Answer:
48 m
Explanation:
Initial speed of laser,u=12 m/s
Acceleration due to gravity=-[tex]1.5m/s^2[/tex]
Because laser goes against to gravity when it is thrown in upward direction.
Final velocity of laser=v=0
We know that
[tex]v^2-u^2=2as[/tex]
Substitute the values then we get
[tex]0-(12)^2=2(-1.5)s[/tex]
[tex]-144=-3s[/tex]
[tex]s=\frac{144}{3}[/tex]
[tex]s=48 m[/tex]
Hence, the maximum height reached by the laser= 48 m
The maximum height reachedby the laser is 48 m.
To calculate the maximu height reached by the laser, we use the formula below.
Formula:
v² = u²+2gH........... Equation 1Where:
u = Initial velocityv = final velocityg = accceleration due to gravity of planet XH = maximum height reached by the laser.Make H the subject of the equation
H = (v²-u²)/2g........... Equation 2From the question,
Given:
v = 12 m/su = 0 m/sg = 1.5 m/s²Substitute these values into equation 2
H = (12²-0²)/(2×1.5)H = 144/3H = 48 mHence, The maximum height reachedby the laser is 48 m.
Learn more about maximum height here: https://brainly.com/question/13665920
7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3 m, at what distance from the pivot point is the small child sitting in order to maintain the balance?
Answer:
Explanation:
Given
mass of children [tex]m_1=20\ kg[/tex]
[tex]m_2=30\ kg[/tex]
distance between two children [tex]L=3\ m[/tex]
suppose small child is at a distance of x m from pivot point
so torque of small child and heavier child must be equal
[tex]20\times (x)=30\times (3-x)[/tex]
[tex]2x=9-3x[/tex]
[tex]5x=9[/tex]
[tex]x=1.8\ m[/tex]
A solar eclipse that occurs when the new moon is too far from Earth to completely cover the Sun can be either a partial solar eclipse or a(n)_____________.
Answer:
ANULAR ECLIPSE
Explanation:
ANULAR ECLIPSE. Because the moon is very far, only a portion of the sun would be obscured, and then only the moon's outer ring will be viewable; this is called the anular eclipse.The ring of fire marks the maximum stage of an annular solar eclipse.
_______ describes the relationship between a stimulus and its resulting sensation by proposing that the JND is a constant fraction of the stimulus intensity.
Answer:
Weber's Law
Explanation:
Weber’s law is an important law in psychology.
This law quantifies the perception of change within a provided stimulus.
This law is also known as Weber-Fechner law as it relates two hypotheses in the field of psychophysics which are Weber law and the Fechner law.
Weber’s law describes that an observed change in a stimulus is a constant ratio of the original stimulus.
Final answer:
Weber's Law describes the relationship between a stimulus and the resulting sensation, stating that the Just Noticeable Difference (JND) is a constant fraction of stimulus intensity. It is a key principle in psychophysics, useful for experiments such as determining the JND for different weights of rice in bags.
Explanation:
The concept you're inquiring about is Weber's Law, which is a fundamental principle in the field of psychophysics, a branch of psychology. Weber's Law describes the relationship between a stimulus and its resulting sensation by proposing that the Just Noticeable Difference (JND) is a constant fraction of the stimulus intensity.
To put this into practice, for example, if one were testing the JND of various weights of rice in bags, they could use incremental percentage increases of the weight to determine the smallest difference that can be perceived. By choosing increments like 10 percent or 20 percent between certain weight thresholds, a researcher can apply Weber's Law to determine the JND in a standardized way.
This law helps explain why when you have a cup of coffee with little sugar, adding a teaspoon can make a noticeable difference, but in a cup with much more sugar, that same teaspoon becomes less detectable, requiring a proportionally larger amount of sugar to notice a change in sweetness.
A Roller Derby Exhibition recently came to town. They packed the gym for twoconsecutive weekend nights at South's field house. On Saturday evening, the68-kg Anna Mosity was moving at 17 m/s when she collided with 76-kg SandraDay O'Klobber who was moving forward at 12 m/s and directly in Anna's path.Anna jumped onto Sandra's back and the two continued moving together atthe same speed. Determine their speed immediately after the collision.
Answer:
14.4 m/s
Explanation:
mass of Anna (Ma) = 68 kg
speed of Anna (Va) = 17 m/s
mass of SandraDay (Ms) = 76 kg
speed of SandraDay (Vs) = 12 m/s
We can find their speed (V) immediately after collision from the conservation of momentum where
(Ma x Va) + (Ms + Vs) = (Ma + Ms) x V
where V = speed immediately after collision
(68 x 17) + (76 + 12) = (68 + 76) x V
2068 = 144 V
V = 2068 / 144 = 14.4 m/s
Suppose a solid uniform sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The angular velocity of the sphere at the bottom of the incline depends on:________.a) the mass of the sphere.
b) the radius of the sphere.
c) both the mass and the radius of the sphere.
d) neither the mass nor the radius of the sphere.
