A 60.0-cm, uniform, 5.00-kg shelf is supported horizontally by two vertical wires attached to a sloping ceiling. The front wire is inset by a distance of 20 cm. A very small 2.50-kg tool is placed on the shelf midway between the points where the wires are attached to it.
Required:
Find the tension in the back wire.
Answers
Image is missing, so i have attached it
Answer:
Tension in back wire = 24.5 N
Explanation:
let the tension in the back wire be Tₒ
let the tension in the front wire be T
Applying static equilibrium in the vertical direction, we have:
T + Tₒ = (5 × 9.8) + (2.5 × 9.8)
T + Tₒ = 73.5
Where Tₒ = (2.5 × 9.8) = 24.5 N
Thus, Tension in back wire = 24.5 N
The tension in the back wire equals 24.5 N.
We can arrive at this answer with the static equilibrium formula, for that, let's consider that the tension in the back wire is represented by the letter "t," while the tension in the front wire is represented by the letter "T."
Therefore, we can use the following formula:
[tex]t= W*g\\t= 2,5*9.8\\t= 24.5 N[/tex]
In this case, the letter "W" refers to the tool weight, while the letter "g" refers to gravity.
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Related Questions
A body on a 20m high cli drops a stone. One second later, he throws
down another stone. Both the stones hit the ground simultaneously. Find the
initial velocity of the second stone g=10m/sec2.
Answers
Taking the downward direction as positive
Stone 1:
Initial velocity (u) = 0 m/s
Acceleration (a) = 10 m/s²
Time taken to reach the ground (t) = t seconds
Distance covered (s) = 20 m
Stone 2:
Initial velocity (u) = u m/s
Acceleration (a) = 10 m/s²
Time taken to reach the ground (t) = (t-1) seconds
[where t is the time taken by stone 1]
Distance covered (s) = 20 m
__________________________________________________________
Time taken by the Stones to reach the ground:Stone 1:
Using the second equation of motion:
s = ut + 1/2*at²
replacing the variables for Stone 1
20 = (0)(t) + 1/2(10)(t)²
20 = 5t²
Dividing both sides by 5
t² = 4
Taking the square root of both the sides
t = 2 seconds
Hence, Stone 1 reaches the ground in 2 seconds
Stone 2:
The time taken by Stone 2 to reach the ground depends on the time taken by stone 1. Since we defined the time taken by stone 2 as:
Time taken by Stone 2 = (Time taken by Stone 1) - 1
replacing the values:
Time taken by Stone 2 = 2 - 1
Time taken by Stone 2 = 1 second
Hence, Stone 2 will reach the ground in 1 second
__________________________________________________________
Initial Velocity of Stone 2:According to the second equation of motion:
s = ut + 1/2 at²
replacing the values for Stone 2
20 = (u)(1) + 1/2(10)(1)²
20 = u + 5
u = 15 m/s
Therefore, the Stone 2 is thrown at a velocity of 15 m/s downwards
A car traveling 13 m/s to the north collides with a car traveling 19 m/s to the
south. Each car has a mass of 1250 kg. If the system is defined as the two
cars, what is the change in momentum of the system due to the collision?
O A. 16,250 kg-m/s
O B. 0 kg•m/s
O C. 40,000 kg•m/s
D. 23,750 kg•m/s
Answers
Answer: B. 0 kg*m/s
Explanation:
In a hydraulic system, a force of 400 N is exerted on a piston with an area of 0.001 m2. The load-bearing piston in the system has an area of 0.2 m^2.
Required:
a. Calculate in kPa the pressure in the hydraulic fluid induced by the applied pressure.
b. What is the magnitude of the force exerted on the load bearing piston by the hydraulic fluid?
Answers
Answer:
Explanation:
Pressure on the hydraulic system is expressed as;
Pressure = Force/Area
Given
Force on the fluid = 400N
Area = 0.001m²
Pressure in the fluid = 400/0.001
Pressure in the fluid = 400,000N/m²
1N/m² = 0.001kPa
400,000N/m² = x
x = 400,000 × 0.001
x = 400kPa
Hence the pressure in kPa is 400kPa
b) Using the formula;
Pa = Pb
Fa/Aa = Fb/Ab
Pa = Fb/Ab
Fb = PaAb
Fb = 400,000(0.2)
Fb = 80,000N
Hence the magnitude of the force exerted on the load bearing piston by the hydraulic fluid is 80,000N
Oil with a density of 890 kg/m3 moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is
P1 = 2.00 ✕ 104 Pa,
and the pipe diameter is 7.00 cm. At another point
y = 0.30 m
higher, the pressure is
P2 = 1.25 ✕ 104 Pa
and the pipe diameter is 3.50 cm.
