The correct option is B) fat molecules contain a lower percentage of hydrogen atoms. Fat molecules, also known as triglycerides, are composed of three fatty acid chains attached to a glycerol backbone. These fatty acid chains are made up of long chains of carbon and hydrogen atoms, with some oxygen atoms as well.
The term "adipose atoms" does not make sense as adipose refers to fat tissue, not a specific element or atom.The percentage of hydrogen atoms in a fat molecule varies depending on the type of fatty acid chain present. Saturated fatty acids, which have no double bonds between their carbon atoms, contain the maximum number of hydrogen atoms possible and therefore have a higher percentage of hydrogen atoms.
Unsaturated fatty acids, on the other hand, have one or more double bonds between their carbon atoms, which reduces the number of hydrogen atoms and lowers the percentage of hydrogen in the molecule.
Overall, fat molecules contain more carbon and hydrogen atoms than oxygen atoms, which is why they are classified as lipids. They serve as a storage form of energy in the body and also play important roles in cell membrane structure and function, hormone production, and insulation .The correct option is B
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Fat molecules contain a lower percentage of oxygen atoms. The correct option is A)
This is because carbohydrates and proteins contain more oxygen atoms in their molecules compared to fats, which have more carbon and hydrogen atoms. The presence of more oxygen atoms in carbohydrates and proteins results in lower energy density compared to fats.
A) Fat molecules contain a lower percentage of oxygen atoms.
Fats contain more energy per gram than carbohydrates or proteins because fat molecules have a lower percentage of oxygen atoms, which leads to a higher proportion of carbon and hydrogen atoms. These carbon-hydrogen bonds store more energy than the bonds found in carbohydrates or proteins.
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What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×10^4v ?
John is talking to his friend about painting a fence. He states that if he uses a roller brush he will cover the fence faster than if he uses a bristle brush. This is an example of a a. Law b. Belief c. Theory d. Hypothesis
Answer:
b
Explanation:
Your best friend weighs 81.5 kg and is a rugby player. In one of his games, he slides to a stop in a phenomenal manner. The coefficient of kinetic friction between the player and the ground is 0.70. His speed at the start of the slide is 8.23 m/s. I a) Calculate his acceleration during the slide. b) How long (in time) does he slide until he stops?
A. The acceleration during the slide is 6.86 m/s²
B. The time taken to slide until he stops is 1.2 s
How to determine the force of frictionMass (m) = 81.5 KgCoefficient of friction (μ) = 0.7Acceleration due to gravity (g) = 9.8 m/s²Normal reaction (N) = mg = 81.5 × 9.8 = 798.7 NFrictional force (F) =?F = μN
F = 0.7 × 798.7
F = 559.09 N
A. How to determine the accelerationMass (m) = 81.5 KgFrictional force (F) = 559.09 NAcceleration (a) =?a = F / m
a = 559.09 / 81.5
a = 6.86 m/s²
B. How to determine the time Initial velocity (u) = 8.23 m/sFinal velocity (v) = 0 m/sDecceleration (a) = -6.86 m/s²Time (t) =?a = (v – u) / t
t = (v – u) / a
t = (0 – 8.23) / -6.86
t = 1.2 s
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Two 5000-kg passenger cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track. They collide, couple, and roll away together with a combined momentum of
The combined momentum of the passengers is 5000 kgm/s.
Combined momentum of the passengerThe combined momentum of the passengers is calculated as follows;
P = mv1 + mv2
where;
m is mass of the passengersv1 is velocity of the first passengerv2 is velocity of the second passengerP = m(v1 + v2)
P = 5000(-1 + 2)
P = 5000 kgm/s
Thus, the combined momentum of the passengers is 5000 kgm/s.
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A fireworks shell is accelerated from rest to a velocity of 60.0 m/s over a distance of 0.200 m. How long did the acceleration last
The acceleration lasted for 0.007s.
To find the answer, we need to know about the Newton's equation of motion.
What's the Newton's equation of motion that relates initial velocity, final velocity, acceleration, time and distance?Newton's equations are
V²-U²= 2aSV²-U²= 2aSV= U+atV= final velocity, U= initial velocity, a= acceleration, t= time and S = distance
What's the acceleration, if the fireshell is accelerated from rest to a velocity of 60.0 m/s over a distance of 0.200 m?Here, U= 0m/s, V= 60.0 m/s, S= 0.200 m60²-0= 2a×0.2= 0.4a
a= 3600/0.4= 9000m/s²What's the time taken by the fireshell to achieve 60m/s with 9000m/s² acceleration from rest?Here, U= 0m/s, V = 60.0 m/s, a= 9000m/s²60= 0+9000tt= 60/9000= 0.007 sThus, we can conclude that the acceleration of the fireshell lasted for 0.007s.
