Answer:
The power generated by the windmill is approximately 1.364 MW
Explanation:
The diameter of the windmill, d = 30 m
The inlet speed of the wind, [tex]V_e[/tex] = 40 mph = 17.88 m/s
The exit stream velocity, [tex]V_i[/tex] = 20 mph = 8.94 m/s
The pressure at the inlet, P₁ = 2 atm
The pressure at the outlet, P₂ = 1 atm
The density of air, ρ = 1.2 kg/m³
The power obtained from the windmill, 'P', is given as follows;
[tex]P =\dfrac{1}{4 \cdot g_c} \cdot \rho \cdot A \cdot (V_i + V_e)\cdot (V_i^2 - V_e^2)[/tex]
Where;
[tex]g_c[/tex] = 1.0 kg/(N·s²)
A = Cross-sectional rea of the the windmill = π·D²/4 = π×(30 m)²/4 = 706.858347 m²
Plugging in the values, we get;
[tex]P =\dfrac{1}{4 \times 1.0} \times1.2 \times 706.858347 \times (17.88 + 8.94)\cdot (17.88^2 - 8.94^2) = 1363668.19438[/tex]
The power generated by the windmill, P ≈ 1363668.19438 W ≈ 1.364 MW.
what type of reaction is being shown in this energy diagram?
X exothermic, because energy is absorbed from the surroundings
O exothermic, because energy is released into the surrounding
X endothermic, because energy is released into the surrounding
X endothermic, because energy is absorbed from the surroundings
best of luck nerds
Answer:
O exothermic, because energy is released into the surrounding
Explanation:
From the diagram the energy of the reactant is higher than the energy of the product, thereby making it exothermic. If you study diagram well, exothermic reaction means that the reactions releases energy into the surroundings.
A vertical wire carries a current straight up in a region of the magnetic field directed north. What is the direction of the magnetic force on the current due to the magnetic field
Answer:
The direction of the force on the vertical wire is towards the East or right.
Explanation:
Using Fleming's right hand rule, the current is the middle finger pointing straight up, the magnetic field is the fore-finger pointing Northwards and then the thumb is the direction of the force on the vertical wire.
Following these conventions, the thumb points towards the East. So, the direction of the force on the vertical wire is towards the East or right.
2) The track for a racing event was designed so that riders jump off the slope at 37 degrees from a height of 1 m. During a race it was observed that the rider remained in mid air for 1.5 seconds. Determine the speed at which he was traveling off the slope, the horizontal distance he travels before striking the ground and the maximum height he attains. Neglect the size of the bike and rider.
Answer:
3.277 m
Explanation:
Given :
Maximum Height (Hmax) = (u²sin²θ) / 2g
Xv = Xh + Uv * t + 0.5gt²
Xv and Xh are vertical and horizontal distances
-1 = 0 + sin37 * 1.5 Uv + 0.5*-9.8*1.5^2
-1 = 0 + 0.903Uv - 11.025
-1 + 11.025 = 0.903Uv
10.025 = 0.903Uv
Uv = 10.025 / 0.903
Uv = 11.10 m/s
Hmax = 1 + (u²sin²θ) / 2g
= (11.10^2 * (sin37)^2) / 2*9.8
= 44.624360 / 19.6
= 2.277
Hmax = 1 + 2.277
Hmax = 3.277 m
Bartender slides a beer mug at 1.1 m/s towards a customer at the end of the bar which is 1.8 m tall. The customer makes a grab for the mug and misses and mug sails at the end of the bar. a) How far away from the end of the bar does the mug hit the floor
Answer:
Δx = 0.7 m
Explanation:
Once the mug is moving in the horizontal direction, it keeps moving at the same speed of 1.1 m/s, due to no other force acts on it in this direction.Since the horizontal and vertical movements are independent each other (due to they are mutually perpendicular), in the vertical direction, the initial speed is just zero.In the vertical direction, the mug is accelerated by the force of gravity at all times, with a constant value of 9.8 m/s2, aimed downward.So, we can use the following kinematic equation in order to get the time passed from the instant that the mug left the bar, until it hit the floor, as follows:[tex]\Delta y = \frac{1}{2} * g* t^{2} = (1)[/tex]where Δy = 0-1.8m = -1.8m, g= -9.8m/s2.Replacing these values in (1) and solving for t, we get:[tex]t = \sqrt{\frac{2*1.8m}{ 9.8m/s2} } = 0.6 s (2)[/tex]
Now, since the mug obviously finishes its horizontal trip at this same time (hitting ground), we can find the horizontal distance traveled, just applying the definition of average speed, as follows:[tex]\Delta x = v_{o} * t = 1.1 m/s* 0.6 s = 0.7 m (3)[/tex]
You are driving a car behind a truck. Both your car and the truck are moving at a speed of 80km/hr. If the driver of the truck suddenly slams on the brakes, what minimum distance betweenyour car and the truck is needed so that your car does not crash into the truck’s rear end? (This is called the "​minimum trailing distance​".) To simplify this problem, assume that the truck andthe car have the same braking acceleration.