A tube is open at both its left and right ends. The tube starts at the left end, extends horizontally to the right, curves up and to the right, and extends horizontally to the right again. The right end is higher than its left end, and the change in height is labeled y. The pressure at the left end is labeled P1, and the pressure at the right end is labeled P2.
a- find the speed of the flow in the lower section (m/s)
b- find the speed of the flow in the higher section (m/s)
c- find the volume flow rate in the pipe (m^3/s)
Answers
Answer:
(a) V₁ = 1.06 m/s
(b) V₂ = 4.24 m/s
(c) Q = 4.08 x 10⁻³ m³/s
Explanation:
(b)
The formula derived for Venturi tube can be used here:
P₁ - P₂ = (ρ/2)(V₂² - V₁²)
where,
P₁ - P₂ = Difference in Pressure = (2x 10⁴ Pa) - (1.25 x 10⁴ Pa) = 0.75 x 10⁴ Pa
ρ = Density of Oil = 890 kg/m³
V₂ = Velocity at Higher End = ?
V₁ = Velocity at Lower End = ?
Therefore,
0.75 x 10⁴ Pa = [(890kg/m³)/2](V₂² - V₁²)
V₂² - V₁² = (0.75 x 10⁴ Pa)/(445 kg/m³)
V₂² - V₁² = 16.85 m²/s² ------------------- equation (1)
Now, we will use continuity equation:
A₁V₁ = A₂V₂
where,
A₁ = Lower End Area = πd₁²/4 = π(0.07 m)²/4 = 3.848 x 10⁻³ m²
A₂ = Higher End Area = πd₂²/4 = π(0.035 m)²/4 = 9.621 x 10⁻⁴ m²
Therefore,
(3.848 x 10⁻³ m²)V₁ = (9.621 x 10⁻⁴ m²)V₂
V₁ = (9.621 x 10⁻⁴ m²)V₂/(3.848 x 10⁻³ m²)
V₁ = 0.25 V₂ -------------------- equation (2)
using this value in equation (1):
V₂² - (0.25 V₂)² = 16.85 m²/s²
0.9375 V₂² = 16.85 m²/s²
V₂² = (16.85 m²/s²)/0.9375
V₂ = √(17.97 m²/s²)
V₂ = 4.24 m/s
(a)
using the value of V₂ in equation (2):
V₁ = 0.25(4.24 m/s)
V₁ = 1.06 m/s
(c)
For fluid flow rate we use the following equation:
Flow Rate = Q = A₂V₂ = (9.621 x 10⁻⁴ m²)(4.24 m/s)
Q = 4.08 x 10⁻³ m³/s
The speed if the fluid increases as pressure decreases.
(a) The speed of the flow in the lower section is 1.06 m/s
(b) The speed of the flow in the higher section is 4.24 m/s
(c) The volume flow rate in the pipe [tex]\bold {4.08 x 10^-^3m^3/s}[/tex].
The relation pressure and volume in constricted pipe can be derived by the formula formula ,
[tex]\bold {P_1 - P_2 = (\rho /2)(V_2^2 - V_1^2)}[/tex]
Where,
P₁ - P₂ -Difference in Pressure = [tex]\bold { 0.75 x 10^4}[/tex]
[tex]\rho[/tex] - Density of Oil = 890 kg/m³
V₂ - Velocity at Higher End = ?