The angular velocity of a solid uniform sphere rolling without slipping down an inclined plane does not depend on the mass or the radius of the sphere; it depends on the height from which it rolls down and gravitational acceleration only.
The question asks what determines the angular velocity of a solid uniform sphere as it rolls without slipping down an inclined plane. It specifically queries whether this angular velocity depends on the mass (M) of the sphere, the radius (R) of the sphere, both, or neither.
The answer is that the angular velocity of a rolling sphere at the bottom of the incline does not depend on the mass of the sphere or the radius of the sphere. According to the principles of conservation of energy and the dynamics of rolling motion, all objects regardless of their mass or radius, will roll down an incline and reach the bottom with the same angular velocity if they start from rest, provided that they do not slip and there is no air resistance. This is because the potential energy lost is completely converted into kinetic energy (both translational and rotational), and the mechanical energy of the system remains constant.
The angular velocity of the sphere at the bottom of the incline depends on d) neither the mass nor the radius of the sphere.
When the sphere rolls without slipping, there is a relationship between its linear velocity (v) and angular velocity (ω) given by the equation v = ωR.
Total mechanical energy of the sphere = translational kinetic energy + rotational kinetic energy + potential energy.
At the bottom of the incline, all potential energy will have converted into kinetic energy:
Initial Potential Energy (U) = mghTranslational Kinetic Energy (Kt) = (1/2)mv²Rotational Kinetic Energy (Kr) = (1/2)Iω², where I for a solid sphere is (2/5)mR²By energy conservation:
mgh = (1/2)mv² + (1/2)Iω²Substituting I and the relation v = ωR gives us:
mgh = (1/2)mv² + (1/2)*(2/5)mR²ω²Solving for ω:
mgh = (1/2)mv² + (1/5)mR²ω²Since v = ωR, ω² = v²/R²mgh = (1/2)mv² + (1/5)m(v²)7gh = v²v = √(7gh)ω = v/R = √(7gh)/RFrom the above steps, it's evident that the angular velocity ω depends on the height h the sphere rolls down, and the acceleration due to gravity g, but not on the mass M or radius R of the sphere.
Therefore, the correct answer is: d) neither the mass nor the radius of the sphere.
Suppose you first walk 12.0 m in a direction 20 owest of north and then 20.0 m in a direction 40.0osouth of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , then this problem asks you to find their sum R = A + B . The two displacements A and B add to give a total displacement R having magnitude R and direction in degrees.
Answer:
R=19.5m
[tex]\theta[/tex] = 4.65° S of W
Explanation:
Refer the attached fig.
displacement of the x and y components
x-component displacement is ([tex]R_{x}[/tex]) = [tex]A_{x}+B_{x}[/tex]
= A [tex]\sin[/tex](20°) + B [tex]\cos[/tex](40°)
= -12.0[tex]\sin[/tex](20°) + 20.0[tex]\cos[/tex](40°)
= -19.425m
x-component displacement is ([tex]R_{y}[/tex]) = [tex]A_{y}+ B_{y}[/tex]
= A [tex]\cos[/tex](20°) - B [tex]\sin[/tex](40°)
= 12.0[tex]\cos[/tex](20°) - 20.0[tex]\sin[/tex](40°)
= -1.579
resultant displacement
∴
R = [tex]\sqrt{R_{x}^{2} +R_{y}^{2} } }[/tex]
=[tex]\sqrt{(-19.425)^{2}+(-1.579)^{2} }[/tex]
=19.5m
[tex]\theta[/tex] = [tex]\tan^{-1}\left | \frac{R_{x}}{R_{y}} \right |[/tex]
[tex]\theta[/tex] = [tex]\tan^{-1}\left | \frac{1.579}{19.425} \right |[/tex]
[tex]\theta[/tex] = 4.65° S of W
This problem involves finding the result of two vector displacements using vector addition. Using the Pythagorean theorem, the magnitude of the overall displacement can be determined. The direction of the displacement can be calculated using the tangent formula.
Explanation:This is a vector addition problem in physics. We can solve it using Pythagorean theorem for finding the magnitude and tangent formula for finding the direction.
For the first leg of the journey(A), you are going 20° W of N, or 70° clockwise from the x-axis. For the magnitude of this vector: Ax = 12.0m cos(70°) and Ay = 12.0m sin(70°).
For the second leg of the journey(B), you are going 40° S of W, or 130° clockwise from the x-axis. For the magnitude of this vector: Bx = 20.0m cos(130°) and By = 20.0m sin(130°).
Summing the x and y components gives: Rx = Ax + Bx and Ry = Ay + By. The magnitude R can then be found with Pythagorean theorem formula: R = sqrt(Rx² + Ry²).
The direction can be found using tangent formula: θ = atan(Ry / Rx). As Rx is negative and Ry is positive, θ will fall in the second quadrant.
Learn more about Vector addition here:https://brainly.com/question/31940047
#SPJ3
The distance between a carbon atom (m = 12 u) and an oxygen atom (m = 16 u) in the CO molecule is 1.13 × 10–10 m. How far from the carbon atom is the center of mass of the molecule?