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Based on illustrations of magnetic field lines, where could an object be placed so it would not experience a magnetic force
Halfway between the like poles of two magnets, because the field lines bend away and do not enter this area.
How does a magnetic field diagram show where the field is strongest?
The magnetic field lines do not ever cross. The lines include arrowheads to indicate the direction of the force exerted by a magnetic north pole. The closer the lines are to the poles, the stronger the magnetic field (thus the magnetic field from a bar magnet is highest closest to the poles).Where is magnetic field the strongest and weakest on a magnet?
The bar magnet's magnetic field is strongest at its core and weakest between its two poles. The magnetic field lines are densest immediately outside the bar magnet and least dense in the core.Which two locations on the magnet would have the greatest attractive forces?
Inside the magnet itself, the field lines run from the south pole to the north pole. The magnetic field is strongest in areas of greatest density of magnetic field lines, or areas of the greatest magnetic flux density.Learn more about magnetic field
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in the photoelectric effect a photon with an energy of 5.3 * 10^-19 J strikes an electron in a metal
The velocity of the photo electron is 6.11 × 105 ms
Given :
Supplied energy, Es = 5.3 x 10-19 J
Minimum energy of the electron to escape from the metal, E. = 3.6 x 10-19 J
To Find :
Velocity of photoelectron
Solution : The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So
5.3 × 10^-19 J = 3.6 × 10^-19 J + K
K = 5.3 x 10^-19 - 3.6 x 10^-19
K = 1.7 × 10^-19 J
The formula of kinetic energy is given by:
K = 1/2 mv^2
v = √2K/m
= √2 x 1.7 x 10^-19
√ 9.1 x 10^-31
v = 6.11 x 10^5 m/s
So, the velocity of the photo electron is 6.11 x 10^5 m/s
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Although centuries ago, astronomers thought that a nova was a new star, appearing for the first time in the heavens, today we know that it is:
Astronomers thought that a nova was a new star, appearing for the first time in the heavens, today we know that it is as a binary star system.
What is the binary star system about?A binary star system is known to be one where one star is known to be called a white dwarf and there is a mass that is said to be transferred to it
A binary star is known to be a kind of a system that is composed of two stars that are known to be gravitationally held together to and in orbit near each other.
Note that Binary stars in the night sky are ones that are often seen as a single object and thus Astronomers thought that a nova was a new star, appearing for the first time in the heavens, today we know that it is as a binary star system.
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A certain frictionless simple pendulum having a length l and mass m swings with period t. If both l and m are doubled, what is the new period?.
If l and m both are doubled then the period becomes √2*T
what is a simple pendulum?
It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.
A pendulum is a weight suspended from a pivot so that it can swing freely.
Here,
A certain frictionless simple pendulum having a length l and mass m
mass of pendulum = m
length of the pendulum = l
The period of simple pendulum is:
[tex]T = 2\pi \sqrt{\frac{l}{g} }[/tex]
Where k is the constant.
Now the length and mass are doubled,
m' = 2m
l' = 2l
[tex]T' = 2\pi \sqrt{\frac{2l}{g} }[/tex]
[tex]T' = \sqrt{2}* 2\pi \sqrt{\frac{l}{g} }[/tex]
[tex]T' = \sqrt{2} * T[/tex]
Hence,
If l and m both are doubled then the period becomes √2*T
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Rank in order, from largest to smallest, the magnitudes of the momentum p1x through p5x of the objects presented in (Figure 1). Rank the magnitudes of the momentum from largest to smallest. To rank items as equivalent, overlap them.
The decreasing order of the momenta are b>a=d=e>c.