V₁ - Velocity at Lower End = ?
put the values in the formula,
[tex]\bold {0.75 x 10^4 Pa = \dfrac{(890kg/m^3)}{2(V_2^2 - V_1^2)}}\\\\\bold {V_2^2 - V_1^2 = \dfrac {0.75 x 10^4 Pa}{445 kg/m^3}}\\\\\bold {V_2^2 - V_1^2 = 16.85 m^2/s^2}[/tex]
The continuity equation,
[tex]\bold {A_1V_1 = A_2V_2}[/tex]
Where,
[tex]\bold {A_1 }[/tex] - Area of lower end = [tex]\bold {3.848 x 10^-^3m^2}[/tex]
[tex]\bold {A_2 }[/tex] - Area of higher end = [tex]\bold {9.621 x 10^-^4m^2}[/tex]
Put the value in the formula above,
[tex]\bold {(3.848 x 10^-^3m^2)V_1= (9.621 x 10^-^4m^2)V^2}\\\\\bold {V_1 = \dfrac {(9.621 x 10^-^4 m^2)V_2}{(3.848 x 10^-^3 m^2)}}\\\\\bold {V_1 = 0.25 \times V_2}[/tex]
Put the value of [tex]\bold {V_1}[/tex], we get
[tex]\bold {V_2= 4.24 m/s}[/tex]
Therefore,
[tex]\bold {V1 = 0.25 (4.24 m/s)}\\\\\bold {V_1 = 1.06 m/s}[/tex]
The flow rate
[tex]\bold {Q = A_2V_2}[/tex]
[tex]\bold {Q = (9.621 x 10^-^4 m^2)(4.24 m/s)}\\\\\bold {Q = 4.08 x 10^-^3m^3/s}[/tex]
Therefore, the flow rate of the oil in the pipe is [tex]\bold {4.08 x 10^-^3m^3/s}[/tex].
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Describe resistance and discuss how it affects current?
Answers
Answer:
Resistance decreases the flow of current in a circuit.
EXPLANATION:
The resistance of an electrical circuit is defined as the opposition to the flow of current in that particular circuit.
From Ohm's law, we know that current flowing through a wire I =
Here,V is the potential across the two ends of a conductor or wire, and R is the resistance of that wire.
Hence, more is the resistance, the less will be the current flow in the current.
What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV. Tries 0/20 In this experiment, what is the minimal potential needed to fully stop the electrons if the wavelength of the incident light is 321 nm?
Answers
Answer:
4.24nm
0.385eV
Explanation:
Maximum wavelength (λmax) :
λmax = ( hc) /Φ
h = plancks constant = 6.63 * 10^-34
c = speed of light = 3*10^8
1ev = 1.6 * 10^-19
Φ = 2.93eV = 2.93* (1.6*10^-19) = 4.688*10^-19
λmax = [(6.63 * 10^-34) * (3 * 10^8)] / 4.688*10^-19
λmax = 19.89 * 10^-26 / 4.688*10^-19
λmax = 4.242 * 10^-7 m
λmax= 4.24nm
B.)
E = hc / eλ eV
λ = 3.75nm = 3.75 * 10^-7m = 375 *10^-9
E = (6.63 * 10^-34) * (3 * 10^8) / (1.6 * 10^-19) * (375 * 10^-9)
E = 19.89 * 10^-26 / 600 * 10^-28
E = 0.03315 * 10^-26 + 28
E = 0.03315 * 10^2
E = 3.315 eV
Stopping potential : (3.315 eV - 2.93eV) = 0.385eV
A force of 1.50 N acts on a 0.20kg trolley so as to accelerate it along an air track
The track and force are horizontal and in line . How fast is the trolley going after acceleration from rest through 30cm , if friction is negligible
Answers
Answer:
2.12m/s
Explanation:
Given parameters:
Force on trolley = 1.5N
Mass of trolley = 0.2kg
Unknown:
Velocity of the trolley = ?
Solution:
To solve this problem, we first find the acceleration of the trolley;
Force = mass x acceleration
Acceleration = [tex]\frac{Force }{mass}[/tex]
Insert the parameters and solve;
Acceleration = [tex]\frac{1.5}{0.2}[/tex] = 7.5m/s²
Now to find the acceleration;
Initial velocity = 0m/s
v² = u² + 2aS
v is the final velocity
u is the initial velocity
a is the acceleration
S is the distance
Distance = 30cm and this is 0.3m
v² = 0² + 2(7.5)0.3 = 4.5
v = √4.5 = 2.12m/s
A monochromatic plane electric-field-wave has a peak value 20 V/m. What is the peak value of the Magnetic field?
Answers
Answer:
The peak value of the Magnetic field is 6.67 x 10⁻⁸ T
Explanation:
Given;
peak value of the electric-field, E₀ = 20 V/m
The peak of the magnetic field, B₀ = ?