a. In order to simplify the calculations for this problem, you are told to assume that the braking acceleration of the car and the truck are the same. What other reasonable assumptions do you need to make in order to solve this problem?
b. For both the truck and the car, draw an acceleration- and velocity-versus-time graph.
c. Find an expression for the minimum trailing distance. (Your expression should only contain symbols of physical quantities. No numbers are needed here.)
d. Find the numerical value for the minimum trailing distance (Plug the values of physical quantities into your expression from part A (do not forget units!))
Answer:
Explanation:
Let the velocity of car and truck be u and breaking acceleration be a .
We shall have to assume the reflex time of the driver of the car . By the time he applies brake , his car will cover some distance . There will be some time tag between the time the truck starts decelerating and the driver of the car responding to that . During this period the car will not start decelerating . It will keep on moving with uniform velocity of u .
Let this time lag be t .
b )
For answer see the attached file
c )
The minimum trailing distance will be the distance covered by car before it starts decelerating in response to truck's deceleration .
minimum trailing distance d = u x t
d ) u = 80 km / h = 22.22 m /s
reflex action time t = 0.1 s ( assumed time )
d = 22.22 x .1
= 2.2 m
A motorcycle and rider have a total mass equal to 300 kg. The rider applies the brakes, causing the motorcycle to decelerate at a rate of -5 m/s^2. What is the net force on the motorcycle?
Answer:
Net force = - 1500 N
Explanation:
We calculate the net force acting using Newton's second Law:
[tex]F_{net}=m*a\\F_{net}=(300 \,kg)*(-5\,m/s^2)\\F_{net}=-1500\,N[/tex]
help please due today
Answer:
equal and opposite
Explanation:
..........
How can you drop two eggs the fewest amount of times, without them breaking?
Answer:
get 2 jugs of water put an egg in each one drop the jugs with parachutes on them in long grass on a sunny non windy day
Explanation:
egg+ground=broken
egg-ground= egg+air
egg+air=unbroken
egg+water= egg+wet
egg+water= unbroken
egg+egg= 2 egg
egg+egg+air= egg+egg+unbroken+unbroken
egg+egg+unbroken+unbroken=(egg+unbroken)2
longgrass+egg= 40%unbroken+60broken+egg
longgrass+egg+egg=20%unbroken+80%broken+2egg
ground+water=mud
mud+egg=unbroken+egg+muddy
air+water=raining
egg+raining+air=wet+egg+slip+50%broken+50%unbroken
ask if need more proof
pls help me this is a major SOS pls help pls btw this is IXL
Explanation:
the object with the higher temperature has greater thermal energy
So the answer is
the stick of butter with less thermal energy.
Hope it will help :)
Answer:
The stick of butter with less thermal energy
Explanation:
I am pretty sure
Suppose one Sherpa uses a force of 980 N to move a load of equipment to a height of 20 meters in 25 seconds. How much power is used?