What is Momentum?Momentum is the result of a particle's mass and velocity. Being a vector quantity, momentum possesses both magnitude and direction.According to Isaac Newton's second equation of motion, the force acting on the particle equals the time rate of change of momentum.According to Newton's second law, if a particle is subjected to a constant force for a specific amount of time, the result of the force and time (referred to as the impulse) is equal to the change in momentum. On the other hand, a particle's momentum represents the length of time needed for a consistent force to bring it to rest.Solution: The momentum of all the options are mentioned below according to the formula:
a. p = 20 × 1 = 20gm/s
b. p = 20 × 2 = 40gm/s
c. p =10 × 1 = 10gm/s
d. p = 10 × 2 = 20 gm/s
e. p =200 × 0.1 = 20gm/s
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Question:
I hope you are trying to ask the question below:
Rank in order, from largest to smallest, the momenta a to e of the objects
a. 20 g & 1 m/s
b. 20 g & 2 m/s
c. 10 g & 1 m/s
d. 10 g & 2 m/s
e. 200 g & 0.1 m/s
A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction acting on the box is a constant 20 N. How much work is done by the 50 N force?
The work is done by the 50 N force is 200 J.
Work done by the 50 N forceWork done = Fd
where;
F is Force applied d is the displacement of the objectWork done = 50 N x 4 m = 200 J
Thus, the work is done by the 50 N force is 200 J.
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20. Q: How long will it take for an apple falling from a 29.4m-tall tree to hit the ground?
A: 1.56 s
B: 2.04 s
C: 2.45 s
D: 3.72 s
Answer:
2.45s
Explanation:
is explanation needed too?
A person stands on the ground with a slingshot and tries to hit a high limb on a nearby tree. The limb is 18.3 feet above the ground. What is the minimum initial velocity in meters/second of a rock leaving the slingshot at 6 feet above the ground that would be able to reach the limb
The minimum initial velocity of the rock is V = 8.56 m/sec.
What is Conservation of Energy?According to the rule of conservation of energy, energy can only be transformed from one form of energy to another and cannot be created or destroyed. This indicates that unless energy is added from the outside, a system always has the same quantity of energy.
Calculate the rock's minimal initial velocity in meters/second as it exits the slingshot at a distance of 6 feet [tex]$\left(h_{2}=6\right.ft.)[/tex] above the ground.
Using the provided information, the net height of the limb is calculated as shown below.
[tex]$$h &=h_{1}-h_{2}$$[/tex]
h = 18.3 - 6
h = 12.3
Now, using the conservation of energy the minimum initial velocity of the rock is computed as shown below,
Kinetic energy of the rock = Potential energy the rock
[tex]$$\begin{aligned}\frac{1}{2} m V^{2} &=m g h \\\frac{1}{2} V^{2} &=g h \\\frac{1}{2} V^{2} &=\left(32.1 \mathrm{ft} / \mathrm{s}^{2}\right)(12.3 \text { feet }) \\V &=\sqrt{2\left(32.1 \mathrm{ft} / \mathrm{s}^{2}\right)(12.3 \mathrm{feet})}\end{aligned}$$[/tex]
V = 28.10 [tex]\frac{ft}{s}[/tex].
The minimal initial velocity in meter/second is 8.56.
Therefore, the minimum initial velocity of the rock is 8.56 m/sec
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An engine has an input of heat energy of 10,750 J and does 2,420 J of work. Which of the following is the heat loss?
Group of answer choices
a. 13,200 J
b. 8,330 J
c. 4.44 J
c. 0.225 J
B. The heat loss by the heat engine is 8,330 J.
Heat loss by the heat engine
The heat loss by the heat engine is calculated as follows;
H = E - W
where;
H is heat lossE is input energyW is work doneH = 10,750 J - 2,420 J
H = 8,330 J
Thus, the heat loss by the heat engine is 8,330 J.
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what is kinematics ?
explain ~
[tex] \\ \\ [/tex]
tysm! :)
Answer:
Kinematics is the study of the motion of mechanical points, bodies and systems without consideration of their associated physical properties and the forces acting on them. The study is often referred to as the geometry of motion, and it models these motions mathematically using algebra
Answer:
the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion is kinematics
A kettle transfers 6,000j of energy electrically. 1,500j of this is wasted. what is the efficiency of this kettle?
Answer:75 percent
Explanation:so in order tro fin d thge efficiency i used the forumla ,efficency=useful output energy/input energyx100%,in order to use this formula i needed the output,which i found by subtracting the input energy with wasted energy,that gave me the output,and after founding the output,i put that into the formula,
output energy=input energy - wasted energy
output energy=6000j-1500j
output energy=4500
put that into the formula
efficiency =output energy/input enrgy x100%
efficiency=4500/6000 multiplied by 100%
efficiency=0.75x100%
efficiency=75%
Which type of galaxy is likely to contain only o-spectral type stars?
