Answer:
Explanation:
when we get out of the water we feel quite cold due to the latent heat of evaporation of water.
Water at our body gets converted into gas and leaves behind the cooling effect which makes us feel cold.
While when we are inside the pool we are in constant touch of water, therefore, we get the warmth of the water and it becomes difficult to bear it.
A player hits a ball with a bat. The action force is the impact of the bat against the ball. The reaction to this force is the
Answer:
Force of the ball against the bat
Explanation:
Newton's third law of motion states that for every action, there is an equal and opposite reaction. When a player hits a ball with a bat, the force the player exerted on the ball through the bat gave the ball a push force in the forward direction. This is an action force. Now that an action has been initiated, there must be a reaction according to Newton's third law of motion. The magnitude of the force on the ball equals the magnitude of the force on the bat. The direction of the force on the ball is opposite to the direction of the force on the ball. The ball will give the bat an equal and opposite force trying to push the bat backward by resisting the impact of the bat. This is the reaction. The ball will only move if the applied force is able to overcome the resisting force by the bat.
[tex]F_{A} = - F_{B}[/tex]
Where;
[tex]F_{A}[/tex] is the action force of the bat against the ball.
[tex]F_{B}[/tex] is the equal and opposite reaction force of the ball towards the ball.
If a player hit a ball with a bat the ore that which the ball is coming towards the player and the force with which the player is moving the bat will impact the ball which will have an equal and opposite reaction.
Thus the reaction of the force will be the force of the ball against the bat.Hence the newtons law of motion will be applied.
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"Q 12.10: A 98 N object is placed 1/3 of the way up a 120 N ladder, which makes an angle of 53 degrees with the horizontal. The lower end is also fastened to the wall by a rope, and both the wall and floor are frictionless. What is the tension in the rope?"
Answer:
68.829 N
Explanation:
The given parameters are:
Weight of ladder, [tex]W_l[/tex] = 120 N
Weight of object, [tex]W_o[/tex] = 98 N
Angle, [tex]\theta[/tex] = 53°
And we also know that, while
Length of ladder = L
Distance the object is placed = L/3
If we apply translational equilibrium horizontally, then [tex]T-N=0[/tex], so [tex]T = N[/tex]
If we apply rotational equilibrium about the distance the object is placed, then
[tex]W_o\frac{L}{3}cos(\theta) + W_l\frac{L}{2}cos(\theta) - NLsin(\theta) = 0\\2NLsin(\theta) = 2W_o\frac{L}{3}cos(\theta)+W_lLcos(\theta)\\2Tsin(\theta) = \frac{2}{3}W_ocos(\theta)+W_lLcos(\theta)\\T = \frac{1}{3}W_ocot(\theta) + \frac{1}{2}W_lcot(\theta)\\T = \frac{1}{3} * 98 * cot(53) + \frac{1}{2} * 120 * cot(53) = 69.829[/tex]
A radio receiver has an effective resistance of 300 Ω to the input signal on the antenna downlead. The signal voltage is 700 µV. What is the signal current flow in the antenna downlead?
Answer:
I = 2.33 µA
Explanation:
given,
Effective resistance (R)= 300 Ω
Voltage of signal (V)= 700 µV
current flow in the antenna = ?
Using ohm's law
V = I R
where I is the current flow
[tex]I = \dfrac{V}{R}[/tex]
[tex]I = \dfrac{700\times 10^{-6}}{300}[/tex]
I = 2.33 x 10⁻⁶ A
I = 2.33 µA
Hence, the current flow in the antenna downloaded is equal to I = 2.33 µA
Answer:I = 2.33 µA
Explanation:
If you push against a wall, the wall pushes back against you with ____ force.
1. equal
2. less
3. more
4. no
Answer:
1. Equal.
Explanation:
Newton's third law of motion;
According to this law every action have its reaction with same magnitude but in the opposite direction.
If we push a wall with some amount of force then wall pushes back with same magnitude but in the opposite direction.This law tell us about action and reaction forces which are in same magnitude but in the opposite direction.
Therefore the answer will be 1.
1. Equal.
When pushing a wall, the wall pushes back with equal force as stated by Newton's Third Law of Motion. This law means forces have equal magnitude, but opposite direction.
