The force result in stretching the spring 10.0 centimeters is 2.5N.
What is Hooke's law?If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.
F = kx
where k is the proportionality constant called the spring constant or force constant.
Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.
Let the spring constant be very low 0.04N/m
The force applied is
F = 10 cm / 0.04
F = 0.1 m / 0.04
F = 2.5 N
Thus, the force result in stretching the spring 10cm is 2.5 N.
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Answer: 2.5 newtons
source- trust me bro
Explain why aircraft are carefully designed so that parts do not resonate.
Answer:
See the answer Explain why aircraft are carefully designed so that parts do not resonate. Expert Answer This virtually takes place, however maximum usually in small piston-engined airplanes, in particular dual-engined airplanes. The resonant frequency of the fuselage of a small plane goes to have numerous nodes, withinside the low loads of hertz.
Entropy is how quickly things get messy.
O A. True
OB. False
Answer : False
Answer:
false
Explanation:
it cant defined the messy and clean states
Jose was out drinking with his friends for nearly the whole night. The next morning he was confused and vomiting, and had a low body temperature.
Answer:
He has a hangover.
Explanation:
Just something I know.
a 10.00mf parallel-plate capacitor is connected to a 24.0v battery. after the capacitor is fully charged ,the
battery is disconnected without loss of any of charge on the plates.
a) a voltmeter is connected across the two plates without discharging them.what does it read
b)what would the voltmeter read if the plate separationwere doubled?
The answer to the first part of the question is 24 V and the answer to the second part of the question is 48 V.
The formula which relates Charge, Capacitance and Voltage is
Charge = Capacitance × Voltage
Q = CV
where Q denotes the charge, C denotes the capacitance and V denote the voltage.
Using the above formula the charge on the capacitance will be
Q = CV
C = 10 mF
V = 24 V
Q = 240 mC
Charge on capacitance is 240 mC.
Now after we disconnect the battery
C = 10 mF (will remain same)
Q = 240 mC
V = Q/C
V = 240/10 V
V = 24 V
So after we removed the battery, The voltage will remain same.
Now we know that parallel plate capacitor formula is
C = ε(A/d)
from here
C ∝ 1/d, and as there is no loss of charge we can say that V ∝ 1/d
Form C ∝ 1/d and V ∝ 1/C we can state that
V ∝ d {where d is the measurement of separation between the plates}
so if we double the distance between the plates then, the voltage will also get double.
Previously our voltage was 24 V, now if we double the distance between the plates the voltage will also get double, and it will become 48 V.
So, the voltmeter will take a reading of 24 V if voltmeter is connected across the two plates without discharging them, and if we double the plate separation then it will take 48 V as reading.
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Can you put this image back together? Type the correct order of letters below.
Answer:
CABD
.....................
The image is cut and jumbled and placed into different places, so the correct order of the image that it can be back together is CABD.
What is an image?A specific piece of something is an image. It can provide information to the optical system in two dimensions, three components, or in another way. An item that mimics a subject, such as a picture or other two-dimensional image, might be considered an image. An image in the context of signal analysis is a dispersed color amplitude.
A graphical image need not make use of the complete visual system. A common illustration of this is a grayscale image, which does not employ color, but instead relies on the visual game's sensitivity to brightness across all wavelengths. Even if it doesn't make proper use of the visual system, a black-and-white and white visual depiction of something is nonetheless an image.
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HELP!! ASAP!!
2. Describe the difference between polar covalent bonds and nonpolar covalent bonds using these two molecules - H2 and HCl. Which molecule contains a polar covalent bond and which molecule contains a nonpolar covalent bond? Explain your reasoning alongside describing the differences between the types of bonds.
3. How is the metallic bonding different than ionic or covalent bonding? What are some properties of metals that result from this type of bonding? Explain/connect how the nature of the bonding leads to the properties of metallic substances.
Answer:
The difference between polar covalent bonds and nonpolar covalent bonds using these two molecules - H2 and HCl are HCl is a polar covalent compound because the chloride ion is extra electronegative than the hydrogen ion.