F = 980 N
h = 20 m
t = 25 s
P=? (power)
W=F*h (work)
P=W*t
P=F*h*t
P=980*20*25 =490000 W = 490 kW = 0.49 MW
Explain the difference in the function of plant and animal cell organelles, including cell membrane, cell wall, nucleus, cytoplasm, mitochondria, chloroplast, and vacuole
Answer:
Plant cell Animal cell
2. Have a cell membrane. 2. Have no chloroplasts.
3. Have cytoplasm. 3. Have only small vacuoles.
4. Have a nucleus. 4. Often irregular in shape.
5. Often have chloroplasts
containing chlorophyll. 5. Do not contain plastids.
A boat is moving in a river with a current that has speed vW with respect to the shore. The boat first moves downstream (i.e. in the direction of the current) at a constant speed, vB , with respect to the water. The boat travels a distance D in a time tOut . The boat then changes direction to move upstream (i.e. against the direction of the current) at a constant speed, vB , with respect to the water, and returns to its original starting point (located a distance D from the turn-around point) in a time tIn .
1) What is tOut in terms of vW, vB, and D, as needed?
2) What is tIn in terms of vW, vB, and D, as needed?
3) Assuming D = 120 m, tIn = 170 s, and vW = 0.3 m/s, what is vB, the speed of the boat with respect to the water?
4) Once again, assuming D = 120 m, tIn = 170 s, and vW = 0.3 m/s, what is tOut, the time it takes the boat to move a distance D downstream?
Answer:
Explanation:
Current has speed vW with respect to the shore and boat has speed vB with respect to water or current so speed of boat with respect to shore
vW + vB .
Distance travelled with respect to shore by boat = D
time ( tout ) = distance / speed with respect to shore
tOut = D / ( vW + vB )
When the boat travels upstream , its velocity with respect to shore
= ( vB - vW ) , vB must be higher .
tin = D / ( vB - vW )
3 ) tin = D / ( vB - vW )
170 = 120 / (vB - 0.3 )
(vB - 0.3 ) = 12 / 17 = .706
vB = 1.006 m / s
4 )
tOut = D / ( vW + vB )
= 120 / ( .3 + 1.006 )
= 92.26 s
Time taken by a body is ratio of the distance traveled by it to the speed.
1)The expression for [tex]t{out}[/tex] is,[tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]
2)The expression for [tex]t{in}[/tex] is,[tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]
3) The speed of the boat with respect to the water is 1.006 m/s. 4) The time it takes the boat to move a distance D downstream is 91.9 seconds.What is upstream and downstream speed?
The net speed of the boat is upstream speed. The difference of the speed of the boat is downstream speed.
Given information-
The speed of the boat with respect to shore is [tex]v_w[/tex].
The speed of the boat in downstream with respect to water is [tex]v_B[/tex].
The distance traveled by the boat is [tex]D[/tex] in time [tex]t_{out}[/tex].
Time taken by a body is ratio of the distance traveled by it to the speed.
1) The net speed of the boat is upstream speed.As the distance traveled by the boat is [tex]D[/tex] in time [tex]t_{out}[/tex]. Thus,[tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]
2) The difference of the speed of the boat is downstream speed.As the distance traveled by the boat is [tex]D[/tex] in time [tex]t_{in}[/tex]. Thus,[tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]
Now the distance is 120 m, the value of [tex]t_{in}[/tex] is 170 s and [tex]v_W[/tex] 0.3 m/s. Thus,
3) The speed of the boat with respect to the water-Put the values in the formula obtains from the 2nd part of the problem,[tex]170=\dfrac{120}{v_B-0.3}\\v_B-0.3=\dfrac{120}{160} \\v_B=0.706+0.3\\v_B=1.006[/tex]
Hence the speed of the boat with respect to the water is 1.006 m/s.
4) The time it takes the boat to move a distance D downstream-Put the values in the formula obtains from the 1st part of the problem,[tex]t_{out}=\dfrac{120}{1.006+0.3}\\t{out}=\dfrac{120}{1.306} \\t{out}=91.9[/tex]
Hence the time it takes the boat to move a distance D downstream is 91.9 seconds.