Explanation:
When you push against a wall, the wall pushes back against you with equal force. This is demonstrated by Newton's Third Law of Motion, which states that 'for every action, there is an equal and opposite reaction'. The force you exert on the wall and the force the wall exerts on you are action-reaction force pairs, meaning the force exerted on the wall is the same magnitude, but in the opposite direction to the force the wall exerts on you.
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What will the temperature of the cosmic microwave background be when the average distances between galaxies are 1.7 times as large as they are today? (Hint: The peak wavelength of photons in the background will then also be 1.7 times as large as it is today.)
Answer:
[tex]T'=1.58K[/tex]
Explanation:
As the Universe expands, the photons of the cosmic microwave background increase its wavelength, making its temperature inversely proportional to the scale factor of the Universe. That is, as the average distances between galaxies increase, the temperature of the cosmic microwave background decreases by the same factor. Therefore, the temperature when the distances between galaxies are 1.7 times as large as they are today will be:
[tex]T'=\frac{T_{now}}{1.73}\\T'=\frac{2.725K}{1.73}\\T'=1.58K[/tex]
When the average distances between galaxies are 1.7 times their present distances, the temperature of the cosmic microwave background (CMB), as a result of Wien's Law, will be about 1.61 Kelvin, which is 1.7 times lower than its current close to 2.73 Kelvin figure.
Explanation:The cosmic microwave background radiation (CMBR), an aftermath of the Big Bang event, exhibits a relationship between wavelength and temperature called Wien's Law. At present, the temperature of the CMBR is approximately 2.73 Kelvin (K), a finding made possible by scientists Penzias and Wilson who associated radiation intensity at different wavelengths to a temperature of roughly 3.5 K.
As the galaxies continue to recede from one another, their emitted radiation wavelengths become redshifted, effectively stretching them out. This movement is also reflected in the increase in the peak wavelength of the CMBR. If the average distances between galaxies were to magnify by 1.7 times, the peak wavelength of the CMBR would equally elongate by the same factor.
Applying Wien's Law, this increase will consequently result in a decrease in the CMB's temperature. Thus, when galaxies have expanded such that their average distances are 1.7 times larger than their current distances, the temperature of the cosmic microwave background will be 1.7 times lower than its current temperature of approximately 2.73 K. Hence, the new temperature would approximately be 1.61 K.
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A fire hose is turned on the door of a burning building in order to knock the door down. This requires a force of 1000 N. If the hose delivers 40 kg per second, what is the minimum velocity of the stream needed, assuming the water doesn't bounce back?
Answer:
Minimum velocity is 25m/s
Explanation:
Given force F =1000N
Mass flow rate m/t= 40kg/s
Velocity=v
Recall Newton second law of motion F =MV/t
1000=40v
V=1000/40
25m/s
What did antoine lavoisier contribute to the atomic theory
Explanation:
Antoine lavoisier is most famous for his role in discovering of oxygen. He recognized and name two important element oxygen(1778) and hydrogen(1779).
In a chemical reaction, Lavoisier observed the mass is retained. A chemical reaction's total mass of the products will always be the same as the total mass of the reactant materials used in the reaction. His results led to one of the basic laws of chemical behavior: the law of material preservation, which states that matter is preserved in a chemical reaction.
Final answer:
Antoine Lavoisier's major contributions to atomic theory include the discovery of the law of conservation of mass and transforming chemistry into a quantitative science by disproving the phlogiston theory and identifying the role of oxygen in combustion. These discoveries paved the way for the future development of the atomic theory.
Explanation:
Contributions of Antoine Lavoisier to Atomic Theory
Antoine Lavoisier, often regarded as the 'father of modern chemistry', made several groundbreaking contributions to atomic theory. He is particularly famous for discovering the law of conservation of mass. This principle, established through careful experimentation, states that mass in an isolated system is conserved over time, regardless of the type of chemical reactions happening within. This fundamental law formed a cornerstone for John Dalton to later develop the modern atomic theory.
Additionally, Lavoisier conducted experiments that disproved the phlogiston theory and instead suggested that oxygen was a key element in combustion. His innovative approach to chemical reactions included the use of precise weighing measures and conducting reactions in sealed environments to accurately track the mass of gases involved. Lavoisier's meticulous quantitative methods transformed chemistry from a qualitative to a quantitative science.
