Answer:
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Explanation:
From the given information:
The elastic potential energy can be calculated by using the formula:
[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]
Making K the subject;
[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]
[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]
k = 2.3 × 10⁻² N/m
Now; the frequency of the small oscillation can be determined by using the formula:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
where;
m = mass of each atom = 1.66 × 10⁻²⁶ kg
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
What is the source of geothermal power
Answer: Geothermal energy is produced by the heat of Earth's molten interior. This energy is harnessed to generate electricity when water is injected deep underground and returns as steam (or hot water, which is later converted to steam) to drive a turbine on an electric power generator.
Explanation: Hope this helps!
1. A box contains 10 blue chips. 5 red chips, and 15 yellow chips. Find the odds of choosing the
following:
blue chip
b. yellow chip
c. yellow chip
Answer:
Explanation:
Blue: 10/30
Red: 5/30
Yellow: 15/30
The probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2
What is the probability?
The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.
Given that
Blue chips =10
Red chips = 5
Yellow chips = 15
Number of the total samples =10+15+5=30
Probability of choosing Blue chips = [tex]\dfrac{10}{30}= \dfrac{1}{3}[/tex]
Probability of Red chips =[tex]\dfrac{5}{30}=\dfrac{1}{6}[/tex]
Probability of Yellow chips =[tex]\dfrac{15}{30}=\dfrac{1}{2}[/tex]
Thus the probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2
To know more about probability follow
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What happens if you move a magnet near a coil of wire?
A) current is induced
B)power is consumed
C)the coil becomes magnetized
D) the magnets field is reduced
What does it mean when work is positive?
O Velocity is greater than kinetic energy
O Kinetic energy is greater than velocity
The environment did work on an object.
O An object did work on the environment.
d. An object did work on the environment.
Explanation:Work is defined in many contexts. Some of these are;
i. Work is the product of force and displacement. In this case, work done is positive if the force applied on an object or body and the displacement caused by the force are in the same direction. If instead the force and displacement are in opposite direction, then the work done will be negative. If it is the case the force and the displacement are perpendicular to each other, the work done is zero.
ii. In the first law of thermodynamics, the internal energy of a system is the sum of the work done and the heat exchanged between the system and the environment. Therefore, work done is the difference between the internal energy of a system and the heat exchanged between the system and the environment.
In this case, work is said to be positive if work is done by the system (object) on the environment. It is negative if work is done by the environment on the system (object).
Answer:
its c
Explanation:
One end of a horizontal spring with the spring constant 1900 N/m is attached to the wall, the other end is attached to a block of mass 1.15 kg. Initially, the spring is compressed 4.5 cm. When released, the spring pushes the block away and is no longer in contact with the block. The block slides along a horizontal frictionless plane.
a/ Compute the maximum speed of the block.
b/ The block goes off the edge of the plane and falls down from the plane to reach the floor with speed of
7 m/s. How high is the plane with respect to the floor?
(a) When the spring is compressed 4.5 cm = 0.045 m, it exerts a restoring force on the block of magnitude
F = (1900 N/m) (0.045 m) = 85.5 N
so that at the moment the block is released, this force accelerates the block with magnitude a such that
85.5 N = (1.15 kg) a ==> a = (85.5 N) / (1.15 kg) ≈ 74.3 m/s²
The block reaches its maximum speed at the spring's equilbrium point, and this speed v is such that
v ² = 2 (74.3 m/s²) (0.045 m) ==> v = √(2 (74.3 m/s²) (0.045 m)) ≈ 2.59 m/s
(b) There is no friction between the block and plane, so the block maintains this speed as it slides over the edge. At that point, it's essentially in free fall, so if y is the height of the plane, then
(7 m/s)² - (2.59 m/s)² = 2gy ==> y = ((7 m/s)² - (2.59 m/s)²) / (2g) ≈ 2.16 m
What is 11 divided by 73370 I need to know how you found the answer
Helppp how a deer gets it’s energy
Answer:
option C is the correct one
Explanation:
I hope it helps u
according to the law of conservation of vhange , what must always be true in a nuclear reaction?
Answer:
The Sum of mass and energy is always conserved in a nuclear reaction. Mass changes to energy, but the total amount of mass and energy combined remains the same
Explanation:
Every single radioactive decay, every single nuclear collision, every single nuclear reaction will conserve mass number and charge.
Answer in your PE notebook
I have learned that
I have realized that
I will apply
Answer:
physical science is important
hety
civil engineering
Where are the nasal cavities found?