Thus,
1)The expression for [tex]t{out}[/tex] is,[tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]
2)The expression for [tex]t{in}[/tex] is,[tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]
3) The speed of the boat with respect to the water is 1.006 m/s. 4) The time it takes the boat to move a distance D downstream is 91.9 seconds.Learn more about the upstream and downstream speed here;
https://brainly.com/question/800251
What is the acceleration of an object going from O m/s to 25 m/s in 5s?
Answer:
5m/s^2 is the acceleration.
Answer:
[tex]\boxed {\boxed {\sf a= 5 \ m/s^2}}[/tex]
Explanation:
Acceleration is the change in speed over time.
[tex]a=\frac{ v_f-v_i}{t}[/tex]
The object accelerates from 0 meters per second to 25 meters per second in 5 seconds.
[tex]v_f= 25 \ m/s\\v_i= 0 \ m/s \\t= 5 \ s[/tex]
Substitute the values into the formula.
[tex]a=\frac{ 25 \ m/s -0 m/s }{ 5 \ s}[/tex]
Solve the numerator.
[tex]a=\frac{25 \ m/s}{5 \ s}[/tex]
Divide
[tex]a= 5 \ m/s/s= 5 \ m/s^2[/tex]
The object's acceleration is 5 meters per square second.
Name the state of matter that diffusion happens the fastest in.
Answer:
Liquids
Explanation:
Diffusion occurs fastest in liquids.
Which of the following does NOT have a positive impact on your position on the
health continuum?
avoiding risk behaviors
having a positive social environment
eating nutritious foods
O having a chronic disease
Answer:
Having a chronic disease
Explanation:
Someone help please
Answer:
it would be downwards due to gravitational force
How much force is needed to accelerate a Kia Soul with a
mass of 1200 kg to 5 m/s2?
Answer:
[tex]\boxed {\boxed {\sf 6,000 \ Newtons}}[/tex]
Explanation:
Force is the product of mass and acceleration.
[tex]F=ma[/tex]
The mass of the Kia Soul is 1200 kilograms and its acceleration is 5 meters per square second.
[tex]m= 1200 \ kg \\a= 5 \ m/s^2[/tex]
Substitute the values into the formula.
[tex]F= 1200 \ kg * 5 \ m/s^2[/tex]
Multiply.
[tex]F= 6000 \ kg*m/s^2[/tex]
1 kilgram meter per square second is equal to 1 Newton. Our answer of 6000 kg*m/s² equals 6000 N[tex]F= 6000 \ N[/tex]
Answer:
Given :-Mass = 1200 kgAcceleration = 5 m/s²To Find :-Force
Solution :-We know that
F = ma
F = Force
m = mass
a = acceleration
F = 1200 × 5
F = 6000 N
[tex] \\ [/tex]
If you stand on a trampoline, it depresses under your weight. When you stand on a hard stone floor, __________. If you stand on a trampoline, it depresses under your weight. When you stand on a hard stone floor, __________. the floor deforms very slightly under your weight only if you are heavy enough does the floor deform at all under your weight the floor does not deform at all under your weight
Answer:
the floor deforms very slightly under your weight
Explanation:
A trampoline is made up of a large piece of strong cloth held by springs on which you jump up and down as a sport. So, If you stand on a trampoline, it depresses under your weight. However, the floor does not deform under your weight as it is too stiff.
Therefore,
when you stand on a hard stone floor, the floor deforms very slightly under your weight.
How long will it take an object traveling at 90 kilometers per hour to travel 910 kilometers?
Explanation:
time = distance / velocity
We know that distance = 910 km and velocity = 90 km/h.
t = d / v
t = 910 km / 90 km/h
t = 10.11 hrs
The object traveled for 10.11 hours long. Hope this helps, thank you !!