The distance between the center of mass of the molecule and carbon atom is 0.65 × 10⁻¹⁰ m
To answwer the question, we need to know the center of mass of CO molecule
What is the center of mass of CO molecule?The center of mass of the CO molecule is given by
x = m₁x₁ + m₂x₂/(m₁ + m₂) where
m₁ = mass of carbon atom = 12 u, x₁ = position of carbon atom = 0 m (assuming its at the origin), m₂ = mass of oxygen atom = 16 u and x₂ = position of oxygen atom = distance between molecules = 1.13 × 10⁻¹⁰ mSubstituting the values of the variables into the equation, we have
x = (m₁x₁ + m₂x₂)/(m₁ + m₂)
x = (12 u × 0 m + 16 u × 1.13 × 10⁻¹⁰ m)/(12 u + 16 u)
x = (0 + 16 u × 1.13 × 10⁻¹⁰ m)/28 u
x = 18.08 × 10⁻¹⁰ um/28 u
x = 0.646 × 10⁻¹⁰ m
x ≅ 0.65 × 10⁻¹⁰ m
The distance between the center of mass of the molecule and carbon atom
Since the carbon atom is at x₁ = 0 m and the center of mass is at x = 0.65 × 10⁻¹⁰ m, the distance between the carbon atom and the center of mass of the molecule is d = x - x₁
= 0.65 × 10⁻¹⁰ m - 0 m
= 0.65 × 10⁻¹⁰ m
So, the distance between the center of mass of the molecule and carbon atom is 0.65 × 10⁻¹⁰ m
Learn more about center of mass here:
https://brainly.com/question/27092359
#SPJ1
What would occur if there was a gain-of-function mutation in the promoter for the cyclin E gene such that cyclin E protein was always made at high levels even under conditions in which cyclin E would not normally be made?
Answer:
b) Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.
Explanation:
In the given problem, if there exists a gain-of-function mutation for the given cell, there would not be the formation of cyclin E when there is the possibility of cells movement via the checkpoint of the G1/S, even when there are non-deal conditions for the division of cell. Thus, the correct option in the lists of options is the option b.
A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 25.0 m/s, make it climb a 2.00×102 m high hill, and then cause it to travel at a constant 25.0 m/s by exerting a 5.00×102 N force for an hour.
Answer:
3894531 coulombs
Explanation:
1 hour = 3600 seconds
Let g = 10m/s2
The distance that the car travel at 25 m/s over an hour is
s = 25 * 3600 = 90000 m
The total mechanical energy of the car is the sum of its kinetic energy to reach 25 m/s, its potential energy to climb up 200m high hill and it work to travel a distance of s = 90000m with F = 500 N force:
[tex] \sum E = E_k + E_p + E_W[/tex]
[tex]\sum E = mv^2/2 + mgh + Fs[/tex]
[tex]\sum E = 750*25^2/2 + 750*10*200 + 500*90000 = 46734375 J[/tex]
This energy is drawn from the battery over an hour (3600 seconds), so its power must be
[tex]P = E / t = 46734375/3600 = 12982 W[/tex]
The system is 12V so its current is
[tex]I = P/U = 12982 / 12 = 1081.8 A[/tex] or 1081.8 Coulombs/s
The the total charge it needs for 1 hour (3600 s) is
C = 1081.8 * 3600 = 3894531 coulombs
The quantity of charge the batteries must be able to move is equal to 3.9 μC.
Given the following data:
Voltage = 12.0 VoltsMass = 750 kgSpeed = 25.0 m/sHeight = 200 meters.Force = 500 Newton.Time = 1 hour = 3600 seconds.Scientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]To find the quantity of charge the batteries must be able to move:
In this scenario, we would calculate the distance traveled and the total energy that is possessed by this battery-operated car.
For the distance.Mathematically, the distance covered by an object is given by this formula:
[tex]Distance = speed \times time\\ \\ Distance = 25 \times 3600[/tex]
Distance = 90,000 meters.
For the total energy:[tex]E = mgh + \frac{1}{2} mv^2 + Fd\\ \\ E=[750\times 9.8 \times 200] + \frac{1}{2} \times 750 \times 25^2 + [500 \times 90000]\\ \\ E=1470000+234375+45000000[/tex]
Total energy = 46,704,375 Joules.
Next, we would calculate the power consumed by this battery-operated car:
[tex]Power = \frac{Energy}{time}\\ \\ Power = \frac{46,704,375}{3600} [/tex]
Power = 12,973.44 Watts.
Also, we would calculate the current:
[tex]Current = \frac{power}{voltage} \\ \\ Current = \frac{12,973.44}{12}[/tex]
Current = 1,081.12 Amperes.
Now, we can calculate the quantity of charge the batteries must be able to move:
[tex]Q = current \times time\\ \\ Q = 1081.12 \times 3600[/tex]
Q = 3,892,032 Coulombs.
Note: 1 μC = [tex]1 \times 10^6 \;C[/tex]
Q = 3.9 μC
Read more on charge here: https://brainly.com/question/14327016
Suppose a 58-N sled is resting on packed snow. The coefficient of kinetic friction is 0.11. If a person weighing 655 N sits on the sled, what force is needed to pull the sled across the snow at constant speed?