Through these contributions, Lavoisier laid down the basis for atomic theory as well as the concept of chemical elements, thereby fundamentally changing our understanding of chemical processes. He also impacted other areas such as explaining the role of oxygen in respiration and the composition of air.
The __________ behind a vehicle is one of the most common causes of collisions while ____________.
The main cause of accidents related to vehicles is rash driving, inexperienced drivers, and bad driving manners. Lack of awareness regarding the driving methods.
Most of the commonly made mistakes are due to drinking and driving. Most people feel drossy at night and some even don't horn when appropriate. Hence the blind spot after the vehicles is the most common cause of collisionsLearn more about the vehicle is one of the most common causes of collisions.
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The blind spot behind a vehicle is a common cause of collisions while driving. Safety measures like properly adjusted mirrors and checking over the shoulder before changing lanes can help. Cars with lighter plastic components crumple upon collision, increasing collision time and reducing the force exerted on occupants for increased safety.
Explanation:The blind spot behind a vehicle is one of the most common causes of collisions while driving. This pertains to the area of the road that cannot be directly seen by the driver while looking forward or through either the rear-view or side mirrors. The risk of such collisions can be minimized by proper adjustment of mirrors, use of blind spot mirrors, and always checking over one's shoulder before making a lane change.
Upon collision, cars with lighter weight are able to offer better protection to their occupants due to longer collision time, because of their plastic components. The crumpling of the car in an event of a collision or the deployment of dashboard padding and airbags increases the collision time and hence, the force exerted on the occupants to bring them to a stop is significantly less. Modern automobiles use this physics principle to enhance safety during collisions.
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2020 NV Cargo’s electronically controlled Multi-Control Valve on the V8 engine helps enhance ________.
Answer:
Power when towing
Explanation:
The V8 engine has Direct Injection Gas (DIG) technology, which provides better wide-open throttle performance and improved fuel consumption and emissions performance (compared to a non-direct-injection system) by reducing engine knock, improving combustion stability and offering precise injection control.
The multi-control valve on the V8 engine helps to enhance power when towing by reducing the amount of torque the engine must expend to move the vehicle, allowing it to commit more to moving the load.
Early LCD backlights use _______________ technology, popular for its low power use, even brightness, and long life.
Answer:
cold cathode fluorescent lamp
Explanation:
A cold cathode fluorescent lamp (CCFL) is a lighting system that uses two phenomena: electron discharge and fluorescence.
CCFLs are mainly used as light sources for backlights, since they are compact and durable than ordinary fluorescent lamps. They exhibit a wide range of brightness and color (color temperature and chromaticity) that can be achieved by varying the pressure and type of the material injected into the glass tube. The thickness and type of the phosphor used to coat the inner wall of the tube also plays a role in altering the color and brightness.
Final answer:
Early LCD backlights used CCFL technology for its efficiency and longevity. LCDs rely on a backlight to illuminate the liquid crystals in pixels. Each pixel can control red, blue, or green filters to create the full color spectrum on the screen.
Explanation:
Early LCD backlights use CCFL (Cold Cathode Fluorescent Lamp) technology, popular for its low power use, even brightness, and long life. LCD, which stands for liquid crystal display, relies on the light modulating properties of liquid crystals. These crystals themselves do not emit light. In flat screen LCD televisions, the backlight illuminates the screen through millions of tiny units called pixels. Each pixel has three cells with red, blue, or green filters, each controlled independently to vary the picture contrast and color representation by adjusting the voltage applied to the liquid crystal.
A person is in a closed room (a racquetball court) with ????=453m3 hitting a ball (m=42.0g) around at random without any pauses. The average kinetic energy of the ball is 2.30 J. (a) What is the average value of ????2x? Does it matter which direction you take to be x? (b) Applying the methods of this chapter, find the average pressure on the walls? (c) Aside from the presence of only one "molecule" in this problem, what is the main assumption in Pressure, Temperature, and RMS Speed that does not apply here?
Answer:
(a) [tex]v^{2} _{x} = 36.508 m^2/s^2[/tex]
(b) 0.1235 Pa
(c) The volume of the ball is significant when compared with the volume of the closed room. Thus, the constant 'b' of the equation of Van der Waal is greater than zero.