O in the forehead
O behind the chin
O between the eyes
O behind the nostrils
A race car starts from rest on a circular track of radius 378 m. The car's speed increases at the constant rate of 0.580 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, find the following.
a. The speed of the race car
b. The distance traveled
c. the elapsed time
Answer:
a) [tex]V=14.904m/s[/tex]
b) [tex]d = 191.49 m[/tex]
c) [tex]t= 25.696 s[/tex]
Explanation:
From the question we are told that:
Radius [tex]r =378m[/tex]
Acceleration [tex]a=0.580[/tex]
a)
Generally the equation for speed of the car is mathematically given by
[tex]a=\frac{v^2}{r}[/tex]
[tex]V=\sqrt{a*r}[/tex]
[tex]V=\sqrt{0.58*383}[/tex]
[tex]V=14.904m/s[/tex]
b)
Generally the equation for distance traveled of the car is mathematically given by
[tex]V^2=u^2+2ad[/tex]
[tex]d=\frac{V^2}{2a}[/tex]
[tex]d=\frac{14.904^2}{2*0.58}[/tex]
[tex]d = 191.49 m[/tex]
c)
Generally the equation for time of the car is mathematically given by
[tex]V=u+at[/tex]
[tex]t=\frac{V}{a}[/tex]
[tex]t=\frac{14.904}{0.58}[/tex]
[tex]t= 25.696 s[/tex]
The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 mm, and a baseball has a mass of 145 g.
Required:
a. Draw a free-body diagram of the ball during the pitch.
b. What force did the pitcher exert on the ball during this record-setting pitch?
c. Estimate the force in part b as a fraction of the pitcher's weight.
Answer:
Following are the solution to the given points:
Explanation:
For point a:
Find the schematic of the empty body and in attachment. Upon on ball during the pitch only two forces act:
The strength of the pitcher F is applied that operates horizontally. Its gravity force acting on an object is termed weight, which value is where m denotes mass, and g the acceleration of gravity.
For point b:
[tex]160.2\ N[/tex]
First, they must find that ball's acceleration. You can use the SUVAT equation to achieve that
where
[tex]v = 47\ \frac{m}{s} \\\\u = 0 \\\\a =?\\\\d = 1.0 \ m \\\\[/tex]
Solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{47^2-0}{2(1.0)}=1104.5 \ \frac{m}{s^2}[/tex]
Calculating the mass:
[tex]m = 145 g = 0.145 kg[/tex]
Calculating the force:
[tex]F=ma=0.145 \times 1104.5= 160.2 \ N[/tex]
For point c:
0.195 times the pitcher's weight
[tex]m = 84 \ kg \\\\g = 9.8\ \frac{m}{s^2}\\\\[/tex]
Solving for W:
[tex]W=84 \times 9.8= 823.2 \ N[/tex]
Now the force of Part B could be defined as the fraction of the mass of the pitcher:
[tex]\frac{F}{W}=\frac{160.2}{823.3}=0.195[/tex]
What is binding energy?
A.' The attractive forces between the protons in the nucleus and the
electrons
B. The energy required to force two nuclei to undergo nuclear fusion
C. The amount of energy stored in the strong nuclear forces of the
nucleus
D. The amount of energy required to overcome an activation energy
barrier
Please help me out.
Answer:
the answer is B i hope it helps :)
[tex]\huge\color{purple}\boxed{\colorbox{black}{♡Answer}}[/tex]
B. The energy required to force two nuclei to undergo nuclear fusion. ✅
They are usually expressed in terms of [tex]\sf\purple{kJ/mole}[/tex] of nuclei or [tex]\sf\pink{MeV's/nucleon}[/tex].[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}[/tex]
A 45-kg skydiver jumps out of an airplane and falls 450 m, reaching a maximum speed of 51 m/s before opening her parachute. How much work, in joules, did air resistance do on the skydiver before she opened her parachute
Answer:
The work done by the friction force is - 139927.5 J.
Explanation:
mass of diver, m = 45 kg
distance falls, h = 450 m
initial speed, u = 0 m/s
final speed = 51 m/s
According to the work energy theorem,
Work done by the gravity + work done by the friction force = change in kinetic energy
[tex]m g h + W' = 0.5 m ()v^2 - u^2)\\\\45\times 9.8\times 450 + W' = 0.5\times 45\times (51^2 - 0)\\\\198450 + W' = 58522.5\\\\W' = - 139927.5 J[/tex]
A body starts from rest and moves with a uniform acceleration of 2 m/s2 in a
straight line. What is the velocity after 5s, How far has it travelled in this time and When will it be 100m from its starting point?