Why HCl is polar and H2 is now no longer?HCl is a polar molecule. This is due to the fact the Chlorine (Cl) atom withinside the HCl molecule is extra electronegative and does now no longer proportion the bonding electrons similarly with Hydrogen (H). But H2 And Cl2 are nonpolar because of comparable electronegativity of each the atoms withinside the molecule H2 And Cl2 .
Hydrogen chloride is a diatomic molecule, such as a hydrogen atom H and a chlorine atom Cl related with the aid of using a polar covalent bond. The chlorine atom is an awful lot extra electronegative than the hydrogen atom, which makes this bond polar.
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C1=10 μF , C2= 50 μF, C3=9 μF, C4=2 μF, C5= 0.1 μF
What is the equivalent capacitance of these capacitors as they are arranged?
The equivalent capacitance of these capacitors as they are arranged in series and parallel will be 0.093 μF and 71.1 μF.
What is a capacitor?A capacitor is a device that can store electrical energy. It is a two-conductor configuration separated by an insulating medium that carries charges of equal size and opposite sign.
If the capacitor is connected in the series the equivalent capacitance is found as;
[tex]\rm \frac{1}{c_{eq}} = \frac{1}{c_1} +\frac{1}{c_2} +\frac{1}{c_3} +\frac{1}{c_4} +\frac{1}{c_5} \\\\\ \rm \frac{1}{c_{eq}} = \frac{1}{10} +\frac{1}{50} +\frac{1}{9} +\frac{1}{2} +\frac{1}{0.1} \\\\\ \frac{1}{c_{eq}} = 0.1+0.02+0.11+0.5+10 \\\\ \frac{1}{c_{eq}} = 10.7 \\\\ c_{eq} = 0.093 \ \mu F[/tex]
If the capacitor is connected in parallel the equivalent capacitance,cₐ is found as;
cₐ= c₁+c₂+c₃+c₄+c₅
cₐ = 10+50+9+2+0.1 μF
cₐ =71.1 μF
Hence, the equivalent capacitance of these capacitors as they are arranged in series and parallel will be 0.093 μF and 71.1 μF.
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Determine the amount of power
used in holding a 25 kg box, 1.5
meters above the floor, for 60
seconds.
[?] W
(answer is not 6.13)
Thank you in advance!
Here is your answer mate,
Question,
[tex]Determine\: the\: amount\\ \: of\: power\:used\: in\\\: holding\: a\: 25\: kg\: box\:\\ , \: 1.5\: meters \: above\: the\: floor\\\: for\: 60\: seconds[/tex]
Answer,
Power is equal to work done per unit timeWork is force × displacement SI UNIT OF WORK Newton meterSI UNIT OF POWER Watt[tex][/tex]
Solution,
[tex][/tex]
Given,
[tex]MASS \: IS\: 25\: KG\: \\ and \: HEIGHTIS\: 1.5m\: [/tex]
[tex][/tex]
WORK DONE (done against gravity) =
mass×acceleration due to gravity ×height
WORK = 25× 10× 1.5
[tex]\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: [/tex]= 375 Nm
[tex][/tex]
Now
POWER =
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\frac{work}{time} [/tex]
POWER
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= \frac{375}{60} Watt [/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =6.25[/tex]
[tex]Therfore\: your \: answer\: is\: 6.25[/tex]
[tex][/tex]
Check this,
[tex]Acceleration\: due\: to \: gravity\\\: can\: be\: 9.8\: m/s²\: \\As\: nothing\: mentioned\\\: in\: question\: \\I \: took \: it \: as \: 10[/tex]
[tex][/tex]
Have a good day
What is the magnetic force on a proton that is moving at 5.2 x 107 m/s to the
right through a magnetic field that is 1.4 T and pointing away from you? The
charge on a proton is 1.6 × 10-19 C. Use F = qvx B sin(e)
Hello!
We can use the following equation for magnetic force on a charged particle:
[tex]F_B = qv \times B[/tex]
[tex]F_B[/tex] = Magnetic force (N)
q = Charge of particle (1.6 × 10⁻¹⁹ C)
v = velocity of particle (5.2 × 10⁷ m/s)
B = Magnetic field strength (1.4 T)
This is a cross-product, so the equation can be rewritten to F = qvBsinφ where φ is the angle between the magnetic field and particle velocity vectors.