Explanation:
In the question, we are given the following variables:
V = 453 m^3
[tex]m_{ball} = m = 42.0 g[/tex]
Kinetic energy (K.E) = 2.30 J
(a) The equation for kinetic energy is:
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
Since the average velocity components of the ball in the three dimensions are equal. Thus, we have:
[tex]v_{x} ^{2} = v^{2} _{y} =v^{2} _{z}[/tex]
In addition:
[tex]v^{2} =[/tex][tex]v_{x} ^{2} + v^{2} _{y} +v^{2} _{z}[/tex] = [tex]3v^{2} _{x}[/tex]
Therefore:
[tex]K.E = \frac{1}{2}m[3v^{2} _{x}][/tex] = [tex]\frac{3}{2}mv^{2} _{x}[/tex]
Thus:
[tex]v^{2} _{x} = \frac{KE}{1.5*m} = \frac{2.3}{1.5*0.042} = 36.508 m^2/s^2[/tex]
It does not matter which direction we take as x-direction because the average velocity component in every direction is the same.
(b) The average pressure on the walls can be calculated using the equation below:
[tex]P =\frac{mv^{2} _{x} }{V}[/tex]
where V = 453 m^3
Thus: P = (0.042*36.508^2)/453 = 0.1235 Pa
(c) The volume of the ball is significant when compared with the volume of the closed room. Thus, the constant 'b' of the equation of van der Waal is greater than zero.
The average value of u² is 54.8 m²/s². It does not matter which direction is taken as x. The average pressure on the walls can be found using the force and area equations.
Explanation:(a) The average value of u² can be found by dividing the average kinetic energy by the mass of the ball: u² = KE / m. In this case, u² = 2.30 J / 0.042 kg = 54.8 m²/s².
It does not matter which direction you take to be x. The average value of u² will be the same regardless of the direction.
(b) The average pressure on the walls can be found using the equation P = F / A, where P is the pressure, F is the force, and A is the area of the walls. The force exerted on the walls can be found using the equation F = ∆p / ∆t, where ∆p is the change in momentum and ∆t is the change in time. Since there are no pauses in the ball's motion, ∆p = m * ∆v, where m is the mass of the ball and ∆v is the change in velocity. The average pressure will be the force divided by the area of the walls.
(c) The main assumption in Pressure, Temperature, and RMS Speed that does not apply here is that there are a large number of gas molecules moving in random directions. In this problem, there is only one molecule, so the assumption of a large number of molecules does not hold.
How can a surface be smooth/polished for some waves and not for others? (This is an important question, refer to the homework on WebAssign and question regarding whether a surface is smooth/polished for some electromagnetic waves verses others)
Answer:
Depending upon the size of the irregularities on the surface and the wavelength of the surface a surface can be smooth for the waves of smaller wavelengths than the size of the irregularities.
Explanation:
Depending upon the size of the irregularities on the surface and the wavelength of the surface a surface can be smooth for some of the waves but not for the others.
More specifically, when a surface has irregularities of the size greater than or equal to the wavelength of a wave then the surface is rough for that particular wave and causes irregular reflection in the incident light.Contrarily, when the surface has the irregularities less than the size of the wavelength then it reflects most of the light in a uniform manner.
A car moves 4 times as fast as another identical car. compared to the slower car, the faster car has:_______
a. 4 times the ke.
b. 8 times the ke.
c.12 times the ke.
d.16 times the ke.
Answer:
Option d
d. 16 times the ke
Explanation:
The kinetic energy is defined as:
[tex]Ke = \frac{1}{2}mv^{2}[/tex] (1)
Where m and v are the mass and velocity of the object.
The kinetic energy represents how much energy have an object as a consequence of its movement.
For the case of a car moving 4 times as fast as another identical car.
By means of equation 1 it is gotten:
[tex]Ke = \frac{1}{2}m\cdot(4v)^{2}[/tex]
[tex]Ke = \frac{1}{2}m\cdot16v^{2}[/tex]
Hence, the faster car has 16 times the kinetic energy.