Answer:
Final velocity = 10 m/s
Time taken to travel 100 meter = 8.16 second (Approx.)
Explanation:
Given:
Acceleration = 2 m/s²
Initial velocity = 0 m/s
Find:
Velocity after 5 seconds
Time taken to travel 100 meter
Computation:
Using first equation of motion
v = u + at
v = 0 + (2)(5)
Final velocity = 10 m/s
Using Second equation of motion
s = ut + (1/2)(a)(t²)
100 = (0)(t) + (1/2)(3)(t²)
100 = (1/2)(2)(t²)
100 = (t²)
t = 10
Time taken to travel 100 meter = 8.16 second (Approx.)
Một vô lăng sau khi bắt đầu quay được một phút thì thu được vận tốc 700
vòng/phút. Tính gia tốc góc của vô lăng
which of the following is the correct description of momentum?
-the product of mass and acceleration -the product of mass and velocity
-velocity divided by mass
-acceleration divided by mass
Answer:
The product of mass and velocity is the correct answer
Explanation:
Momentum is defined as mass × velocity
p = mv
Answer:
The product of mass and velocity
Explanation:
I just did it and got it right with a 100%
Posted 1/3/23
What happens in the gray zone between solid and liquid?-,-
A Child stands on the bus Remains Still When The bus is at rest. When the bus moves forward AndeaThe bus is at rest. When the bus moves forward And then slows down, the children the Contnues moving forward at the original speed. This is an example of
Answer:
inertia
Explanation:
A 0.25 kg mass is placed on a vertically oriented spring that is stretched 0.56 meters from its equilibrium position. If the spring constant is 105 N/m, how fast will the mass be moving when it reaches the equilibrium position? Hint: You cannot ignore the change in gravitational potential energy in this problem. Please give your answer in units of m/s, however, do not explicitly include units when typing your answer into the answer box.
Answer:
The speed of the ball when it reaches equilibrium position is 3.31 m/s
Explanation:
Given;
mass of the object, m = 0.25 kg
initial displacement of the object, h₁ = 0.56 m
spring constant, k = 105 N/m
displacement at equilibrium position, h₂ = 0
initial velocity of the object, v₁ = 0
velocity of the object at equilibrium position = v₂
The change in gravitational potential energy at the equilibrium position is given as;
ΔP.E = mg(h₂ - h₁)
The change in kinetic energy of the object at the equilibrium position is given as;
ΔK.E = ¹/₂m(v₂² - v₁²)
Apply the principle of conservation of mechanical energy;
ΔK.E + ΔP.E = 0
¹/₂m(v₂² - v₁²) + mg(h₂ - h₁) = 0
¹/₂m(v₂² - 0) + mg(0 - h₁) = 0
¹/₂mv₂² - mgh₁ = 0
¹/₂mv₂² = mgh
¹/₂v₂² = gh
v₂² = 2gh
v₂ = √2gh
v₂ = √(2 x 9.8 x 0.56)
v₂ = 3.31 m/s
Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s
reviews the general principles in this problem. A projectile is launched from ground level at an angle of 13.0 ° above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?
Answer: θ would equal approximately 28.7°
This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.
Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°
Now if we multiply the range by 2, we get:
2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:
2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ
Thus, θ = 28.67780425
It's been awhile since I did this; though I hope it helped!
A certain microscope is provided with objectives that have focal lengths of 20 mm , 4 mm , and 1.4 mm and with eyepieces that have angular magnifications of 5.00 × and 15.0 × . Each objective forms an image 120 mm beyond its second focal point.
Answer:
Explanation:
Given that:
Focal length for the objective lens = 20 mm, 4 mm, 1.4 mm
For objective lens of focal length f₁ = 20 mm
s₁' = 120 mm + 20 mm = 140 mm
∴
Magnification [tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{140}{20}[/tex]
[tex]m_1 = 7 \ m[/tex]
For objective lens of focal length f₁ = 4 mm
s₁' = 120 mm + 4 mm = 124 mm
[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{124}{4}[/tex]
[tex]m_1 = 31 \ m[/tex]
For objective lens of focal length f₁ = 1.4 mm
s₁' = 120 mm + 1.4 mm = 121.4 mm
[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{121.4}{1.4}[/tex]
[tex]m_1 = 86.71 \ m[/tex]
The magnification of the eyepiece is given as:
[tex]m_e = 5X \ and \ m_e = 15X[/tex]
Thus, the largest angular magnification when [tex]m_1 \ and \ m_e \ are \ large \ is:[/tex]
[tex]M_{large}= (m_1)_{large} \times (m_e)_{large}[/tex]
= 86.71 × 15
= 1300.65
The smallest angular magnification derived when [tex]m_1 \ and \ m_e \ are \ small \ is:[/tex]
[tex]M_{small}= (m_1)_{small} \times (m_e)_{small}[/tex]
= 7 × 5
= 35
The largest magnification will be 1300.65 and the smallest magnification will be 35.