The peak value of the magnetic field is given by;
B₀ = E₀/C
Where;
C is the speed of light = 3 x 10⁸ m/s
Substitute the givens and solve for B₀
B₀ = E₀ / C
B₀ = (20) / (3 x 10⁸)
B₀ = 6.67 x 10⁻⁸ T
Therefore, the peak value of the Magnetic field is 6.67 x 10⁻⁸ T
B₀ =
a ball falls 0.35m to the floor. it lands 3m from the edge of the table. what was the velocity of the ball before it left the table?
Answers
The ball rolls off the table with speed v from a height of 0.35 m, so that it covers a horizontal distance x with height y at time t of
x = v t
y = 0.35 m - 1/2 g t ²
where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.
Solve for t when y = 0, i.e. the time it takes for the ball to reach the ground:
0 = 0.35 m - 1/2 g t ²
t ² = (0.70 m) / g
t ≈ 0.267 s
Now solve for v given that the ball falls 3 m away from the table:
3 m = v (0.27 s)
v = (3 m) / (0.27 s)
v ≈ 11.2 m/s
3. A player catches a ball. The action force is the impact of the ball against the player's glove. The
reaction force is...
a. The force the glove exerts on the ball
b. The player's grip on the glove
c. The friction of the ground on the player's shoes
Why???
Answers
Answer:
Answer: A
Explanation:
Why?? Two objects meaning they have the same force pair must remain the same
Using Newton's third law, we find that the correct answer to the reaction force is a: the force of the glove on the ball
Given parameters
The force of the ball on the gloveto find
The reaction force
Newton's third law say that forces are produced in pairs, that is, when two bodies interact, one body feels the action of the other and responds with an interaction of equal magnitude but in the opposite direction. In general, these two forces are called action and reaction, they have the following characteristics:
They are equal in magnitude has the opposite direccion each one is applied to one of the interacting bodies
let's analyze the given options using Newton's third law:
a. True. The force of the ball is on the glove, it must respond with a force of equal magnitude on the ball, therefore these are two forces of action and reaction
b. False. The grip or the player's hand does not receive the action of the ball, it receives the action of the glove
c) False. the player's shoes do not interact with the ball, interact with the ground
In conclusions by using Newton's third law the correct answer is a: the force of the glove on the ball
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Which of the following is a form of potential energy? O A. Sound energy O B. Elastic energy O C. Light energy O O O D. Kinetic energy
Answers
Answer:
O C. Light energy
Explanation:
it conducts energy in it and is an energy itself.
A ray diagram shows that an object is placed in front of a plane mirror. What are the characteristics of the image produced by the object? inverted, larger than object, real upright, smaller than object, virtual inverted, same size as object, real upright, same size as object, virtual
Answers
Answer:
D. upright, same size as object, virtual
Explanation:
it is a plane mirror, it has no changes
The characteristics of the image produced by the object placed in front of a plane mirror will be upright, the same size as the object, and virtual.
What is the law of reflection?The law of reflection states that when a ray of light reflects back from a surface the angle of incidence of light on a surface is equal to the angle of reflection.
The image produced by an object in front of a plane mirror will be upright, the same size as the object, and virtual.
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A piano string having a mass per unit length equal to 4.70 10-3 kg/m is under a tension of 1 400 N. Find the speed with which a wave travels on this string.
Answers
Answer:
The speed of the sound wave on the string is 545.78 m/s.
Explanation:
Given;
mass per unit length of the string, μ = 4.7 x 10⁻³ kg/m
tension of the string, T = 1400 N
The speed of the sound wave on the string is given by;
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
where;
v is the speed of the sound wave on the string
Substitute the given values and solve for speed,v,
[tex]v = \sqrt{\frac{T}{\mu} }\\\\v = \sqrt{\frac{1400}{4.7*10^{-3}} }\\\\v = \sqrt{297872.34}\\\\v = 545.78 \ m/s[/tex]
Therefore, the speed of the sound wave on the string is 545.78 m/s.