Answer:
You have to apply force of 78.43 N to pull the sled or person across the snow at constant speed
Explanation:
Given data
Sled force F₁=58 N
Person weight F₂=655 N
Coefficient of kinetic friction u=0.11
To find
Friction Force
Solution
To pull the sled across the snow means the force you apply must be equal to and opposite the force of frictional.
First we need to find the Total normal force
So
[tex]F_{Normal-Force}=F_{1}+F_{2}\\F_{Normal-Force}=58N+655N\\F_{Normal-Force}=713N[/tex]
Now the force of friction
[tex]F_{Friction-Force}=u_{coefficient}*F_{Normal-Force}\\F_{Friction-Force}=0.11*713N\\F_{Friction-Force}=78.43N[/tex]
So you have to apply force of 78.43 to pull the sled or person across the snow at constant speed
A rocket is continuously firing its engines as it accelerates away from Earth. For the first kilometer of its ascent, the mass of fuel ejected is small compared to the mass of the rocket. For this distance, which of the following indicates the changes, if any, in the kinetic energy of the rocket, the gravitational potential energy of the Earth-rocket system, and the mechanical energy of the Earth-rocket system? System Gravitational System Potential Energy Rocket Kinetic Energy Mechanical Energy
(A) Increasing IncreasingIncreasing
(B) Increasing Increasing Constant IncreasingDecreasing Decreasing Decreasing Increasing Constant
Answer:
A) Increasing, Increasing, Increasing
Explanation:
The greater the rate of fuel ejection higher will be the kinetic energy of the rocket. As rocket is fire upward its fuel rejection rate increases and hence it's kinetic energy increases.
Gravitational potential energy,
as the rocket moves further it's r from the Earth increases. Hence the Gravitational potential energy increases as its height from the Earth increases.
Therefore, mechanical energy of the rocket must also increase as it is sum of kinetic and potential energy.
Hence Option A is correct.
For the first kilometer of the rocket's ascent, the kinetic energy, gravitational potential energy, and mechanical energy of the Earth-rocket system increase.
Explanation:The changes in the kinetic energy, gravitational potential energy, and mechanical energy of the Earth-rocket system for the first kilometer of the rocket's ascent can be determined.
Since the mass of fuel ejected is small compared to the mass of the rocket, the rocket's kinetic energy will increase as it accelerates away from Earth. The gravitational potential energy of the Earth-rocket system will also increase as the rocket moves higher against the pull of gravity. Therefore, the changes in the three quantities are as follows:
Kinetic Energy: IncreasingGravitational Potential Energy: IncreasingMechanical Energy: IncreasingWhat amount of heat is required to raise the temperature of 25 grams of copper to cause a 15ºC change? The specific heat of copper is 0.39 J/gºC. A. 115 J B. 150 J C. 250 J D. 300 J
The amount of heat required is B) 150 J
Explanation:
The amount of heat energy required to increase the temperature of a substance is given by the equation:
[tex]Q=mC\Delta T[/tex]
where:
m is the mass of the substance
C is the specific heat capacity of the substance
[tex]\Delta T[/tex] is the change in temperature of the substance
For the sample of copper in this problem, we have:
m = 25 g (mass)
C = 0.39 J/gºC (specific heat capacity of copper)
[tex]\Delta T = 15^{\circ}C[/tex] (change in temperature)
Substituting, we find:
[tex]Q=(25)(0.39)(15)=146 J[/tex]
So, the closest answer is B) 150 J.
Learn more about specific heat capacity:
brainly.com/question/3032746
brainly.com/question/4759369
#LearnwithBrainly
50 J of work was performed in 20 seconds. How much power was used to perform this task? A. 0.4 W B. 2.5 W C. 4 W D. 24.5 W
Answer:
B. 2.5 W
Explanation:
Power: This can be defined as the rate at which energy used or it i the rate at which work is done. The S.I unit of power is Watt (W).
Mathematically,
Power = Energy or work/Time
P = W/t ........................Equation 1
Where P = power used to perform the task, W = work, t = time.
Given: W = 50 J, t = 20 s.
Substitute into equation 1
P = 50/20
P = 2.5 W.
Hence the power is 2.5 W.
The right option is B. 2.5 W
The power used to perform the work of 50J in 20 seconds is 2.5 watts. This much power should be generated by it. The correct answer is B.
Given that 50 J of work was performed in 20 seconds,
The power used to perform a task, you can use the formula:
Power = Work / Time
Power = 50 J / 20 s
Power = 2.5 W
Therefore, the power used to perform the task is 2.5 W.
Therefore, the correct answer is B) 2.5 W.
To know more about the power:
https://brainly.com/question/19382734
#SPJ6
You break a piece of Styro foam packing material, and it releases lots of little spheres whose electric charge makes them stick annoyingly to you. If two of the spheres carry equal charges and repel with a force of 22 { mN} when they're 16 { mm} apart, what's the magnitude of the charge on each?
