13a value of the hanging mass does the block accelerate at 1.8 m/s2
For 8 kg mass,
m 1 g - T = m 1 a
8g - T = 8a —————(I)
For 5 kg mass,
T - m 2 g = m 2 a
T - 5g = 5a —————(II)
Adding both equations, we get
8g - 5g = 13a
On a surface with no friction, is there acceleration?The constant velocity of the body on the frictionless surface is represented by the frictionless surface acceleration. No net force on the moving object can stop it from going when the surface is frictionless. As a result, the body keeps moving at the same speed. The skater has no net acceleration if no forces are operating on him and he is on a frictionless surface. The acceleration must be zero since the skier must have mass. If there is no acceleration when an item is moving, its velocity stays constant.
We must understand that acceleration always occurs in areas where there is friction. There are several types of forces at work on moving objects.
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a 0.230-kg block along a horizontal track has a speed of 1.40 m/s immediately before colliding with a light spring of force constant 3.75 n/m located at the end of the track. (a) what is the spring's maximum compression if the track is frictionless? m (b) if the track is not frictionless, would the spring's maximum compression be greater than, less than, or equal to the value obtained in part (a)? greater less equal
The spring's maximum compression is 0.346 when mass of the block is 0.230 Kg and Speed is 1.40 m/s.
given,
mass of block = 0.230 Kg
speed = 1.40 m/s
spring constant = k = 3.75 N/m
using conservation of energy
a) K.E = P.E
1/2mv^2=1/2kx^2
1/2*0.23*1.40^2=1/2*3.75*x^2
X^2=0.120
X=0.346
b) The maximum compression will be equal to the compression in part an if the track is not friction-less.
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a uniformly charged conducting sphere of 1.22 m radius has a surface charge density of 8.13 c/m*. (a) find the charge on the sphere. (4) what ts the total electric flux leaving the surface of the sphere? (c) calculate the electric field at the surface of the sphere.
We select a spherical-shaped Gaussian surface that is concentric with the conducting sphere and has a little bigger radius. Gauss's law determines the flux:
ϕ= ϵ 0 q \s = 8.85×10 −12 C 2 /N.m 2
3.66×10 −5 C =4.1×10 6 N.m 2 /C
What makes it the Gauss law?Gauss' Law, which bears the name of German physicist Carl Friedrich Gauss, is the third of Maxwell's four equations. According to Gauss' Law, an electric charge, qv, or static electricity, produces an electric field, E. (voltage). According to this equation, adding more charge will result in a larger electric field.
According to Gauss's Law, the flux of an electric field through a closed surface equals the contained charge divided by a constant.
Gauss's Law may be used to determine the field due to four different types of equally charged objects: a straight wire, an infinite plate sheet, a thin spherical shell, and a low evenly charged sphere.
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Equal masses are suspended from two separate wires made of the same material. The wires have identical lengths. The first wire has a larger cross-sectional area than the second wire. Which wire will stretch the least?.
The original lengths of both wires are equal, the thicker wire will stretch less as a result since Young's Modulus predicts that its original length will change less.
Young's modulus is a measurement of a material's capacity to endure changes in length when subjected to compression or tension along its length. Young's modulus, also known as the modulus of elasticity, is determined by dividing the longitudinal stress by the strain. i.e.Young's modulus = stress/strain
We can assume that the thicker wire must have less strain if it suffers less stress than the thinner wire because strain is defined as the percentage change in length.
Therefore we can said that, For the Young's modulus to remain constant, which it does, it must also drop correspondingly as strain rises. The thicker wire will stretch less as a result since its original length will change less because both wires started off the same length.
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what is the principal difference between a radio wave and light? what is the principal difference between a radio wave and light? radio waves have significantly higher frequencies than light. a radio wave is a mechanical vibration of matter and light is a vibration of electric and magnetic fields. radio waves have significantly lower frequencies than light. light is a mechanical vibration of matter and radio wave is a vibration of electric and magnetic fields.
The principal difference between a radio wave and light is that radio waves have significantly lower frequencies than light.
Explanation:
Both radio waves and light are electromagnetic waves; their main difference is their frequency. Radio waves have a lower frequency and longer wavelength than visible light waves. Light waves have a lower frequency and longer wavelength than X-rays.