Suppose you want to heat a 600 mL cup of water in the microwave from 20° to 100°C. Water has a specific heat capacity of 4.184 J/g°C and a density of 1.00 g/mL in this temperature range. How much energy is required to heat the cup of water?
Answer:
The quantity of heat required to to heat the cup of water through a temperature change of 80°C = 200.83 kJ
Explanation:
Density = mass/Volume.
∴ mass = Density × volume
Where Density of water = 1.00 g/mL, volume of water = 600 mL
Mass = 1 × 600 = 600 g.
(Q) = cm(ΔT)...................... equation 1
Where Q = quantity of heat, c = specific heat capacity of water, m = mass of water, ΔT = T₁-T₂ = change in temperature.
Where c = 4.184 J/g°C, m = 600 g, ΔT = T₁-T₂ = 100 - 20 = 80°C
Substituting these values into equation 1,
Q = 4.184 × 600 × 80
Q = 200832 J = 200.83 kJ
Therefore the quantity of heat required to to heat the cup of water through a temperature change of 80°C = 200.83 kJ
The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?
Answer:20 cal/s
Explanation:
Given
Heat transfer rate is [tex]\dot{Q}=10 cal/s[/tex]
Also heat rate is given by
[tex]\dot{Q}=kA\frac{dT}{dx}[/tex]
where [tex]k=thermal\ conductivity[/tex]
[tex]A=area\ of\ cross-section[/tex]
[tex]dT=change\ in\ temperature[/tex]
[tex]dx=change\ in\ length[/tex]
[tex]10=k\frac{\pi }{4}d^2\frac{dT}{L}----1[/tex]
For [tex]d'=2d, Length L'=2L[/tex]
[tex]\dot{Q}=k\frac{\pi }{4}(2d)^2\frac{dT}{2L}---2[/tex]
dividing 1 and 2 we get
[tex]\frac{10}{\dot{Q}}=\frac{2d^2}{4d^2}[/tex]
[tex]\dot{Q}=20 cal/s[/tex]
The power rate of the steel is mathematically given as
P=20cal/s
What rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?Question Parameter(s):
The rod conducts heat from one end to the other at a rate of 10 cal/s
Generally, the equation for the Power is mathematically given as
Power = (kA∆T)/l
Therefore
Power = [k × 4(πd^2)/4 × ∆T]/2l
Therefore
Power = 2(kA∆T)/l
Hence, Initial Power
iP = (kA∆T)/l
iP= 10cal/s
Where
dT is the same
2[(kA∆T)/l] = 2 × 10
P=20cal/s
In conclusion, The power rate is
P=20cal/s
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A golfer starts with the club over her head and swings it to reach maximum speed as it contacts the ball. Halfway through her swing, when the golf club is parallel to the ground, does the acceleration vector of the club head point straight down, parallel to the ground, approximately toward the golfer's shoulders, approximately toward the golfer's feet, or toward a point above the golfer's head? Explain.
Answer:
approximately toward the golfer's feet
Explanation:
The golf club experiences two accelerations. when the club is horizontal, centripetal acceleration acts on it toward the center parallel to the ground. The tangential acceleration acts towards the ground in downward direction. So when the golf club is horizontal, the acceleration vector is given by the combination of tangential and centripetal acceleration. the acceleration vector thus points in between horizontal and vertical direction . hence towards the golfer's feet.
Answer:
approximately toward the golfer's feet
Explanation:
The club head undergoes circular motion so it has a centripetal acceleration. As the speed of the club head is increasing, it also has an acceleration component pointed vertically down. Thus, the acceleration vector is pointed as described in (c)
Laser light of wavelength lambda passes through a thin slit of width a and produces its first dark fringes at angles of +/- 30 degree with the original direction of the beam. The slit is then reduced in size to a circle of diameter a. When the same laser light is passed through the circle, the first dark fringe occurs at:________
A) +/- 66.9 degree.
B) +/- 45.0 degree.
C) +/- 37.6 degree.
D) +/- 36.6 degree.
E) +/- 15.0 degree.