Answer:
The magnitude of charge on each sphere is [tex]q=2.50\times 10^{-8}\ C[/tex].
Explanation:
Given that,
Force of repulsion between the charges, F = 22 mN
The distance between spheres, r = 16 mm = 0.016 m
It is mentioned that both the spheres carry equal charges. The force between charges is given by :
[tex]F=\dfrac{kq^2}{r^2}[/tex]
[tex]q=\sqrt{\dfrac{Fr^2}{k}}[/tex]
[tex]q=\sqrt{\dfrac{22\times 10^{-3}\times (0.016)^2}{9\times 10^9}}[/tex]
[tex]q=2.50\times 10^{-8}\ C[/tex]
So, the magnitude of charge on each sphere is [tex]q=2.50\times 10^{-8}\ C[/tex]. Hence, this is the required solution.
A revolutionary war cannon, with a mass of 2240 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 131 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?Answer in units of m/s.
Answer:
v' = -0.0906 m/s
Explanation:
given,
mass of cannon, M = 2240 Kg
mass of the ball, m = 15.5 Kg
speed of cannon ball, v = 131 m/s
speed of he cannon = ?
initial speed of cannon and the cannon ball is equal to 0 m/s
using conservation of energy
(M+m)V = M v' + m v
(M+m) x 0= 22400 v' + 15.5 x 131
22400 v' = -2030.5
v' = -0.0906 m/s
negative sign represent the canon will move in opposite direction of the ball.
hence, speed of cannon is equal to 0.0906 m/s
Using the law of conservation of momentum and the given values, the recoil velocity of the cannon is found to be approximately 0.9058 m/s after firing a cannonball.
According to the law of conservation of momentum, the total momentum before an event must equal the total momentum after the event, if no external forces act on the system. In this case, the cannon and the cannonball system is isolated and free to recoil, so the momentum before the cannon fires must be equal to the momentum after the cannon fires.
Momentum is the product of mass and velocity. Before the cannon fires, both the cannon and the cannonball are at rest and have a momentum of zero. After the cannon fires, the momentum of the cannonball is mcannonball × vcannonball, which must be equal and opposite to the momentum of the cannon. Therefore, mcannonball × vcannonball = mcannon × vcannon.
Using the given values for mass and velocity of the cannonball (15.5 kg and 131 m/s), and the mass of the cannon (2240 kg), we can calculate the cannon's recoil velocity using the equation:
mcannonball × vcannonball = mcannon × vcannon
(15.5 kg × 131 m/s) / 2240 kg = vcannon
The calculated recoil velocity of the cannon is approximately 0.9058 m/s.
You have a set of calipers that can measure thicknesses of a few inches with an uncertainty of ± 0.005 inches. You measure the thickness of a deck of 52 cards and get 0.590 in: a. If you now calculate the thickness of 1 card, what is your answer, including its uncertainty?
Answer:
The thickness of 1 card is between 0.006in and 0.016in
Explanation:
Thickness of a deck of 52 cards = 0.590in
Thickness of 1 card = 0.590in/52 = 0.011in
Uncertainty = + or - 0.005in
Lower limit of thickness of 1 card = 0.011in - 0.005in = 0.006in
Upper limit of thickness of 1 card = 0.011in + 0.005in = 0.016in
Therefore, thickness of 1 card is between 0.006in and 0.016in
A sheet of custom-size copy paper measures 3.5 in. by 14 in. If a ream (500 sheets) of this paper has a volume of 98 in. cubed, how thick is the ream?
Answer:
2 in
Explanation:
Area of a sheet = 3.5 x 14 in²
Volume of ream, V = 98 in³
Let t be the thickness of ream
So, Volume of ream Area of a sheet x thickness of ream
98 = 3.5 x 14 x t
t = 2 in
Thus, the thickness of ream is 2 in.
Answer:
t = 2 in
Explanation:
given,
width of sheet, W = 3.5 in
Length of the sheet, L = 14 in
thickness of the ream = ?
volume of paper = 98 in³
we now,
Volume = Length x width x thickness
V = l w t
98 = 14 x 3.5 x t
[tex]t = \dfrac{98}{14\times 3.5}[/tex]
t = 2 in
thickness of the ream is equal to 2 in.
This traction loss occurs in the rear wheels of a vehicle. a) Braking-induced traction loss b) Acceleration-induced traction loss c) Front wheel traction loss (skid) d) Rear wheel traction loss (skid) e) Driver-induced skids
Traction loss in a vehicle's rear wheels is most likely due to either acceleration-induced traction loss or rear wheel traction loss (skid). The former is caused by rapid acceleration, while the latter can be due to turning or braking.
Explanation:The traction loss described in your question occurs in the rear wheels of a vehicle and is likely due to acceleration-induced traction loss or rear wheel traction loss (skid). Acceleration-induced traction loss happens when rapid acceleration causes the tires to lose grip on the road. This is common in high-powered, rear-wheel drive vehicles. Rear wheel traction loss or a rear wheel skid, on the other hand, generally occurs when the rear wheels lose grip during turning or braking. In this scenario, the vehicle's rear end can swing out in either direction, causing the vehicle to spin.