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The skater eventually comes to a stop.
What has happened to all of the skater's
original potential energy?
Neptune has a mass that is about 17 times the mass of earth. the distance between the sun and neptune is about 30.1 times the distance between the sun and earth. if the gravitational force between the sun and earth is 3.5 x 1028 n, which is closest to the force between neptune and the sun? 6 x n 6 x n 6 x n 6 x n
Answer: 6 x 10^26 N
Explanation:
What is the BEAT FREQUENCY when 512 Hz and a 515 Hz tuning forks are sounded at the same time?
The beat frequency is 3 beats/sec.
Beat frequency(n) can be found by using the following formula:
n=f2-f1, where f2 and f1 are the given frequencies
n= 515-512
n= 3 beats/sec
Therefore the beat frequency when 512 Hz and 512Hz tuning forks are sounded together is 3 beats/sec.
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The separation distance between two 2.0 kg masses is decreased by two-thirds. How is the gravitational force between them affected?
Given:
The mass of two objects is: m1 = m2 = m = 2 kg
The distance between the object is decreased by two-thirds
To find:
How on reducing the distance, the gravitational force between them affects them.
Explanation:
Let the distance between two objects each having mass "m" be "r". The gravitational force between them is given as:
[tex]F_1=G\frac{m_1\times m_2}{r^2}[/tex]Here, G is the universal gravitational constant.
Substituting the values in the above equation, we get:
[tex]\begin{gathered} F_1=G\times\frac{2\text{ kg}\times2\text{ kg}}{r^2} \\ \\ F_1=G\times\frac{4\text{ kg}^2}{r^2}..........(1) \end{gathered}[/tex]Now, the distance between the mass is reduced by two-thirds. Thus, the new distance between them will be "R" which is given as:
[tex]R=r-\frac{2}{3}r=r(1-\frac{2}{3})=\frac{r}{3}[/tex]Now, the gravitational force between two masses with their distance of separation reduced by two-thirds is given as:
[tex]F_2=G\frac{m_1\times m_2}{R^2}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} F_2=G\times\frac{2\text{ kg}\times2\text{ kg}}{(\frac{r}{3})^2} \\ \\ F_2=G\times\frac{4\text{ kg}^2}{\frac{r^2}{9}} \\ \\ \begin{equation*} F_2=G\times\frac{9\times4\text{ kg}^2}{r^2} \end{equation*} \\ \\ F_2=9\times(G\times\frac{4\text{ kg}^2}{r^2})..........(2) \end{gathered}[/tex]Substituting equation (1) in equation (2), we get:
[tex]\begin{gathered} F_2=9\times(G\times\frac{4\text{ kg}^2}{r^2}) \\ \\ F_2=9F_1 \end{gathered}[/tex]From the above equation, we observe that the new gravitational force F2 between two masses when their distance of separation is reduced by two-thirds will be nine times that of the original value of gravitational force F1.
Final answer:
The new gravitational force F2 between two masses when their distance of separation is reduced by two-thirds will be nine times that of the original value of gravitational force F1.
A 10 kg box is pulled across a level floor, where the coefficient of kinetic friction is
0.35. What horizontal force is required for an acceleration of 2.0 m/s2?
A 10 kg box is pulled across a level floor, where the coefficient of kinetic friction is 0.35. The horizontal force is required for an acceleration of 2.0 m/s2 will be 54.3 N
Newton’s second law states that the acceleration of an object depends upon two variables – the net force acting on the object and the mass of the object. The acceleration of the body is directly proportional to the net force acting on the body and inversely proportional to the mass of the body.
equation of motion will be
F(net) = mass * acceleration
F - fr = m * a
where
F = force by which the box is moving
fr = force due to friction
m = mass of the object
a = acceleration by the object is moving
F = fr + m * a
= (mu * N) + ma where N = mg
= (mu * mg) + ma
= (0.35 * 10 * 9.8) + 10 * 2
= 54.3 N
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Study the graph shown below. It is a graphic of a car trip taken by Amber and Ken. The starting position (0,0) is their home.