Answer:
D) θ₂= 36. 6º
Explanation:
In this diffraction experiment it is described by the equation
sin θ = m λ
The first dark strip occurs for m = 1 and since the angle is generally small we can approximate sine to the value of the angle
θ₁ = λ/ a
This equation is valid for linear slits, in the case of a circular slit the problem must be solved in polar coordinates, so the equation changes slightly
θ₂ = 1.22 λ / a
In the proposed exercise we start with a linear slit of width a, where tes1 = 30º and end with a circular slit of the same diameter
θ₂ = 1.22 λ / a
Let's clear (Lam/a) of equalizing the two equations
θ₁ = θ₂/ 1.22
θ₂ = 1.22 θ₁
θ₂ = 1.22 30
θ₂= 36. 6º
When reviewing the correct results is D
What is the volume of water in 150ml of the 35% w/w of sucrose solution with a specific gravity of 1.115?
Answer:108.71 mL
Explanation:
Given
Volume of sample V=150 mL
concentration of sucrose solution 35 % w/w i.e. In 100 gm of sample 35 gm is sucrose
specific gravity =1.115
Density of solution [tex]\rho _s=1.115\times density\ of\ water[/tex]
Thus
[tex]\rho _s=1.15\times 1 gm/mL=1.115\ gm/mL[/tex]
mass of sample [tex]M=1.115\times 150=167.25\ gm[/tex]
mass of sucrose [tex]m_s=0.35\times 167.25=58.53\ gm[/tex]
mass of Water [tex]m_w=108.71 gm[/tex]
Volume of water [tex]=108.71\times 1=108.781 mL[/tex]
A 35-kg girl is standing near and to the left of a 43-kg boy on the frictionless surface of a frozen pond. The boy throws a 0.75-kg ice ball to the girl with a horizontal speed of 6.2 m/s. What are the velocities of the boy and the girl immediately after the girl catches the ice ball?
Answer:
The velocity of the boy is 0.11 m/s to the right.
The velocity of the girl is 0.13 m/s to the left.
Explanation:
Hi there!
The momentum of the system is conserved. That means that the initial momentum is equal to the final momentum (after throwing the ice ball for the boy and after catching the ball for the girl).
Let´s calculate the momentum of the system boy-ball:
The equation of the momentum is the following:
p = m · v
Where:
p = momentum.
m = mass of the object.
v = velocity.
Initially, the ball and the boy are at rest, so that the momentum of the system boy-ball is zero:
initial momentum = mib · vib + mb · vb
Where:
mib = mass of the ice ball.
vib = velocity of the ice ball.
mb =mass of the boy.
vb = velocity of the boy.
The initial momentum is zero because both velocities are zero.
After throwing the ball, the momentum of the system will be:
final momentum = mib · vib + mb · vb
final momentum = 0.75 kg · (- 6.2 m/s) + 43 kg · vb (Let´s consider the left as negative).
Since final momentum = initial momentum
0 kg · m/s =- 4.65 kg · m/s + 43 kg · vb
Solving for vb:
4.65 kg · m/s/43 kg = vb
vb = 0.11 m/s
The velocity of the boy after releasing the ball is 0.11 m/s. Since there is no friction, this velocity is constant. When the girl catches the ball, the velocity of the boy will be 0.11 m/s (to the right).
Now let´s do the same for the system girl-ball.
initial momentum = mg · vg + mib · vib
initial momentum = 35 kg · 0 m/s + 0.75 kg · (-6.2 m/s)
initial momentum = -4.65 kg · m/s
The final momentum is the momentum of the girl plus the ball:
final momentum = (mg + mb) · vgb
final momentum = (35 kg + 0.75 kg) · vgb
Since initial momentum = final momentum
-4.65 kg · m/s = (35 kg + 0.75 kg) · vgb
-4.65 kg · m/s / (35 kg + 0.75 kg) = vgb
vgb = -0.13 m/s
The velocity of the girl after catching the ball is -0.13 m/s (0.13 m/s to the left).
The velocity of the boy after catching the ice is 0.108m/s right and the velocity of the girl after catching the ice is 0.132m/s left.
Data;
mass of girl = 35kgmass of the boy = 43kgmass of the ice ball - 0.75kgspeed of the ball = 6.2m/sConservation of linear momentumThe conservation of linear momentum expects that the momentum of the boy or girl must be equal to the momentum of the ball.
for the boy
[tex]m_1v_1 = m_2v_2\\43*v = 0.75*6.2\\v = 0.108m/s[/tex]
The velocity of the boy after catching the ice is 0.108m/s right
for the girl
[tex]m_1v_1 = m_2 v_2\\0.75*6.2 = 35 * v\\v = 0.132 m/s[/tex]
The velocity of the girl after catching the ice is 0.132m/s left.