Learn more about Rear Wheel Traction Loss here:https://brainly.com/question/31667096
#SPJ6
Final answer:
Rear wheel traction loss (skid) occurs when the rear tires lose grip on the road surface. In a front-wheel-driven car, friction at the rear wheel is in the opposite direction of motion. This concept is important for analysis in vehicle dynamics and skid prevention.
Explanation:
The traction loss referred to in the question occurs in the rear wheels of a vehicle and is specifically known as rear wheel traction loss (skid). This is a type of skid that happens when the rear tires of a vehicle lose grip on the road surface, often as a result of oversteering, sudden acceleration, or slippery conditions. It's essential to distinguish this from other types of traction loss, such as braking-induced traction loss, which occurs when the tires can no longer grip the road during heavy braking, or driver-induced skids, which are the result of driver error.
With respect to the problem provided, in a front-wheel-driven car, the friction at the rear wheel is generally in the opposite direction of motion. This is because the rear wheels are not powered in this configuration; instead, they follow the rotation induced by the car's movement, which is primarily driven by the front wheels. Therefore, while the front wheels are pulled forward by the engine, creating friction in the direction of motion, the rear wheels experience a kind of rolling resistance and the friction present there acts in the opposite direction of the car's movement.
Understanding the dynamics of friction and traction is critical for analyzing cases like where a vehicle skids and comes to rest after traveling a certain distance or determining the force required to maintain a constant speed in the presence of kinetic friction. These analyses involve concepts like travel reduction ratio, torque transferred to the wheel axle, tractive efficiency of the wheel, and the actual velocity of the wheel.
heather and matthew walk with an average velocity of 98 m/s eastward. If it takes them 34 min ro walk to the store, what is their displacement?
Answer:
The answer is 199920 meters.
Explanation:
First we need to find how many seconds they walked to the store:
[tex]34*60=2040[/tex]
If they walked 2040 seconds to eastward with velocity of 89 m/s, the displacement will be:
[tex]2040*98=199920[/tex]
Force = mass x acceleration : __________.a. for any force, there is an equal and opposite reaction force. b. the orbit of each planet about the Sun is an ellipse with the Sun at one focus. c. a planet moves faster in the part of its orbit nearer the Sun and slower when farther from the Sun, sweeping out equal areas in equal times. d. an object moves at constant velocity if there is no net force acting upon it. e. more distant planets orbit the Sun at slower average speeds, obeying the precise mathematical relationship p2 =a3.
Answer:
a. for any force, there is an equal and opposite reaction force.
Explanation:
Newton third law of motion states that " for every action, there is an equal and opposite reaction.
Action force= reaction force =
M₁ * a₁ = M₂ * a₂
where M₁ = mass of object 1
a₁ = acceleration due to object 1
M₂ = mass of object 2
a₂ = corresponding acceleration due to object 2
this happen in everyday life of an individual. an example is seen in a groups of three man exerting a push force on a static car, the push force on the car make the car to exhibit some motion equivalent to the force applied by the three man.
If you are caught outdoors in a thunderstorm, why shouldn’t you stand under a tree? Can you think of a rea- son why you should not stand with your legs far apart? Or why lying down can be dangerous? (Hint: Consider the electric potential difference.)
Answer:
Because its dangerous.
Explanation:
During lighting strikes, there is discharge or energy transfer(electrons) from the clouds to the earth, these electrons flow through the path with least resistance between the cloud and the earth. Also the electric field around the tip of the leaves are strong, which makes trees a great target.
we as humans have lower resistance than trees, that is to say, the lighting may leave the tree and flow trough the body to the earth.
The tree and the ground around it are then raised to a high potential relative to the ground some distance away.
If you stand with your legs far apart, one leg on a higher-potential part of the ground than the other, or if you lie down with a potential difference between your head and your feet, you may find yourself a conducting path.
If it is also raining, the electricity may transfer down the wet tree to the wet ground and shock anyone standing near the tree.
During a thunderstorm, it is advised not to stand under a tree to avoid being struck by lightning, not to stand with legs far apart to reduce the electric potential difference, and not to lie down due to increased risk. Staying inside a car provides safety as it acts as a Faraday cage.
Explanation:Standing under a tree during an electrical storm is dangerous since lightning tends to strike the tallest object in an area, which could be the tree you're under, potentially causing serious injury or death. As for not standing with legs apart, this is because, in the event of a ground strike, electricity can travel through the ground. If your legs are far apart, there could be a significant electric potential difference between them, which can result in a stronger current passing through your body, leading to severe harm. Lying down increases your contact with the ground, and consequently, the risk of current flowing through your body from a ground strike is greater.
During thunderstorms, your car acts as a Faraday cage, which shields you from electric fields if a lightning strike occurs nearby. It's safest to remain inside the car with windows closed. However, it's critical to refrain from touching metal parts inside the car as lightning can transfer its charge through the car's metal frame.