What is magnification?Magnification is defined as the ratio of the size of the image of an object to the actual size of the object.
Now for objective lens and eyepieces, it is defined as the ratio of the focal length of the objective lens to the focal length of the eyepiece.
It is given in the question:
Focal lengths for the objective lens is = 20 mm, 4 mm, 1.4 mm
now we will calculate the magnification for all three focal lengths of the objective lens.
Also, each objective forms an image 120 mm beyond its second focal point.
(1) For an objective lens of focal length [tex]f_1=20 \ mm[/tex]
[tex]s_1'=120\ mm +20 \ mm =140\ mm[/tex]
Magnification will be calculated as
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{140}{20} =7[/tex]
(2) For an objective lens of focal length [tex]f_1= \ 4 \ mm[/tex]
s₁' = 120 mm + 4 mm = 124 mm
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{124}{4} =31[/tex]
(3) For an objective lens of focal length [tex]f_1=1.4\ mm[/tex]
s₁' = 120 mm + 1.4 mm = 121.4 mm
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{121.4}{1.4} =86.71[/tex]
Now the magnification of the eyepiece is given as:
[tex]m_e=5x\ \ \ & \ \ m_e=15x[/tex]
Thus, the largest angular magnification when
[tex]m_1 = 86.17\ \ \ \ m_e=15x[/tex]
[tex]m_{large}= (m_1)_{large}\times (m_e)_{large}[/tex]
[tex]m_{large}=86.71\times 15=1300.65[/tex]
The smallest angular magnification derived when
[tex]m_1=7\ \ \ \ m_e=5[/tex]
[tex]m_{small}=(m_1)_{small}\times (m_e)_{small}[/tex]
[tex]m_{small}=7\times 5=35[/tex]
Thus the largest magnification will be 1300.65 and the smallest magnification will be 35.
To know more about magnification follow
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A laser beam enters one of the sloping faces of the equilateral glass prism (n=1.42) and refracts through the prism. Within the prism the light travels horizontally. What is the angle between the direction of the incident ray and the direction of the outgoing ray?
Answer:
30.5°
Explanation:
Since the light travels horizontally through the prism, it undergoes minimum deviation. So, the angle between the direction of the incident ray and that of the outgoing ray D is gotten from
n = [sin(D + α)/2]/sin(α/2) where n = refractive index of prism = 1.42 and α = angle of prism = 60° (since it is a n equilateral glass prism).
Making D subject of the formula, we have
n = [sin(D + α)/2]/sin(α/2)
nsin(α/2) = [sin(D + α)/2]
(D + α)/2 = sin⁻¹[nsin(α/2)]
D + α = 2sin⁻¹[nsin(α/2)]
D = 2sin⁻¹[nsin(α/2)] - α
So, substituting the values of the variables into the equation, we have
D = 2sin⁻¹[nsin(α/2)] - α
D = 2sin⁻¹[1.42sin(60°/2)] - 60°
D = 2sin⁻¹[1.42sin(30°)] - 60°
D = 2sin⁻¹[1.42 × 0.5] - 60°
D = 2sin⁻¹[0.71] - 60°
D = 2(45.23°) - 60°
D = 90.46° - 60°
D = 30.46°
D ≅ 30.5°
When a ball rolls down a hill, what energy conversion occurs?
A. Kinetic energy to potential energy
ОО
B. Chemical energy to kinetic energy
c. Potential energy to kinetic energy
D. Potential energy to chemical energy
Answer:
C
Explanation:
When a ball rolls down a hill, Potential Energy Conversion takes place to Kinetic Energy.
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
Answer:
[tex]\lambda=1.39\times 10^{-4}\ s^{-1}[/tex]
Explanation:
Given that,
The half-life of Barium-139 is [tex]4.96\times 10^3[/tex]
A sample contains [tex]3.21\times 10^{17}[/tex] nuclei.