An 11.8 eV photon is absorbed by an electron in an atom, causing the electron to be excited from level 1 to level 3. The electron then emits a 4.1 eV photon and shifts to level 2. What photon will be emitted when the electron moves back to level 1?
a. 15.9 eV
b. 11.8 eV
c. 7.7 eV
d. 4.1 eV
Answers
Answer:
c. 7.7 eV
Explanation:
The computation of the photon emitted is shown below:
Given that
From level 1 to level 3, the energy absorbed by electrons is 11.8eV
And, when from level 3 to level3, the energy released by electron is 4.1eV
So, from level 2 to level 1, the energy released is
= 11.8 - 4.1
= 7.7
= 7.7eV
Hence, the third option is correct
The same is to be considered
And, the other options are wrong
if a car is moving down the highway at a constant velocity what does this mean about the cars acceleration?
Answers
Suppose someone travels 50 mi at 50 mi/h, then travels 50 mi at 25 mi/h, then travels 50 mi at 10 mi/h.
Answers
a). My estimate is 18.75 mph.
b). The average speed is 18.75 mph.
I nailed it.
Why were Brittany and Michael told that nobody wanted to continue working with them?
In the survey, other participants indicated that they were the ones with whom they did not want to continue to work.
They were randomly preassigned to the condition of social rejection.
They were chosen for the rejection condition because they belong to an ethnic minority.
They were confederates of the experimenter and acted in a way that guaranteed the other participants would dislike them.
Answers
Answer: They were randomly preassigned to the condition of social rejection.
Explanation:
Humans are said to be social beings which means that we survive in part due to our social connections. The experiment from the video was meant to research and evaluate what happens to human beings when they experience social rejection instead of acceptance.
There were four participants being Alexandria, Mike, Brittany and Michael. For the purpose of the research, Brittany and Michael were preassigned to the condition of social rejection whilst the other two were preassigned to acceptance with the goal being to observe the reactions of the parties after they find that they were either accepted or rejected.
Attached is the transcript to the video of the research summary.
Bru this was a challenge..........
Answers
Answer:
um
Explanation:
Answer: Dang thats tough g but u did it
Explanation:
A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s and lands 70.0 m from the base of the ramp. How high is the end of the ramp from the ground? m
Answers
Answer:
The end of the ramp is 38.416 m high
Explanation:
Horizontal Motion
When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.
The maximum horizontal distance traveled by the object can be calculated as follows:
[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]
If the maximum horizontal distance is known, we can solve the above equation for h:
[tex]\displaystyle h=\frac {d^2g}{2v^2}[/tex]
The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:
[tex]\displaystyle h=\frac {70^2\cdot 9.8}{2\cdot 25^2}[/tex]
[tex]\displaystyle h=\frac {4900\cdot 9.8}{2\cdot 625}[/tex]
h= 38.416 m
The end of the ramp is 38.416 m high
if something is shot straight up in the air at a speed of 17 m/s, how much time before I it will hit the ground
Answers
(20 points) An engineer is tasked with developing a model to study a cylindrical heat exchanger in a steam system. The prototype cylinder is 2.54 cm in diameter and the steam properties are: velocity
Answers
This question is incomplete, the complete question is;
An engineer is tasked with developing a model to study a cylindrical heat exchanger in a steam system. The prototype cylinder is 2.54 cm in diameter and the steam properties are: velocity = 30 m/s; density = 0.6 kg/m³; and absolute viscosity = 1.197 X 10⁻⁵ N-s/m², respectively. The model is going to be tested in a water tunnel where the water temperature is 20°C and the velocity is 3 m/s
a) what is the prototype Reynolds number, based on using the cylinder diameter as the characteristic length?
b) what should be the diameter of the model be to ensure dynamic similitude?
Answer:
a) the prototype Reynolds number is 38195.488
b) 0.01278 m or 1.278 cm should be the diameter of the model be to ensure dynamic similitude
Explanation:
Given that;
1 prototype ; d = 2.54 cm = 0.0254 m, Vp = 30 m/s, Sp = 0.6 kg/m³, Up = 1.197 × 10⁻⁵ N-s/m²
Model{ water at 20°C}; dm = ?, Vm = 3 m/s, Pm = 998.23 kg/m³, Um = 1.002 × 10⁻³ N-s/m²
a) what is the prototype Reynolds number,
to calculate prototype Reynolds number we use the expression;
(Re)p = SpVpdP / Up
we substitute our value
(Re)p = (0.6 × 30 × 0.0254) / 1.197 × 10⁻⁵
(Re)p = 38195.488
Therefore the prototype Reynolds number is 38195.488
b)
what should be the diameter of the model be to ensure dynamic similitude?
