end{gathered}[/tex]therefore, she must exerts a force of 58.8 Newtons on each arm
a)58.8 N
Step 2
now, if she holds the box with only one arm:
in this case, the weight of the box must be supported by only one arm, so
the force must be equal to the weigth of the box
[tex]\begin{gathered} arm\text{ force= weigth} \\ arm\text{ force=mg} \\ armforce=12\text{ kg*9.8 }\frac{m}{s^2} \\ arm\text{ force= 117.6 N} \end{gathered}[/tex]so
b)117.6 N
I hope this helps you
Find the mass of the meter stick (100 cm). For this problem you do not have to convert the grams to kilograms ornewtons.Remember:Only respond with a numerical answer. No units are required.
The center of the stick from the left edge of the stick is at,
[tex]\begin{gathered} d=\frac{100}{2} \\ d=50\text{ cm} \end{gathered}[/tex]The mass 125 g is at the distance of 25 cm from the hanging.
Thus, the distance of center of the stick from the hanging is,
[tex]\begin{gathered} d^{\prime}=50-25 \\ d^{\prime}=25\text{ cm} \end{gathered}[/tex]As the stick is balanced,
Thus, the net moment due to the mass of the stick and given 125 g mass is zero.
[tex]125\times g\times25-m\times g\times25=0[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} 125\times9.8\times25-m\times9.8\times25=0 \\ (125-m)\times9.8\times25=0 \\ (125-m)=0 \\ m=125\text{ g} \end{gathered}[/tex]Thus, the mass of the stick is 125 grams.
In the terms of kilogram, the value of the mass is,
[tex]\begin{gathered} m=125\times10^{-3}\text{ kg} \\ m=0.125\text{ kg} \end{gathered}[/tex]The value of the mass in newton is,
[tex]\begin{gathered} W=m\times g \\ W=0.125\times9.8 \\ W=1.225\text{ N} \end{gathered}[/tex]Thus, the value of the weight of the stick is 1.225 N.
How much force is required to achieve an acceleration of 55 m/s2 if the mass of an object is 125 kg?
ANSWER:
6875 N
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 125 kg
Acceleration (a) = 55 m/s^2
We can calculate the value of the force by means of the following formula:
[tex]\begin{gathered} F=m\cdot a \\ \\ \text{ We replacing} \\ \\ F=125\cdot55 \\ \\ F=6875\text{ N} \end{gathered}[/tex]The force is 6875 newtons.
A coin placed at a constant radius R from the center of a rotating turntable experiences a frictional force F when the turntable rotates at a constant rate. If the coin's period of rotation is reduced to 1/5 its original amount, then what frictional force is required to keep the coin moving in the same circle?
The radius of rotation of the coin is R.
The frictional force is F.
The period of rotation further reduces to 1/5 of its initial value.
for the first case, let the period of rotation is,
[tex]\omega[/tex]For not slipping of the coin we can write,
[tex]F=m\omega^2R[/tex]Here, m is the mass of the coin.
for the further case,
[tex]\begin{gathered} F_2=m(\frac{\omega}{5})^2R_{} \\ F_2=\frac{m\omega^2R}{25} \\ F_2=\frac{F}{5} \end{gathered}[/tex]Hence the frictional force is F/5.