Graph of position versus time, with position on the y axis from 0 to 75 miles and time on the x axis from 0 to 8.5 hours. Segment 1 goes from 0,0 to 1,30. Segment 2 goes from 1,30 to 3,50. Segment 3 goes from 3,50 to 4,50. Segment 4 goes from 4,50 to 5,70. Segment 5 goes from 5,70 to 6,70. Segment 6 goes from 6,70 to 7,30. Segment 7 goes from 7,30 to 8,0.
How far had they traveled after four hours of travel?
a
30 miles
b
40 miles
c
50 miles
d
60 miles
The distance that they had traveled after four hours is given by:
c 50 miles.
How to find the distance traveled?We are given the a set of points (x,y) in this problem, in which the input and the output are defined as follows:
Input: x is the number of hours.Output: y is the distance traveled, in miles, after a time of x hours.The endpoints of the segments are as follows:
(0,0): Initially, 0 miles were traveled.(1,30): 30 miles were traveled in one hour.(3, 50): 50 miles were traveled in three hours.(4,50): 50 miles were traveled in four hours.Hence, the distance that they had traveled after four hours is given by 50 miles, and option c is the correct option in the context of this problem (how long they had traveled).
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Find the acceleration of the 1kg block m the following diagram 5n 4n 5n
Explanation:
56777^7744 hold the 4 nd add 2
32. heavy rain reduces your ability to see and be seen; in daytime turn on your a. windshield wipers. b. low beam headlights. c. windshield defroster. d. all of the above
Heavy rain reduces your ability to see and be seen. So in daytime turn on all of the above ( d ) mentioned options.
Windshield wipers and windshield defroster are for the ability to see and the low beam headlights are for the ability to be seen by others. If it is night time drive below the speed limit because it is extremely hard to maneuver under those conditions.
Windshield wipers wipes the water off your windshield consistently. Low beam headlights indicate the presence of your car. Windshield defroster condenses the frosted windshield.
Therefore, in daytime turn on your windshield wipers, low beam headlights and windshield defroster.
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light enters glass from air. the angle of refraction will be: group of answer choices less than the angle of incidence. equal to the angle of incidence. greater than the angle of incidence
Less than the angle of incidence, the angle of refraction will exist.
What is angle of refraction?Angle of refraction refers to the angle formed by the refracted ray and the normal at the point of incident.
In a mirror, the angle of incidence and angle of reflection are the same. If the incident ray falls along the normal, the angle of incidence for a plane mirror is 0 degrees rather than 90 degrees.
The behavior of light is described by Fermat's principle, which states that a light ray always chooses the shortest route to its destination. On reflection, the exact same behavior is seen. The angle of incidence and the angle of reflection are equal when a ray is reflected from a plane.
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The deepest section of ocean in the world is the Mariana Trench, located in the Pacific Ocean. Here,the ocean floor is as low as 10,918m below the surface. If the index of refraction of water is 1.33, howlong would it take a laser beam to reach the bottom of the treach?
Given:
The depth of the trench, d=10,918 m
The refractive index of the water, n=1.33
To find:
The time it takes for the laser beam to reach the bottom of the ocean.
Explanation:
The refractive index of water is given by,
[tex]n=\frac{c}{v}[/tex]Where c is the velocity of the laser in vacuum and v is the velocity of the laser in water.
On substituting the known values,
[tex]\begin{gathered} 1.33=\frac{3\times10^8}{v} \\ \implies v=\frac{3\times10^8}{1.33} \\ =2.3\times10^9\text{ m/s} \end{gathered}[/tex]The velocity is given by the equation,
[tex]v=\frac{d}{t}[/tex]Where t is the time it takes for the laser to reach the bottom of the trench.
On substituting the known values,
[tex]\begin{gathered} 2.3\times10^8=\frac{10,918}{t} \\ \implies t=\frac{10,918}{2.3\times10^8} \\ =47.5\times10^{-6}\text{ s} \\ =47.5\text{ }\mu\text{s} \end{gathered}[/tex]Final answer:
The laser beam would reach the bottom in 47.5 μs
i would appreciate it if you could help me if would help a bunch on getting caught up with my work.