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In a ballistics test, a 28-g bullet pierces a sand bag that is thick. If the initial bullet velocity was and it emerged from the sandbag moving at what was the magnitude of the friction force (assuming it to be constant) that the bullet experienced while it traveled through the bag?
A. 130 N
B. 1.3 N
C. 1.4 N
D. 38 N
E. 13 N
Answer:
A) 130N
Explanation:
We know that:
[tex]E_i -E_f = W_f[/tex]
where [tex]E_i[/tex] is the initial energy, [tex]E_f[/tex] is the final energy and [tex]W_f[/tex] is the work of the friction force.
so:
[tex]\frac{1}{2}MV_i^2 -\frac{1}{2}MV_f^2 = Fd[/tex]
where M is the mass, [tex]V_i[/tex] is the initial velocity, [tex]V_f[/tex] is the final velocity, F the force and d the distance.
Replacing values, we get:
[tex]\frac{1}{2}(0.028)(55m/s)^2 -\frac{1}{2}(0.028)(18m/s)^2 = F(0.3m)[/tex]
solving for F:
F= 126N
the closest answer is A. 130 N
If another driver does not follow the right-of-way rules at a YIELD sign, what should you do?
Answer:Stop
Explanation:
To avoid accident even thought you in the right
An object is lifted from the surface of aspherical planet to an altitude equal to the radius of the planet.As a result, what happens to the mass and wieght of the object?a) massincreases; weight decreasesb) massdecreases; weight decreasesc) massincreases; weight increasesd) massincreases; weight remains the samee) mass remainsthe same; weight decreasesf) mass remainsthe same; weight increasesg) mass remainsthe same; weight remains the same
Answer: Mass remains the same and weight decrease
Explanation: mass is always constant but weight varies. Weight is affected by height. The higher you go the lesser your weight because you are moving away from gravity
If Air Traffic Control advises that radar service is terminated when the pilot is departing Class C airspace, the transponder should be set to code?
Explanation:
1200 feet.
In the given scenario we would be operation under VFR.
VFR is visual flight rules are set of regulations under which the pilot operates aircraft in a weather which generally clear enough that facilitates pilot to see the flight where it is going.
A man plans to make a rectangular garden with one side adjoining a neighbor’s yard. The garden is to be 675 ft2. If the neighbor agrees to pay half of the dividing fence, what should the dimensions of the garden be to ensure a minimum cost for the man?
Answer:
L = W = 26 ft.
Explanation:
Let the length of the garden is L and the width is W.
area of garden , A = L x W
675 = L x W ... (1)
Costing is minimum when the perimeter is minimum.
Perimeter, P = 2 ( L + W)
[tex]P=2\left ( L+\frac{675}{L} \right )[/tex]
For maxima and minima, differentiate perimeter with respect to L.
[tex]\frac{dP}{dL}=2\left ( 1-\frac{675}{L^{2}} \right )[/tex]
It should be zero for maxima and minima
L² = 675
L = 26 ft
W = 675/26 = 26 ft
Now, [tex]\frac{d^{2}P}{dL^{2}}=4\frac{675}{L^{3}}[/tex]
It is positive, so the costing is minimum.
So, length and width of the garden is 26 ft.
When does the most significant drop in physical activity typically occur in people's lives?
Answer:
Later high school years and freshman year of college
Explanation:
The transition from high school to college is an important developmental milestone that holds the potential for personal growth and behavioral change. A cohort of 2,025 students was recruited during the summer before they matriculated into college and completed Internet-based surveys about their participation in a variety of behavioral risks during the last three months of high school and throughout the first year of college. Alcohol use, marijuana use, and sex with multiple partners increased during the transition from high school to college, whereas driving after drinking, aggression, and property crimes decreased. Those from rural high schools and those who elected to live in private dormitories in college were at highest risk for heavy drinking and driving after drinking.