2. (9 points) A car starts from 10 mph and accelerates along a level road, i.e., no grade change. At 500 ft from its starting point, a radar gun measures its speed as 50 mph. Assuming the car had a constant rate of acceleration, (a) calculate the time elapsed between when the car started at 10 mph to when its speed was measured and (b) what will the speed of the car be another 500 ft downstream of this point
Answer:
a) t = 11.2 s
b) v = 70.5 mph
Explanation:
a)
Since we need to find the time, we could use the definition of acceleration (rearranging terms) as follows:[tex]t = \frac{v_{f} - v_{o}}{a} (1)[/tex]
where vf = 50 mph, and v₀ = 10 mph.However, we still lack the value of a.Assuming that the acceleration is constant, we can use the following kinematic equation:[tex]v_{f} ^{2} - v_{o} ^{2} = 2*a* \Delta x (2)[/tex]
Since we know that Δx = 500 ft, we could solve (2) for a.In order to simplify things, let's first to convert v₀ and vf from mph to m/s, as follows:[tex]v_{o} = 10 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 4.5 m/s (3)[/tex]
[tex]v_{f} = 50 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 22.5 m/s (4)[/tex]
We can do the same process with Δx, from ft to m, as follows:[tex]\Delta x = 500 ft *\frac{0.3048m}{1ft} = 152.4 m (5)[/tex]
Replacing (3), (4), and (5) in (2) and solving for a, we get:[tex]a = \frac{v_{f} ^{2} - v_{o}^{2}}{2*\Delta x} = \frac{(22.5m/s) ^{2} - (4.5m/s)^{2}}{2*152.4m} = 1.6 m/s2 (6)[/tex]
Replacing (6) in (1) we finally get the value of the time t:[tex]t = \frac{v_{f} - v_{o}}{a} = \frac{(22.5m/s) - (4.5m/s)}{1,6m/s2} = 11.2 s (7)[/tex]
b)
Since the acceleration is constant, as we know the displacement is another 500 ft (152.4m), if we replace in (2) v₀ by the vf we got in a), we can find the new value of vf, as follows:[tex]v_{f} = \sqrt{v_{o} ^{2} +( 2*a* \Delta x)} = \sqrt{(22.5m/s)^{2} + (2*1.6m/s2*152.4m)} \\ v_{f} = 31.5 m/s (8)[/tex]
If we convert vf again to mph, we have:[tex]v_{f} = 31.5m/s*\frac{1mi}{1609m} *\frac{3600s}{1h} = 70.5 mph (9)[/tex]
Q5. Use Superposition to V. in the circuit below? (5 points)
4 mA
12V
2 ΚΩ
2 mA
1 ΚΩ
2 ΚΩ
Answer:
4va
12va
2jk
1jk
2jk
Concept Simulation 4.1 reviews the central idea in this problem. A boat has a mass of 4490 kg. Its engines generate a drive force of 4520 N due west, while the wind exerts a force of 890 N due east and the water exerts a resistive force of 1210 N due east. Take west to be the positive direction. What is the boat's acceleration, with correct sign
Answer:
-0.54m/s²
Explanation:
According to Newton's second law of motion
F = ma
Force = mass * acceleration
Given
Mass m = 4490kg
Take the sum of forces
Sum of force along the east = 890+1210 = 2100N
Sum of forces along the west = -4520N
Net force = -4520+2100
Net force = -2420N
Acceleration = Net force/Mass
Acceleration = -2420/4490
Acceleration = -0.54m/s²
Hence the boat acceleration is -0.54m/s²
g Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of magnitude 46.0 N is exerted and the magnitude of the astronaut's acceleration is measured to be 0.834 m/s2. Calculate her mass.
Dereck is looking at how electrically charged objects can attract other objects without touching. What control would he need to use?
An electrically charged object
An uncharged object
A positively charged object
A negatively charged object
Answer:
its An uncharged object.
if its not charged the electrically wont go on it
Answer:
uncharged object
Explanation:
Which of the following statements is true?
A. Friction primarily affects objects that contain iron.
B. Friction pulls objects toward the center of the Earth
C.
Friction does not affect objects in motion.
D.
Friction slows down or stops objects in motion.
Answer:
D. Friction slow down or stop objects in motion.
Displacement vector A points due east and has a magnitude of 1.9 km. Displacement vector B points due north and has a magnitude of 2.08 km. Displacement vector C points due west and has a magnitude of 2.4 km. Displacement vector D points due south and has a magnitude of 2.8 km. Find the magnitude and direction (relative to due east) of the resultant vector A + B + C + D
Answer:
Explanation:
We shall represent all the four displacement in vector form in terms of unit vector i and j where i represents unit vector towards east , j represents unit vector towards north .