2. Experiment:
We need to make a circuit to test which of the materials are conductors and insulators.
Build the circuit as shown below.
If the light bulb is on then it is a conductor otherwise it is an insulator.
Keep replacing the material one by one to test all the materials.
alena is performing a floor routine. in a tumbling run she spins through the air, increasing her angular velocity from 43.00 to6 5.00 rev/s while rotating through one-half of a revolution. how much time does this maneuver take?
The tumbling run by Alena lasted for 0.009 seconds.
The time of tumbling can be calculated using the formula -
time = (2 × theta) ÷ (initial angular velocity + final angular velocity)
Keep the values in formula to find the amount of time took while tumbling.
Time = (2 × 0.5) ÷ (43 + 65)
Performing multiplication in numerator and addition in denominator of Right Hand Side of the equation
Time = 1 ÷ 108
Performing division on Right Hand Side of the equation
Time = 0.009 seconds
Thus, Alena took 0.009 seconds in tumbling while performing floor routine.
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a boy and a girl are at rest on separate skis 12 m apart on a horizontal icy surface. the boy and his skis have twice the inertia of the girl and her skis. determine the location of the center of mass in meters measured from the position of the girl.
With the position of the girl as a reference point, the center of mass is 8 m from the girl position.
If there are two objects as shown on the attached picture, the center of mass is:
xc = m₁ / (m₁ + m₂) . x₁ + m₂ / (m₁ + m₂) . x₂
From the problem, we know that the boy and his skis have twice the inertia of the girl and her skis. We can replace the mass in the equation with inertia, hence:
xc = I₁ / (I₁ + I₂) . x₁ + I₂ / (I₁ + I₂) . x₂
Let:
I₁ = the girl's inertia
I₂ = the boy's inertia
Hence,
I₂ = 2 I₁
Let make the position of the girl as a reference point, then,
x₁ = 0
x₂ = 12
Plug these parameters into the formula:
xc = I₁ / (I₁ + I₂) . 0 + 2 I₁ / (I₁ + 2 I₁) . 12
xc = 2/3 . 12 = 8 m
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Isaac is practicing his volleyball skills by volleying a ball straight up and down, over and over again. His teammate
Marie notices that after one volley, the ball rises 3.6 m above Isaac's hands. What is the speed with which the ball left
Isaac's hand? (8.4 m/s)
Answer:
the velocity of a ball at highest point is 8.5m/s
Explanation:
: The Freefall Problem
Xue,is standing on the roof of a building. Emily is standing below and tosses a ball straight upwards to Xue. It travels up past
Xue, comes back down and Xue reaches out and catches it. Xue catches the ball 6.0 m above Emily’s hands. The ball was
travelling at 12.0 m/s upwards the moment it left Emily’s hand. We would like to know how much time this trip takes.
1. Represent. Complete part A below. Indicate the y-origin for position measurements and draw a sign convention where
upwards is positive. Label the important events.
2. Represent. Complete part C below. Make sure the two graphs line-up vertically. Draw a single dotted vertical line
through the graphs indicating the moment when the ball is at its highest.
A: Pictorial Representation
Sketch, coordinate system, label givens, conversions, describe events
Event 1:
Event 2:
Event 3:
C: Physics Representation
Motion diagram, motion graphs, key events
The total length of the path traveled by an object is the distance. The change in position, from one event to another is the
displacement. Distance is a scalar quantity and displacement is a vector quantity.
3. Reason. Explain why this in this example it is relatively easy to find the displacement, but harder to find the distance.
4. Reason. The BIG 5 equations are valid for any interval of motion where the acceleration is uniform. Does the ball
accelerate uniformly during events 1 and 3? Explain D: Mathematical Representation
Describe steps, complete equations, substitutions with final statement isThere are multiple ways of solving…one
will require the use of the quadratic formula.