We need to find the decay constant for this decay. The formula for half life is given by :
[tex]T_{1/2}=\dfrac{0.693}{\lambda}\\\\\lambda=\dfrac{0.693}{T_{1/2}}[/tex]
Put all the values,
[tex]\lambda=\dfrac{0.693}{4.96\times 10^3}\\\\=1.39\times 10^{-4}\ s^{-1}[/tex]
So, the decay constant is [tex]1.39\times 10^{-4}\ s^{-1}[/tex].
Which pair of magnets has the strongest attraction between them?
A tennis ball of mass of 0.06 kg is initially traveling at an angle of 47o to the horizontal at a speed of 45 m/s. It then was shot by the tennis player and return horizontally at a speed of 35 m/s. Find the impulse delivered to the ball.
Answer:
The impulse delivered to the ball is [tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex].
Explanation:
By Impulse Theorem, the motion of the tennis ball is modelled after the following expression:
[tex]Imp = m\cdot (\vec v_{f} - \vec v_{o})[/tex] (1)
Where:
[tex]m[/tex] - Mass of the ball, in kilograms.
[tex]\vec v_{o}[/tex] - Vector of the initial velocity, in meters per second.
[tex]\vec v_{f}[/tex] - Vector of the final velocity, in meters per second.
[tex]Imp[/tex] - Impulse, in meters per second.
If we know that [tex]m = 0.06\,kg[/tex], [tex]\vec v_{o} = \left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})[/tex] and [tex]\vec v_{f} = \left(35\,\frac{m}{s} \right)\cdot (-1, 0)[/tex], then the impulse delivered to the ball is:
[tex]Imp = (0.06\,kg)\cdot \left[\left(35\,\frac{m}{s} \right)\cdot (-1,0) -\left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})\right][/tex]
[tex]Imp = (0.06\,kg)\cdot (-65.670, -32.911)\,\left[\frac{m}{s} \right][/tex]
[tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex]
The impulse delivered to the ball is [tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex].
Need an answer in hurry u can make the pic big
Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds of 5.00 m/s, four have speeds of 8.00 m/s, three have speeds of 9.00 m/s, and two have speeds of 15.0 m/s. Find:
a. the average speed
b. the rms speed
c. the most probable speed of these particles.
Answer:
a) [tex]V=7.5m/s[/tex]
b) [tex]rms=8.4m/s[/tex]
c) Generally the most probable speed is 8m/s as it the most posses by particles being the average
Explanation:
From the question we are told that:
Sample size N=15
[tex]Speed 1 v_1=2m/s\\\\Speed 2 v_2=3m/s\\\\Speed 3 v_2=5m/s\\\\Speed 4 v_4=8m/s\\\\Speed 3 v_5=9m/s\\\\Speed 2 v_6=15m/s\\\\[/tex]
Generally the equation for Average speed is mathematically given by
[tex]V_{avg}=\frac{\sum(nv)}{N}[/tex]
Therefore
[tex]V_{avg}=\frac{(2+2(3)+3(5)+4(8)+3(9)+2(15))}{15}[/tex]
[tex]V=7.5m/s[/tex]
b)
Generally the equation for RMS speed of the particle is mathematically given by
[tex]rms=\sqrt{\frac{\sum(nv^2)}{N}}[/tex]
[tex]rms=\sqrt{\frac{2^2+2(3)^2+3(5)^2+4(8)^2+3(9)^2+2(15)^2}{15}}[/tex]
[tex]rms=\sqrt{69.73}[/tex]
[tex]rms=8.4m/s[/tex]
c
Generally the most probable speed is 8m/s as it the most posses by particles being the average
You have just landed on Planet X. You take out a ball of mass 101 g , release it from rest from a height of 16.0 m and measure that it takes a time of 2.91 s to reach the ground. You can ignore any force on the ball from the atmosphere of the planet. How much does the ball weigh on the surface of Planet X?
Answer:
0.3817 N
Explanation:
Remark
One thing is certain: the ball has a mass of 101 grams wherever it is in the universe. That is not true of the force. The force on the moon is a whole lot less than it is on earth, and maybe planet x as well.
Givens
m = 101 g
vi = 0 That's what at rest means.
t = 2.91 s
d = 16 m
F= ?
Formulas
d = vi*t + 1/2*a * t^2
Force = m * a
Solution
16 = 0 + 1/2 a * 2.91^2
16 = 4.234 a Divide by 4.234
16/4.234 = a
a = 3.779
F = m * a
a = 3.779
m = 101 g = 1 kg / 1000 grams
m = 0.101 kg
F = 0.101 * 3.779
F = 0.3817N