How much power will be required to force a current of 4.13 amps to flow through a conductor whose resistance is 113 ohms? Use two decimals for your answer. Round your answer to two decimals.
The power required to force the current of 4.13 A to flow through the conductor is 1927.43 watts
What is power?This is defined as the rate in which energy is consumed. Electrical power is expressed mathematically as:
Power (P) = square current (I²)× resistancet (R)
P = I²R
How to determine the powerCurrent (I) = 4.13 AResistance (R) = 113 ohmsPower (P) =?P = I²R
P = 4.13² × 113
P = 1927.43 watts
Thus, the power required is 1927.43 watts
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When operating an aircraft at cabin pressure altitudes above 12,500 feet msl up to and including 14,000 feet msl, supplemental oxygen shall be used during?
Supplemental oxygen shall be used in that flight time in excess of 30 minutes at those altitudes.
Most plane cabins are pressurized to 8,000 feet above sea level, an altitude that lowers the amount of oxygen inside the blood by approximately 4 percentage points, researchers say.
The higher the altitude, the less oxygen there is in the air and the lower the overall air pressure is. If flights were not pressurized, passengers could be prone to various physiological illnesses. Because of this, federal policies require that all commercial flights over 8,000 feet be pressurized.
The flight crew must use supplemental oxygen for the entire duration of flight operations above a cabin pressure altitude of 14,000 feet MSL.
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1. A 500 g piece of silver at 250°C is submerged in 1000 g of water at 5°C to be cooled. Determine the
final temperature of the silver and water. Given Cwater = 4180 J/kg°C and Csilver = 240 J/kg°C.
Taking into account the definition of calorimetry, the final temperature of the silver and water is 11.84 °C.
CalorimetryCalorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
So, the equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where:
Q is the heat exchanged by a body of mass m. c is the specific heat substance. ΔT is the temperature variation.Final temperature of the silver and waterIn this case, you know:
For silver:Mass of silver= 500 g= 0.5 kg (being 1000 g= 1 kg)Initial temperature of silver= 250 °CFinal temperature of silver= ?Specific heat of silver= 240 [tex]\frac{J}{kgC}[/tex] For water:Mass of water = 1000 g= 1 kgInitial temperature of water= 5 ºCFinal temperature of water= ?Specific heat of water = 4180 [tex]\frac{J}{kgC}[/tex]Replacing in the expression to calculate heat exchanges:
For silver: Qsilver= 240 [tex]\frac{J}{kgC}[/tex]× 0.5 kg× (Final temperature of silver - 250 C)
For water: Qwater= 4180 [tex]\frac{J}{kgC}[/tex] × 1 kg× (Final temperature of water - 5 C)
If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.
Then, the heat that the silver gives up will be equal to the heat that the water receives. Therefore:
- Qsilver = + Qwater
And the final temperature of the silver is equal to the temperature of the water (Final temperature of silver= Final temperature of water= Final temperature). Then:
- 240 [tex]\frac{J}{kgC}[/tex]× 0.5 kg× (Final temperature - 250 C)= 4180 [tex]\frac{J}{kgC}[/tex] × 1 kg× (Final temperature - 5 C)
Solving:
- 120 [tex]\frac{J}{C}[/tex]× (Final temperature - 250 C)= 4180 [tex]\frac{J}{C}[/tex]× (Final temperature - 5 C)
120 [tex]\frac{J}{C}[/tex]× (250 C - Final temperature) = 4180 [tex]\frac{J}{C}[/tex]× (Final temperature - 5 C)
120 [tex]\frac{J}{C}[/tex]×250 C - 120 [tex]\frac{J}{C}[/tex]×Final temperature = 4180 [tex]\frac{J}{C}[/tex]×Final temperature - 4180 [tex]\frac{J}{C}[/tex]×5 C
30,000 J - 120 [tex]\frac{J}{C}[/tex]×Final temperature = 4180 [tex]\frac{J}{C}[/tex]×Final temperature - 20,900 J
50,900 J= 4300 [tex]\frac{J}{C}[/tex]×Final temperature
50,900 J÷ 4300 [tex]\frac{J}{C}[/tex]= Final temperature
11.84 °C= Final temperature
Finally, the final temperature of the silver and water is 11.84 °C.
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A mountain skier has begun descending the 30o slope. if the coefficient of kinetic friction is 0.10, what is her acceleration?