Since the proton's velocity vector and the magnetic field vector are perpendicular, sin(90) = 1. We can reduce the equation to:
[tex]F_B = qvB[/tex]
Plug in the known values.
[tex]F_B = (1.6*10^{-19})(5.2*10^7)(1.4) = \boxed{1.1648 *10^{-11} N}[/tex]
Jason while driving on Kukum highway at 70 m.s – 1 seeing the traffic light turn red, he applies the brake and comes to rest in a time of 2 seconds. What is his deceleration?
Answer:
35 m/s^2
Explanation:
Decceleration = change in velocity / change in time
= ( 70 m/s ) / 2 seconds = 35 m/s^2
DEcceleration = 35 m/s^2
Jason's speed changes by - 35 m/s^2
a 1kg ball is bening pushed by the rod to move in horizontal groved smooth slot if it startes from angle teta = zero degree . determaine the force the rod exertes on the ball at teta is =15 dgree if ai this instant the rod moves at angular speed of teta = 1 rad per sec end with angular acceleration theta = 2 rad persec and square the ball is only in contact with the outer side of the slot
The force the rod exerts on the ball at the given angle is determined as 3.94 N.
Force exerted on the rod by the ball
The force exerted is calculated as follows;
F = ma
F = mv²/r
F = mω²r
where;
m is mass of the ballω is angular speed of the ballr is radius of the pathr = 2cosθ
Angular speed when the ball moves 15 degreesωf² = ωi² + 2αθ
where;
θ is angular displacement in radians, 15⁰ = 15 x π/180 radωf² = (1)² + 2(2)(15 x π/180)
ωf² = 2.04
ωf = √2.04
ωf = 1.428 rad/s
F = mω²(2cosθ)
F = (1)(1.428)²(2 x cos15)
F = 3.94 N
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if you were to draw a 3rd harmonic of a tube open at both ends, what would you draw at the ends of the tube?
A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.
In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.
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You are watching a television show about Navy pilots. The narrator says that when a Navy jet takes off, it accelerates because the engines are at full throttle and because there is a catapult that propels the jet forward. You begin to wonder how much force is supplied by the catapult. You look on the Web and find that the flight deck of an aircraft carrier is about 90.0 m long, that an F-14 has a mass of 24800 kg, that each of the two engines supplies 27000 lb of thrust, and that the takeoff speed of such a plane is about 158 mi/h. Estimate the average force on the jet due to the catapult.
You are watching a television show about Navy pilots. The narrator says that when a Navy jet takes off, the average force on the jet is due to the catapult is mathematically given as
What is the average force on the jet is due to the catapult?
Generally, the equation for acceleration is mathematically given as
[tex]a=\frac{vf^2-vi^2}{2s}\\\\\Therefore\\\\a=\frac{69.29^2-0^2}{2(90}\\\\a=26.673m/s^2[/tex]
In conclusion, The force
F=m*a
F=15100*26.673
F=40272.3N
F 402 KN
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The cutoff frequency for a certain element is 1.22 x 10^15 Hz. What is its work function in eV?
The work function in eV for the given cutoff frequency is 5.05 eV.
What is cutoff frequency?The work function is related to the frequency as
Wo = h x fo
where, fo = cutoff frequency and h is the Planck's constant
Given is the cutoff frequency for a certain element is 1.22 x 10¹⁵ Hz
Wo = 6.626 x 10⁻³⁴ x 1.22 x 10¹⁵ Hz / 1.6 x 10⁻¹⁹
Wo = 5.05 eV
Thus, the work function is 5.05 eV
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A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 100-kilogram weight 2-decimeters above the ground with an energy efficiency of 25%. How many repetitions can she do with the energy supplied from a single Oreo cookie? What happens to the number of repetitions that can be done if the efficiency increases?
Answer:
Approximately [tex]325[/tex] (rounded down,) assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].
The number of repetitions would increase if efficiency increases.
Explanation:
Ensure that all quantities involved are in standard units:
Energy from the cookie (should be in joules, [tex]{\rm J}[/tex]):
[tex]\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}[/tex].
Height of the weight (should be in meters, [tex]{\rm m}[/tex]):
[tex]\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}[/tex].