i.e dm = ?
so dynamic similarity [ viscous flow]
(Re)m = (Re)p
[PmVmdm / Um] = [SpVpdP / Up]
we substitute
[998.23 × 3 × dm / 1.002 × 10⁻³] = (0.6 × 30 × 0.0254) / 1.197 × 10⁻⁵
2994.69dm / 1.002 × 10⁻³ = 38195.488
2994.69dm = 38.2718
dm = 38.2718 / 2994.69
dm = 0.01278 m or 1.278 cm
Therefore 0.01278 m or 1.278 cm should be the diameter of the model be to ensure dynamic similitude
in the space below derive two equations one in the y direction and one in the x direction expressing newton’s second law using symbols
Answers
Answer:
Fₓ = maₓ
Fy = may
Explanation:
The Newton's Second Law tells us about the force. It says whenever an unbalanced force is applied to a body, an acceleration is produced in that body. The direction of this acceleration is in the direction of the force itself. Hence, the law can be written as:
F = ma
where,
F = Unbalanced force
m = mass of body
a = acceleration magnitude
We, can write two equations separately for x-direction and y-direction as follows:
Fₓ = maₓ
Fy = may
where,
Fₓ = x-component of unbalanced force
aₓ = x-component of acceleration
Fy = y-component of unbalanced force
ay = y-component of acceleration
Bru this was a challenge......but
Answers
Answer:
Free points
Explanation:
Answer:
wdym??
Explanation:?
what forces are used to move a 20000 kilogram block of stone
Answers
Answer:
196200N
Explanation:
Fg=mg
Fg=?
m=20000
g=9.81
Fg= (20000kg)(9.81[tex]m/s^{2}[/tex])=196200N
A 0.964 kg radio-controlled car is driving 4.38 m/s. What is its momentum?
Unit: kg*m/s
Answers
P= 0.964 * 4.38
P= 4.222 kg/ms
Aria is riding her bike. She is bent slightly over, holding onto her handlebars. She wants to go faster. Which best explains what she should do? She should bend her hips so her head is closer to the handlebars. She should sit straight up on the bike, keeping her arms tight to her body. She should spread her arms wide while sitting straight in her seat. She should flap her arms up and down like a bird.
Answers
Answer:
She should bend her hips so her head is closer to the handlebars
Explanation:
Answer:
Yea its A
Explanation:
jose has a mass of 70 kg, what is his weight
Answers
Which shows the correct order of steps during the formation of an ionic bond?
Ions are attracted to each other → Electrons are transferred → An ionic compound forms
An ionic compound forms → Ions are attracted to each other → Electrons are transferred
Electrons are transferred → Ions form → Ions are attracted to each other
Ions form → Electrons are transferred → Ions are attracted to each other
Answers
Answer:
c
Explanation:
1- First electrons are transferred.
2-The ions are formed
3-The ions are attracted to form the compound
Answer:
answer is c.........
Explanation:
Which state of matter does this model represent?
Image of particles loosely packed that fill the bottom of the container
Solid
Liquid
Gas
Plasma
Answers
Answer:
Liquid
Explanation:
A liquid will fill its container and are neither tightly packed like a solid nor very loosely packed like a gas.
Answer: Its Solid
Explanation:
Object 1 is undergoing uniform circular motion at a radius of 10 cm. Object 2 is undergoing uniform circular motion at a radius of 0.2 m. If the objects have the same mass and are traveling with the same speed, which object is experiencing the larger centripetal force and why
Answers
Answer:
The first object is experiencing a larger centripetal force because its radius is smaller compared to the second object.
Explanation:
Given;
radius of the first object, r₁ = 10 cm = 0.1 m
radius of the second object, r₂ = 0.2 m
let the mass of the two objects, = m
let the speed of the two objects = v
The centripetal force of the first object is given by;
[tex]F_c_1 = \frac{mv^2}{r_1} \\\\F_c_1 = \frac{mv^2}{0.1}\\\\F_c_1 = 10mv^2[/tex]
The centripetal force of the second object is given by;
[tex]F_c_2 = \frac{mv^2}{r_2} \\\\F_c_2 = \frac{mv^2}{0.2}\\\\F_c_1 = 5mv^2[/tex]
Therefore, the first object is experiencing a larger centripetal force because its radius is smaller compared to the second object.