The electromagnetic interaction _______.A. applies only to charges at rest B. applies only to charges in motion C. is responsible for sliding friction and contact forces D. all of the above E. none of the above
Answer:
C. is responsible for sliding friction and contact forces
30 seconds of exposure to 115 dB sound can damage your hearing, but a much quieter 94 dB may begin to cause damage after 1 hour of continuous exposure. You are going to an outdoor concert, and you'll be standing near a speaker that emits 50 W of acoustic power as a spherical wave. What minimum distance should you be from the speaker to keep the sound intensity level below 94 dB?
Answer:
39.8 m ≈ 40 m
Explanation:
power (P) = 50 W
sound intensity level ([tex]p[/tex]) = 94 dB
the distance (r) can be gotten from the equation I = [tex]\frac{power}{4nr^{2} }[/tex] (take not that π is shown as [tex]n[/tex])
making r the subject of the formula we have r = [tex]\sqrt{\frac{power}{4nI} }[/tex] (take not that π is shown as [tex]n[/tex])
But to apply this equation we need to get the value of the intensity (I)
we can get the intensity (I) from the formula sound intensity level ([tex]p[/tex]) = 10 log₁₀[tex](\frac{I}{I'})[/tex] rearranging the above formula we have intensity (I) = [tex]I' x 10^{\frac{p}{10} }[/tex]I' = reference intensity = 1 x[tex]10^{-12} W/m^{2}[/tex]now substituting all required values into the formula for intensity (I) I = [tex]1 x 10^{-12} x 10^{\frac{94}{10} }[/tex] = 0.00251 [tex]W/m^{2}[/tex]now that we have the value of the intensity (I) we can substitute it into the formula for the distance (r)
distance (r) = [tex]\sqrt{\frac{power}{4nI} }[/tex]
r = [tex]\sqrt{\frac{50}{4x3.142x0.00251} }[/tex] = 39.8 m ≈ 40 m
A bowl contains 7 red balls and 10 blue balls. A woman selects 4 balls at random from the bowl. How many different selections are possible if at least 3 balls must be blue?
Answer:
1050 possible selections
Explanation:
Number of red balls = 7
Number of blue balls = 10
Let blue balls be B and red balls be R
four balls are to be selected. At least three must be blue.
the first combination is 4blue 0red
second combination is 3blue 1red
(10C4*7C0 + 10C3*7C1)
210 *1 +120 *7
210+840
=1050 possibilities
There are 71 different selections possible if at least 3 balls must be blue.
Explanation:To find the number of different selections possible, we can consider the different scenarios where at least 3 balls are blue:
Scenario 1: Exactly 3 balls are blue and 1 ball is red. In this case, there are 10 ways to select the 3 blue balls (from the 10 blue balls available) and 7 ways to select the red ball (from the 7 red balls available). So, the total number of selections for this scenario is 10 * 7 = 70.Scenario 2: All 4 balls are blue. In this case, there is only 1 way to select all 4 blue balls. So, the total number of selections for this scenario is 1.Summing up the number of selections from both scenarios, we get 70 + 1 = 71 different selections possible.
Learn more about Different selections of balls here:https://brainly.com/question/34323846
#SPJ3
In 1977, Kitty O’Neil drove a hydrogen peroxide–powered rocket dragster for a record time interval (3.22 s) and final speed (663 km/h) on a 402-m-long Mojave Desert track. Determine her average acceleration during the race and the acceleration while stopping (it took about 20 s to stop). What assumptions did you make?
Answer:
57.19461 m/s²
-9.20833 m/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Equation of motion
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{663}{3.6}-0}{3.22}\\\Rightarrow a=57.19461\ m/s^2[/tex]
The acceleration during the race is 57.19461 m/s²
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-\dfrac{663}{3.6}}{20}\\\Rightarrow a=-9.20833\ m/s^2[/tex]
The acceleration while stopping is -9.20833 m/s²
The assumptions made here are:
The initial velocity of the dragster at the start of the race is zero
The acceleration is constant in both cases of acceleration.
The final velocity when it stops is zero
Motion is in a straight line path
A +7.5 nC point charge and a -2.0 nC point charge are 3.0 cm apart. What is the electric field strength at the midpoint between the two charges?
Answer:
Ep= 3.8 10⁵ N/C
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
Equivalence
1nC= 10⁻⁹C
1cm= 10⁻²m
Data
k= 9*10⁹ N*m²/C²
q₁ =+7.5 nC = +7.5*10⁻⁹C
q₂ = -2.0 nC = -2.0*10⁻⁹C
d₁ =d₂ = 1.5cm = 1.5 *10⁻²m = 0.015 m
Calculation of the electric fieldsat the midpoint (P) between the two charges
Look at the attached graphic:
E₁: Electric Field at point ;Due to charge q₁. As the charge q₁ is positive negative (q₁+), the field leaves the charge .