The other way will take additional steps.
x = -b+-under root b^2-4ac/2a
What is the magnitude of the velocity of the plane
We will have the following:
First, we are told that the wind accelerates directly northwest, that means that the angle of inclination is 45°, so the following is true:
[tex]\begin{gathered} \sum F_x=115-75cos(45) \\ \\ \sum F_y=75sin(45) \end{gathered}[/tex]Then:
[tex]\begin{gathered} v_f=\sqrt{(115-75cos(45))^2+(75sin(45))^2}\Rightarrow v_f=81.56229536... \\ \\ \Rightarrow v_f\approx81.6 \end{gathered}[/tex]So, the magnitude of the final velocity is approximately 81.6 m/s.
a pipe that is 0.46 m long and open at both ends vibrates in the second overtone with a frequency of 1150 hz. in this situation, the distance from the center of the pipe to the nearest antinode is closest to group of answer choices zero. 7.7 cm. 15 cm. 12 cm. 3.8 cm.
The distance of the center pipe to the nearest antinode at the second harmonic frequency is closest to 12 cm.
We need to know about the harmonic frequency to solve this problem. When the string is vibrating at its second harmonic frequency, it should follow the rule
λ = L
where λ is wavelength and L is the length of the string
From the question above, we know that
f = 1150 Hz
L = 0.46 cm
Find the wavelength
λ = L
λ = 0.46 m
The distance from the center of the pipe to the nearest antidote is equal to 1/4 of the wavelength. Hence,
s = 1/4 λ
s = 1/4 . 0.46
s = 0.115 m
s ~ 12 cm
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a 60- kg mountain climber moves 10 m up a vertical slope. if the muscles in her body convert chemical energy into gravitational potential energy with an efficiency of no more than 5%, what is the chemical energy used to climb the slope?
120000 Joules of Chemical Energy is used by the climber to climb the rope.
As the climber is moving up, its gravitation potential energy energy (G) will be given by,
Gravitational potential Energy = MgH
G = MgH
Where,
M is mass of climber which is 69 Kg,
g is acceleration due to gravity which is 10m/s².
H is the height above the earth's surface which is 10 meters.
It is provided that the chemical energy is converted to gravitational energy at a rate of 5%.
So, we can write,
5% Chemical energy (C) = Gravitational potential energy (G)
So,
5%C = MgH
Putting all the values,
5C/100 = 60(10)(10)
C = 120000 Joules.
The used chemical energy is 120000 Joules.
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a propeller is modeled as five identical uniform rods extending radially from its axis. the length and mass of each rod are 0.711 m and 2.99 kg, respectively. when the propellor rotates at 569 rpm (revolutions per minute), what is its rotational kinetic energy?
The rotational kinetic energy of five identical uniform rods is 9436.21 joule.
We need to know about rotational kinetic energy to solve this problem. Rotational kinetic energy can be defined as the energy of a rotating object. The magnitude of rotational energy is
KE = 1/2 . I . ώ²
where KE is rotational kinetic energy, I is inertia and ώ is angular velocity.
From the question above, the given parameters are
L = 0.711 m
m = 2.99 kg
N = 5
ώ = 569 rpm = 59.59 rad/s
Find the total inertia of five identical rods
I = N . 1/2 . m . L²
I = 5 . 1/2 . 2.99 . 0.711
I = 5.31 kg.m²
Calculate the rotational kinetic energy
KE = 1/2 . I . ώ²
KE = 1/2 . 5.31 . 59.59²
KE = 9436.21 joule
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HELP!
A box is sitting on a frictionless surface when a net force of 15 N is applied to the right. If the acceleration of the box is 0.33 m/s2 to the right, what is the mass of the box?
a
2.0 kg
b
5.0 kg
c
18 kg
d
45 kg
Answer:
F=m × a - answer is D
Explanation:
F=15N
m=m
a=0.33
15÷0.33=45.45...
How long would it take an airplane to travle 1089 miles if its flying at a speed of 375 miles per hour
Answer:
2.904
Explanation:
It would take an airplane 2.904 hours to travel 1089 miles.
As a fraction, this would equal 363/125
As a percentage, this would equal 290.4%
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A football player kicks a football off a tee with a speed of 21 m/s at an angle of 51°.
a. What is the horizontal component of the velocity?
b. What is the vertical component of the velocity?
c. How long does it take the ball to reach its highest point?
d. What is the maximum height of the ball?
e. What is the total amount of time that the ball is in the air?
f. How far is the ball from the football player when it lands?