Energy required to lift the weight by [tex]\Delta h = 0.2\; {\rm m}[/tex] without acceleration:
[tex]\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}[/tex].
At an efficiency of [tex]0.25[/tex], the actual amount of energy required to raise this weight to that height would be:
[tex]\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}[/tex].
Divide [tex]2.551 \times 10^{5}\; {\rm J}[/tex] by [tex]784\; {\rm J}[/tex] to find the number of times this weight could be lifted up within that energy budget:
[tex]\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}[/tex].
Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.
A 200-gram liquid sample of Alcohol Y is prepared at -6°C. The sample is then added to 400 g of water at 20°C in a sealed styrofoam container. When thermal equilibrium is reached, the temperature of the alcohol-water solution is 12°C. What is the specific heat capacity of the alcohol? Assume the sealed container is an isolated system. The specific heat capacity of water is 4.19 kJ/kg · °C. 3.14 kJ/kg \xe2\x88\x99 °C 4.14 kJ/kg \xe2\x88\x99 °C 3.72 kJ/kg \xe2\x88\x99 °C 4.88 kJ/kg \xe2\x88\x99 °C
The specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
What is the specific heat capacity?The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."
Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.
Given data;
Mass of liquid sample of Alcohol m₁ = 200-gram
The temperature of alcohol, T₁ = -6°C.
Mass of liquid sample of water m₂ = 400-gram
The temperature of the water, T₂= 20°C.
The specific heat capacity of the alcohol, S₁=?
The specific heat capacity of water is, S₂=4.19 kJ/kg.°C
As we know that;
[tex]\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72 \ kJ /kg ^0 C[/tex]
Hence the specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
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Consider the baby being weighed in Figure 4.25.
Figure 4.25
(a) What is the mass of the child and basket if a scale reading of 104 N is observed?
kg
(b) What is the tension T in the cord attaching the child to the scale?
N
(c) What is the tension T' in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg?
N
(d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible. (Do this on paper. Your instructor may ask you to turn in this work.)
The mass and tension due to the system are as follows:
The mass of the child and scale = 10.6 kgThe tension T, in the cord attaching the child to the scale = 104N The tension T', in the cord attaching the scale to the ceiling T' = 108.9 NWhat is tension?Tension is a type of pulling force due transmitted by means of a string or cable.
Force = mass * acceleration due to gravitya) The mass of the child and scale = 104/9.81 = 10.6 kg
b) The tension T, in the cord attaching the child to the scale = scale reading = 104N
c) The tension T', in the cord attaching the scale to the ceiling = scale reading + weight of scale
T' = 104 + (0.5 * 9.81)
T' = 108.9 N
d) The sketch is attached in the picture
In conclusion, the tension is force exerted on the cord due to the weight of the scale and the baby.
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22. On a day the wind is blowing toward the south at 3m/s, a runner jogs west at 4m/s. What is the velocity of the air relative to the runner?
The velocity of the air relative to the runner is 5 m/s.
What is the relative velocity?We must recall that velocity is a vector quantity and the relative velocity must be obtained vectorially. Thus we know that;
Velocity of the runner = 4m/s. due west
Velocity of the wind = 3m/s due south
The relative velocity is;
Vr = √(4)^2 + (3)^2
Vr = 5 m/s
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If the distance between two objects is cut in half, what happens to the
gravitational force between them?
A. It decreases to 1/2 its original magnitude.
B. It decreases to 1/4 its original magnitude.
O
C. It increases to 4 times its original magnitude.
D. It increases to 2 times its original magnitude.
5.00 μF, 10.0 μF, and 50.0 μF capacitors are connected in series across a 12.0-V battery.
(a) How much charge is stored in the 5.00-μF capacitor?
(b) What is the potential difference across the 10.0-μF capacitor?
(a) The charge stored in the 5.00-μF capacitor is 37.2 μC.
(b) The potential difference across the 10.0-μF capacitor is 3.72 V.
What is capacitor?The capacitance of a capacitor is defined as the ratio of the charge stored and the potential difference between the capacitor.
The capacitance of a capacitor is denoted by C and expressed as
C = Q/V
Given, 5.00 μF, 10.0 μF, and 50.0 μF capacitors are connected in series across a 12.0-V battery.