The acceleration of the body is found to be -4.05 m/s^2
What is the acceleration?We define the acceleration as the rate of change of the velocity with time. Hence, we can write that;
a = (μcosθ - sinθ)g
a = acceleration
μ = coefficient of kinetic friction
θ = angle of inclination
g = acceleration due to gravity
Thus;
a = (0.10cos 30 - sin 30) 9.8
a = -4.05 m/s^2
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As the electric potential near the ground increases during a thunderstorm, a positive charge current can move up pointed objects, such as masts of ships, producing a luminous halo or glow known as ____.
Answer:
St. Elmo's Fire - well known to old time sailors
A proton exits the cyclotron 1. 0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this 1. 0 ms ?.
The diameter of the largest orbit just before the protons exit the cyclotron is 39 cm.
The number of orbits completed by the proton during this 1.0 ms is 14000 revolutions.
The kinetic energy for the protons can be computed by using the formula:
K.E = 1/2mv²mv² = 2 K.E
v = sqrt( 2*KE / M)
the kinetic energy of the medical isotopes = 6.5 MeV
substituting the values,
v = sqrt( 2* 6.5* 1.6* 10^-13 / 1.67* 10^-27 )
v = 3.53 × 10⁷ m/s
The radius of the orbit can be estimated by using the formula:
mv² / R = qvB
r = q*v / mv²
r = ( 1.67* 10^-27 * 3.53 × 10⁷ ) / ( 1.9* 1.6* 10^-19 )
r = 0.19415 m
Since diameter (D) = 2r,
D= 2(0.19415 m)
D= 0.39 m
D≅ 39 cm
The time period to complete a revolution around the spiral trajectory is:
T = 2πr / v
T = 2*3.14* 0.1941 / 3.53*10^7
T = 0.7 × 10⁻⁷ s
Finally, the number of orbits that the proton does to complete the revolution in 1 ms is:
n = t / T
n = 10^-3 / (0.7*10^-7)
n = 14285.71
n ≅ 14000 revolutions
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Your question is incomplete, but most probably full question was:
A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotron is 1.9 T.
(a) What is the diameter of the largest orbit, just before the protons exit the cyclotron? Express your answer with the appropriate units. d = 57 cm Previous
(b) A proton exits the cyclotron 1.0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this 1.0 ms?
a pulley is assumed massless and frictionless and rotates freely about its axle. the blocks have masses m1=40g and a block m2= 20g, and block m1 is pulled to the right by a horizontal force magnitude f=0.03n. find the magnitude of acceleration of a block m2 and the tension in the cord.....?
The magnitude of acceleration of a block m₂ is 0.05 m/s² and the tension in the cord is 0.01 N.
Given:
mass of block 1, m₁ = 40 gm = 40×10⁻³ kg
mass of block 2, m₂ = 20 gm = 20×10⁻³ kg
Applied force, F = 0.03 N
Calculation:
Consider the free-body diagram of the system as shown below. Using Newton's second law of motion we get:
F = ma
where F is the applied force
m is the total mass of the system
a is the acceleration of block 2 (as it is pulled by horizontal force)
From the above equation we get:
0.03 N = (m₁+m₂) a
⇒ a = (0.03 N) / (m₁+m₂)
⇒ a = (0.03 N) / (40×10⁻³ kg + 20×10⁻³ kg)
⇒ a = 0.5 m/s²
Now, from the free-body diagram of block 2 as shown in figure 3, we get:
Balancing the forces along the horizontal:
∑Fₓ = 0
∴ T = m₂ a
where T is tension in the string
a is the acceleration of block 2
Applying values in the above equation we get:
T = (20×10⁻³ kg) × (0.5 m/s²)
= 0.01 N
Therefore, the acceleration of block 2 due to the applied horizontal force is 0.5 m/s² and the tension in the cord is 0.01 N.
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question is down below
The vertical components of velocity is 10.35 m/s and the horizontal component of velocity is 38.6 m/s
What are the components of velocity?We know that velocity is a vector quantity, a vector often can be resolved into its components. The vertical components is V sinθ while the horizontal component is vcosθ.
Hence;
Vertical component = 40 m/s sin 15 degrees = 10.35 m/s
Horizontal component = 40 cos 15 degrees = 38.6 m/s
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A 15.2 kg mass has a gravitational potential energy of -342 J. How high from the ground is it? Group of answer choices GPE cannot be negative 2.3 m 22.5 m 530 m
A. The height above the ground reached by the mass is 2.3 m.
Height traveled by the mass
P.E = mgh
where;
m is mass h is height reached by the objectg is acceleration due to gravityh = P.E/mg
h = (342)/(15.2 x 9.8)
h = 2.3 m
Thus, the height above the ground reached by the mass is 2.3 m.