A football player kicks a football off a tee with a speed of 21 m/s at an angle of 51°, then the horizontal component of the velocity would be 15.58 m/s and the vertical component of the velocity would be 14.07 m/s.
What is a projectile motion?
It can be defined as the motion of any object or body when thrown from the earth's surface and follows any curved path under the effect of the gravitational force of the earth.
The horizontal component of the velocity = 21 × cos51 = 15.58 m/s
The vertical component of the velocity = 21 × sin51 = 14.07 m/s
Thus, the horizontal component of the velocity would be 15.58 m/s and the vertical component of the velocity would be 14.07 m/s.
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The atomic number of aluminum is 13. What is the correct designation of the electron configuration of aluminum?.
Aluminum has the following electron configuration: 1s2, 2s2, 2p6, 3s2, and 3p1.
According to the information supplied, aluminum has an atomic number of 13. That makes it abundantly evident that the nucleus of any atom that is an isotope of aluminum will contain 13 protons. The proton number will be equal to the number of electrons surrounding the nucleus. There will be 13 electrons as a result.
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If a car can go from 0-69mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if it’s starting speed was 50mi/hr
If a car can go from 0 - 69 mi/hr in 8.0 seconds, then the final speed after 5.0 seconds would be 41.6 m/s, if its starting speed was 50 mi/hr.
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.
The acceleration of the car If a car can go from 0-69mi/hr in 8.0 seconds,
69 miles / hour = 69 × 1609 / 3600
= 30.83 m / s
v = u + at
30.83 = 0 + a × 8.0
a = 30.83 / 8.0
= 3.85 meters / second²
50 miles per hour = 50 × 1609 / 3600
= 22.35 m / s
The velocity of the car after 5 seconds,
v = 22.35 + 3.85 × 5
= 41.6 m / s
Thus, the velocity of the car after 5 seconds would be 41.6 m / s.
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larry uses a lever to list a box. where does inefficiency occur in the lever
Answer:
low internal resistance.
Explanation:
A lever has a high efficiency due to the fact that it has low internal resistance. The work it puts out is almost equal to the work it receives, because energy used up by friction is quite small. On the other hand, an a pulley might be relatively inefficient due to a considerably greater amount of internal friction.
Find the total translational kinetic energy of0.5 L of oxygen gas held at a temperature of3◦C and a pressure of 1.5 atm.Answer in units of J.
Given
Volume of the oxygen gas is
[tex]V=0.5L=0.0005m^3[/tex]The temperature is
[tex]T=3^oC[/tex]Pressure is
[tex]\begin{gathered} P=1.5atm \\ \Rightarrow P=1.5\times101325=151,987.5Pa \end{gathered}[/tex]To find
The translational kinetic energy
Explanation
The translational kinetic energy is given by
[tex]K=\frac{3}{2}nRT[/tex]For ideal gas,
[tex]PV=nRT[/tex]Thus,
[tex]\begin{gathered} K=\frac{3}{2}PV \\ \Rightarrow K=\frac{3}{2}\times151987.5\times0.0005 \\ \Rightarrow K=113.99J \end{gathered}[/tex]Conclusion
The translational kinetic energy is 113.99J
Four different balls with different masses (measured in grams) were dropped from the same height on Earth. Assuming no other forces acted on the balls, match each ball with the gravitational force (measured in newtons) it experienced as it fell.gravitational force = 9.8 m/s2
We will have the following:
1. 143 g = 0.143 kg:
[tex]F=(0.143kg)(9.8m/s{}^2)\Rightarrow F=1.4014N[/tex]2. 110 g = 0.110 kg:
[tex]F=(0.110kg)(9.8m/s^2)\Rightarrow F=1.078N[/tex]3. 410 g = 0.410 kg:
[tex]F=(0.410kg)(9.8m/s^2)\Rightarrow F=4.018N[/tex]4. 46 g = 0.046 kg:
[tex]F=(0.046kg)(9.8m/s^2)\Rightarrow F=0.4508N[/tex]From this we can see that the given weights do not correspond to any of the values of column B, since the calculations to obtain the values of column B were done using grams, instead of kg, thus those values do not correspond to the unit of force Newtons.