(a) The equivalent capacitance is
1 / Ceq = 1 / C₁ +1 / C₂ + 1/ C₃
Substitute the values, we get
Ceq = 3.1 μF
The charge stored in 5.00-μF capacitor is
Q = Ceq x V
Q = 3.1 μF x 12 V
Q = 37.2 μC
Thus, the charge stored in the 5.00-μF capacitor is 37.2 μC
(b) The potential difference across the 10.0-μF capacitor is given by
V = Q/C₂
Put the values, we get
V = 37.2 / 10
V = 3.72 V
Thus, the potential difference across the 10.0-μF capacitor is 3.72 V.
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Jon and Jim push a dead car forward to the gas station. Jon pushes with 90N while Jim pushes 80 N.
The resultant of the two forces is about 170 N.
What is the resultant force?The term resultant force has to do with the single force that has the same magnitude and direction as two or more forces acting together.
In this case, the both forces are acting in the forward direction. This implies that the resultant force is 90 N + 80 N = 170 N.
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A bumblebee
is flying towards a flower in a
straight line at 4.09 m/s when it begins to
accelerate at 1.01 m/s².
How long does it take the bee to reach the
flower if it is 23.4 m away?
Answer:
given -
initial velocity = 4.09 m/s
acceleration = 1.01 m/s²
distance = 23.4 m
time = ?
using second formula of motion,
s = ut + 1/2 at².
where, s = distance
u = initial velocity
t = time
a = acceleration
23.4 = 4.09(t) + 1/2(1.01)(t) ²
23.4 = 4.09t + 2.02t²
2.02t² + 4.09t - 23.4 = 0
solving the equation by using quadratic formula
Use the standard form, ax² + bx + c = 0 , to find the coefficients of our equation, :
a = 2.02
b = 4.09
c = -23.4
we get t=2.539 or t= -4.563
time cannot be negative so
t= 2.539 sec = 2.6 Sec is the answer
Which picture correctly shows the path of refracted light rays given an object outside the focal point? Select one: a. A b. B c. C d. D
Answer:
Answer is C because light travels in a sight line but when light pass through a refractor the light from the source changes direction when passes through a refractor
A grocery cart is pushed with a force of 21.4 N. If 1,974 J of work is done in pushing the grocery cart through the store, what is the total distance that the grocery cart has traveled?
Answer:
Given - Force = 21.4 N
Work done = 1974 J
To find - Distance
Solution -
Work done = Force * displacement
1974 j = 21.4 N * displacement
1974/21.4 = displacement
92.24
[tex]\\ \rm\Rrightarrow Work=Force\times Displacement [/tex]
[tex]\\ \rm\Rrightarrow 1974=21.4×Displacement[/tex]
[tex]\\ \rm\Rrightarrow Displacement=92.24m[/tex]
Which of the following causes a car's passenger to lean in the direction
opposite the direction in which the car is turning?
OA. Friction
OB. Inertia
OC. Centripetal acceleration
OD. Centripetal force
Explanation:
This case is inertia.
Mass is directly proportional to inertia
Helium is the second most abundant element on Jupiter (and in the Universe overall). But it's rare on Earth, being only 0.0005% of the atmosphere. Why is this?
A) Jupiter's stronger gravity results in a faster escape velocity, and almost no helium atoms are moving fast enough to escape its gravity. Earth's escape velocity is lower, so a few helium atoms at a time are moving fast enough to escape, causing Earth to slowly lose its helium.
B) Jupiter's core continually generates helium, while the Earth's core does not.
C) Jupiter has a powerful magnetic field that can attract helium atoms. Earth has a magnetic field as well, but it's not strong enough.
Answer:
Explanation:
Comment
I think the best answer is probably A. It's not the best reason for Earth loosing it's helium, but it is the only one that is close. The Earth derives some of its Helium (most) by the decomposition process of radio active high weight chemical that break down. Helium combines with practically nothing so once it gets moving, very little will stop it. That's the way we loose our Helium.