Displacement of A
D₁ = 1.9 i
Displacement of B
D₂ = 2.08 j
Displacement of C
D₃ = - 2.4 i
Displacement of D
D₄ = -2.8 j
Resultant displacement
= 1.9 i + 2.08 j - 2.4 i - 2.8 j
= - 0.5 i - 0.72 j
magnitude of resultant vector
= √ ( .5² + .72² )
=√ ( .25 + .5184 )
= √ .7684
= .876 km
Both i and j are negative of resultant displacement
hence its direction is towards south of west . Angle with west is Ф .
TanФ = .5184 / .25 = 2.0736
Ф = 64.25° .
From east direction is = 180 + 64.25 = 244.25° .
Which of the following describes the products of a chemical reaction?
A. The original materials
B. The substances that are changed
C. The chemicals on the left side of a chemical equation
Ο Ο
D. The chemicals on the right side of a chemical equation
Answer:
D The chemicals on the right side of a chamical equation
Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag force on a prop-driven transport flying at one-fourth the speed and half the altitude of the jet. At 10 km the density of air is 0.38 kg/m3 and at 5.0 km it is 0.67 kg/m3. Assume that the airplanes have the same effective cross-sectional area and the same drag coefficient C. (drag on jet / drag on transport)
Answer:
[tex]2.267[/tex]
Explanation:
Drag force is given by
[tex]F=\dfrac{1}{2}\rho Av^2C[/tex]
C = Drag coefficient is constant
A = Area is constant
[tex]v_1[/tex] = Velocity of the passenger jet = 1200 km/h = [tex]\dfrac{1200}{3.6}\ \text{m/s}[/tex]
[tex]v_2[/tex] = Velocity of the prop plane = [tex]\dfrac{1}{4}v_1[/tex]
[tex]\rho_1[/tex] = Density of the air where the jet was flying = [tex]0.38\ \text{kg/m}^3[/tex]
[tex]\rho_2[/tex] = Density of the air where the prop plane was flying = [tex]0.67\ \text{kg/m}^3[/tex]
[tex]F\propto \rho v^2[/tex]
[tex]\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267[/tex]
The ratio of the drag forces is [tex]2.267[/tex].
You are trying to push a 30 kg canoe across a beach to get it to a lake. Initially, the canoe is
at rest, and you exert a force over a distance of 3 m until it has a speed of 1.2 m/s.
a. How much work was done on the canoe?
b. The coefficient of kinetic friction between the canoe and the beach is 0.2. How much work was done by friction on the canoe?
c. How much work did you perform on the canoe?
d. What force did you apply to the canoe?
Answer:
m = 30, g = 9.8, coefficient = 0.2, so force due to friction = 30 x 9.8 x 0.2 = 58.8 N, so work done by friction = 58.8 x 1.2 = 70.56 J
Explanation:
Mischievous Joey likes to play with his family's lazy susan (this drives Mom crazy because it is an antique). He puts the salt shaker near the edge and tries to spin the tray at a speed so that the shaker just barely goes around without slipping off. Joey finds that the shaker just barely stays on when the turntable is making one complete turn every two seconds. Joey's older sister measures the mass of the shaker to be 79 grams. She also measures the radius of the turntable to be 0.23 m, and she is able to calculate that the speed of the shaker as it successfully goes around in a circle is 0.7222 m/s.
Required:
What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?
Answer:
0.179 N
Explanation:
What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?
The horizontal part of the constant force of the turntable on the shaker is the centripetal force of the turntable on the shaker, F.
So, F = mv²/r where m = mass of shaker = 79 g = 0.079 kg, v = speed of shaker = 0.7222 m/s and r = radius of turntable = 0.23 m
So, substituting the values of the variables into the equation, we have
F = mv²/r
F = 0.079 kg (0.7222 m/s)²/0.23 m
F = 0.0412 kgm/s² ÷ 0.23 m
F = 0.179 kgm/s²
F = 0.179 N