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A proud deep-sea fisherman hangs a 69.5-kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.130 m. (a) Find the force constant of the spring.
Force constant = k = 6811 N/m
Given:
mass = m = 69.5 kg
L(initial) = ideal spring length = 12 cm= 0.120 m
L(final) = final spring length = 13 cm= 0.130 m
Two forces acting on the object is k*x and mg
so,
kx-mg=0
kx = mg
k = mg/x
We know that x= displacement = L(final)- L(initial)
0.130-0.120=0.10
k= 69.5×9.8/0.10
k =6811 N/m
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A convex lens of focal length 25 cm is in contact with a concave lens of focal length 50 cm. The equivalent focal length of this combination-lenses should be Group of answer choices
The equivalent focal length of this combination-lenses should be +50 cm.
Given:
Focal length of convex lens, f₁ = +25 cm
Focal length of concave lens, f₂ = -50 cm
Calculation:
We know that the power of a lens is given as:
P = 1 / f
where, f is the focal length of lens
Now, the power of convex lens can be calculated as:
P₁ = 1/f₁
= 1/(25×10⁻² m)
= 100/25
= +4 D
Similarly, the power of concave lens can be calculated as:
P₂ = 1/f₂
= (1/-50×10⁻² m)
= -100/50
= -2 D
Thus, the power of the combined lenses can be calculated as:
P₍₁₊₂₎ = P₁ + P₂
= +4 D - 2 D
= +2 D
Now, the combined focal length of the lens can be calculated as:
f₍₁₊₂₎ = 1 / P₍₁₊₂₎
= 100 / 2 D
= +50 cm
Therefore, the equivalent focal length of this combination-lenses will be +50 cm.
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The image above shows a crash test dummie's head traveling at -1.005 m/s while striking a headrest from a car traveling 4.524 m/s with a mass of 2005.6 kg. If the crash test dummie's head head bounces off the headrest with a 9.965 m/s and the car continues traveling at 4.487 m/s, calculate the mass of the crash test dummie's head?
Answer:
From the calculation, the mass of the dummies head is 1647.44 Kg
What is the mass of the crash test dummies head?We know that the momentum after collision is equal to the momentum before collision.
Mass of the headrest = 2005.6 kg
Initial velocity of the head rest = 4.524 m/s
Final velocity of the head rest = 4.487 m/s
Mass of the dummy = m
Initial velocity of the dummy = -1.005 m/s
Final velocity of the dummy = 9.965 m/s
Then;
(m * -1.005) + (2005.6 * 4.524 ) = (2005.6 * 4.487) + (m * 9.965)
-1.005m + 9073.33 = 8999.13 + 9.965m
9073.33 - 8999.13 = 9.965m + 1.005m
18072.46 = 10.97m
m = 18072.46/ 10.97
m = 1647.44 Kg
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________ occurs when a person is holding an object that is directly hit or splashed by lightning.
Contact injury happens when a person is holding an object that is directly hit or splashed by lightning.
To find answer, we need to know more about the lightning and injuries.
What are the injuries caused by lightning?The phenomenon of natural electrical discharge happens in a short time with a bright flash is called lightning.It carries a huge amount of energy.The injuries caused by lightning can be classified as Direct strikes, Side splash and the Contact injury.Direct injuries happens when a person directly hit by lightning.Side splash occurs when a current splashes from a nearby object.Contact injury happens when a person touching an object that is hit by lightning.Thus, we can conclude that, contact injury occurs when a person is holding an object that is directly hit or splashed by lightning.
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When someone is holding something that has been struck or splashed by lightning, contact damage occurs.
We need additional information concerning lightning and injuries in order to identify the solution.
What types of injuries are brought on by lightning?Lightning is the name for a natural electrical discharge that occurs quickly and with a dazzling flash.It has a tremendous amount of energy.Lightning-related injuries can be divided into three categories: direct strikes, side splashes, and contact injuries.When someone is struck by lightning directly, they can get direct injury.When a current splashes from a neighboring object, it is called a side splash.When someone touches a lightning-hit object, contact harm results.In light of this, we can say that contact injuries happen when a person is holding an object that has been struck by lightning or splashed by it.
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