Still answer A.
a. Recent studies have raised concern about 'heading' in youth soccer (i.e., hitting the ball with the head). A soccer player 'heads' a size 3 ball deflecting it by 40.0°, and keeps its speed of 13.20 m/s constant. A size 3 ball has a mass of approximately 2.000 kg. What is the magnitude of the impulse which the player must impart to the ball?
b. If the player's head has a mass of 2.90 kg, what is the magnitude of the average acceleration of the player's head during the impact. Assume that over the brief time of the impact, 30.40 ms, the player's head can be treated separately from the player's body.
a. The magnitude of the impulse which the player must impart to the ball is 17.25 kg.m/s.
b. The magnitude of average acceleration of the player's head during the impact is 195.66 m/s².
What is impulse?The change in momentum is equal to the product of impact force applied while colliding and time for that impact.
Impulse F. t = m (Vf -Vi)
where, Vf is the final velocity and Vi is the initial velocity.
Given, a soccer player 'heads' a size 3 ball deflecting it by 40.0°, and keeps its speed of 13.20 m/s constant. A size 3 ball has a mass of approximately 2.000 kg.
a. Substitute the values into the expression, we get
Impulse (x) = 2 x (13.20 - 13.20cos40°)
Impulse (x) = 3.088 kg.m/s
Impulse in y direction, is
Impulse (y) = 2 x 13.20sin40°
Impulse (y) = 16.97 kg.m/s
so, the magnitude of impulse is
I = sq rt(Ix² + Iy²)
Put the values, we get
I = 17.25 kg.m/s
Thus, the magnitude of the impulse which the player must impart to the ball is 17.25 kg.m/s
b. The player's head has a mass of 2.90 kg. The brief time of the impact, 30.40 ms,
F.t = I
Put the given values, we have
F = 17.25 / 30.40 x 10⁻³
F = 567.434 N
From the Newtons second law of motion,
F = ma
Plug the values from the question, we have
567.434 N = 2.90 x a
a = 195.66 m/s²
Thus the magnitude of average acceleration of the player's head during the impact is 195.66 m/s².
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Which force is reasonable for making fusion possible in the sun?
The force is reasonable for making fusion possible in the Sun is heat energy.
What is nuclear fission and fusion?When the slow moving neutrons are bombarded with the heavy radioactive nuclei, the product is the more number of neutrons are produced with the large amount of energy. This multiplying process is called nuclear fusion.
The amount of energy produced in such a reaction can be calculated using the equivalence of mass and energy relationship.
E = mc²
The same happens in nuclear fusion where large amount of energy is needed to make more heavy nuclei.
Thus, fusion requires heat energy to continue the reaction.
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What work do you think is done when you carry a 20 N weight backpack for a 1000 m walk? will the work be positive, negative, or potentially zero?
The work done is positive and is equal to 20000 J
What is work done?Work done is defined as the product of force and the distance moved by the force.
Mathematically:
Work done = force * distanceThe work done by the force = 20 * 1000 = 20000J
The work done is positive and is equal to 20000 J
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A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1 = 18 V and V2 = 12 V. The positive terminals are indicated with a + sign, The values for the resistors are: R1 = R5 = 70 Ω, R2 = R6 = 106 Ω R3 = 59 Ω, and R4 = 83 Ω. The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows. What is I3?
The I3 will be 158 A.
How to find the current through the circuit?The foundation of circuit analysis is Kirchhoff's circuit laws.We have the fundamental instrument to begin studying circuits with the use of these principles and the equation for each individual component (resistor, capacitor, and inductor).These rules aid in calculating the current flow in various network streams as well as the electrical resistance of a complicated network, or impedance in the case of AC.To calculate I3 firstly, V4 has to be calculated,
[tex]V_{4} =I_{4} R_{4}[/tex]
[tex]V_{4} = V_{2} / R_{4} + R_{5} * R_{4}[/tex]
[tex]V_{4} = 12 * 135 / 135+61[/tex]
[tex]V_{4} = 8.26V[/tex]
For I3,
[tex]I_{3} = R_{1} /(R1+R3 + (R1+R3)(R2+R6) * (V2 - V1 (R1+R2+R6/R1)[/tex]
[tex]I3=(61)/((61)(50)+(61+50)(141+141)) (12 -18 (1+(141+141)/61)) = -.158 A[/tex]
Hence, the current through I3 will be 158 A.
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