A particle with a positive charge q is moving in the positive x direction with velocity of magnitude v0. If there were an electric field in the positive y direction with magnitude E0, and the velocity of the particle is unchanged, find the magnitude and direction of magnetic field.
The magnitude of the magnetic field in the given scenario is zero.
Explanation:The magnitude and direction of the magnetic field can be determined by analyzing the force experienced by the positively charged particle q as it moves in the positive x direction and encounters an electric field in the positive y direction. The force experienced by the particle, due to the Lorentz force law, is given by:
F = qvB sin
Where F is the force, q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. Since the velocity of the particle remains unchanged, the force experienced by the particle in the y direction must be zero. This means that the angle θ between the velocity and the magnetic field must be 90 degrees. Therefore, sin θ = 1. Substituting these values into the equation, we get
0 = qv0
Solving for B, we find that
B = 0
Therefore, the magnitude of the magnetic field in this scenario is zero.
When a person strikes and lights a match, potential energy in the match is transformed into which types of energy?
a. electrical and thermal
b. thermal and radiant
c. radiant and nuclear
d. nuclear and kinetic
The eel has a certain amount of rotational kinetic energy when spinning at 14 spins per second. If it swam in a straight line instead, about how fast would the eel have to swim to have the same amount of kinetic energy as when it is spinning
Answer:
[tex]v=\frac{28\pi R}{\sqrt{2} }[/tex] m/sec
Explanation:
According to the question
[tex]\frac{1}{2}I\omega^2=\frac{1}{2} mv^2[/tex]
\omega= angular velocity = 14 spins per sec
I= MOI of the spinning eel= 0.5mR^2
m= mass of eel
v= linear velocity of eel.
[tex]\frac{1}{2}0.5mR^2\omega^2=\frac{1}{2} mv^2[/tex]
solving for v in the above equation we get
[tex]v=\sqrt{\frac{R^2\omega^2}{2} }[/tex]
now \omega in radians = 14 rev/sec×2π radian/rev= 28π radian/sec
value of radius is not provided in the question
therefore we get
[tex]v=\frac{28\pi R}{\sqrt{2} }[/tex]
With each bounce off the floor, a tennis ball loses 18% of its mechanical energy due to friction. When the ball is released from a height of 2.9 m above the floor, what height will it reach after the fourth bounce?
170 cm
130 mm
130 cm
13 cm
Answer:
130 cm
Explanation:
After each bounce, the mechanical energy decreases by 18%, which means the maximum height decreases by 18%. Modeling this as exponential decay:
h = 290 (1 − 0.18)ⁿ
h = 290 (0.82)ⁿ
h = 290 (0.82)⁴
h = 130
The ball reaches a height of 130 cm.
Answer:
130 cm
Explanation:
Gradpoint
A particle on a string is whirled in a circle at a constant speed, in the absence of a gravitational field. a top view of the circular motion at one instant is
Answer:
This is an incomplete question. The diagram and a part of the question is missing.
Explanation:
In the context, it is given that the particle tight to a string is whirled in a circular direction in a constant speed. And there is no gravitational field.
Now since the particle's speed is constant, hence the tangential acceleration is zero. The particle exhibits radial acceleration only which will act along the radius towards the center of the circle (towards the left).
Therefore the correct representation of the component of the acceleration is given below --
A particle being whirled in a circle at a constant speed demonstrates uniform circular motion. The changing direction of velocity results in a centripetal acceleration, determined by an external centripetal force. Similar motion is observed in centrifuges and orbiting satellites.
Explanation:A top view of a particle on a string whirled in a circle at a constant speed would show uniform circular motion. In this simplest form of curved motion, the particle moves along a fixed radius at a constant speed. Although the magnitude of its velocity doesn't change, the direction of the velocity is constantly changing, leading to what's referred to as centripetal acceleration. This is caused by a net external force - known as the centripetal force.
This type of motion is similar to other examples of circular motion, like a satellite orbiting Earth, a race car moving around a racetrack, or a ball being swung in a circle. Whenever there's no centripetal force to cause circular motion, inertia carries the object along a line tangent to the circle observed as outward motion in a centrifugal frame of reference. This inertial effect is seen in various physical phenomena.
Learn more about Uniform Circular Motion here:https://brainly.com/question/29345640
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