E₂: Electric Field at point : Due to charge q₂. As the charge q₂ is negative (q₂-) ,the field enters the charge
E₁ = k*q₁/d₁² = 9*10⁹ *7.5 *10⁻⁹/ ( 0.015 )² = 3*10⁵ N/C
E₂ = k*q₂/d₂²= 9*10⁹ *2*10⁻⁹/( 0.015 )² = 0.8*10⁵ N/C
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Ep= E₁ + E₂
Ep= 3*10⁵ N/C + 0.8*10⁵ N/C
Ep= 3.8 10⁵ N/C
The electric field strength at the midpoint between a +7.5 nC point charge and a -2.0 nC point charge that are 3.0 cm apart is 1.56 *10^6 N/C (Newtons per Coulomb) away from the positive charge.
Explanation:The question is asking for the electric field at the midpoint between two point charges. The formula for the electric field created by a point charge is given by E=kQ/r^2, where E is the electric field, k is Coulomb's constant (approximately 9.0 * 10^9 N*m^2/C^2), Q is the charge, and r is the distance from the charge. In this scenario we have two charges, so we calculate the electric field created by each charge and sum the results.
Applying this to both point charges and summing the electric fields we obtain: E_total = |E1| + |E2| = k*|Q1|/d^2 + k*|Q2|/d^2 = ((9.0*10^9 N*m^2/C^2) * 7.5 *10^-9 C/0.015m^2) + ((9.0*10^9 N*m^2/C^2) * 2 *10^-9 C/ 0.015m^2) = 1.8 *10^6 N/C and -0.24 * 10^6 N/C respectively. The total electric field strength at the midpoint is the sum which equals 1.56 *10^6 N/C away from the positive charge.
Learn more about Electric Field Strength here:
https://brainly.com/question/1812671
#SPJ3
The time in hours it takes a satellite to complete an orbit around the earth varies directly as the radius of the orbit (from the center of the earth) and inversely as the orbital velocity. If a satellite completes an orbit 660 miles above the earth in 13 hours at a velocity of 32,000 mph, how long would it take a satellite to complete an orbit if it is at 1500 miles above the earth at a velocity of 38,000 mph?
To find the new orbital period of a satellite at 1500 miles above Earth with a velocity of 38,000 mph, we use a proportional relationship comparing it to a known orbit. We set up the ratio using the direct variation with the orbit's radius and inverse variation with the orbital velocity, then solve for the unknown orbital period.
Explanation:The duration of a satellite's orbit around Earth can be described as varying directly with the orbit's radius and inversely with the orbital velocity. To calculate the orbital period for a new satellite trajectory, we need to compare it with a known situation using a proportional relationship. Given that a satellite orbits 660 miles (or 1062 kilometers) above Earth in 13 hours at 32,000 mph, we can set up a ratio to determine the orbital period at 1500 miles (or 2414 kilometers) above Earth with a velocity of 38,000 mph.
The ratio for the original orbit is T1 / (r1 / v1) = T2 / (r2 / v2), where T is time, r is radius (distance from Earth's center to the satellite), and v is velocity. Using the provided values, we get:
13 / ((3963 + 660) / 32000) = T2 / ((3963 + 1500) / 38000)
Solving for T2 gives us the new orbital period. Considering that we should convert the altitude to the same units and use Earth's radius in miles (3963 miles), the final calculation will provide us with the new orbital period in hours.
Describe the weather in orlando today and how is it different than describing florida’s climate.
Answer:it is cold
Explanation:because of a winter storm it is very cold which doesn’t
Match Florida’s tropical warm climate
Weather refers to short-term atmospheric conditions while climate refers to long-term predictable conditions. Weather in Orlando today depicts the present conditions, whereas Florida's climate represents the long-term averages and patterns across the state.
Explanation:The weather in Orlando today simply refers to the current atmospheric conditions in Orlando, such as temperature, humidity, precipitation, wind, etc. It's a mere snapshot of what's happening right now in the atmosphere around Orlando. Now, when we're talking about Florida's climate, we're referring to the long-term atmospheric conditions in the state of Florida. This includes consistent seasonal temperature and rainfall patterns over many years. Therefore, the primary difference between describing the weather in Orlando today and Florida's climate lies in the timeframe – weather explains temporary conditions, while climate refers to long-term, predictable conditions.
Learn more about Weather vs Climate here:https://brainly.com/question/33384928
#SPJ12
Most active galactic nuclei are at large distances from us; relatively few nearby galaxies harbor active galactic nuclei. What does this imply?
Answer:
The milky way does not have an active galactic nuclei and active galactic nuclei reduce their activity as time pass.
Explanation:
An active galactic nucleus (AGN) is a region that is compact and at the galaxy's center. It has a extra high luminosity in which some observations indicate that the luminosity is not as a result of the presence of stars.
Sir Lance a Lost new draw bridge was designed poorly and stops at an angle of 20o below the horizontal. Sir Lost and his steed stop when their combined center of mass is 1.0 m from the end of the bridge. The bridge is 8.0 m long and has a mass of 2000 kg; the lift cable is attached to the bridge 5.0 m from the castle end and to a point 12 m above the bridge. Sir Lost’s mass combined with his armor and steed is 1000 kg.
Determine
(a) the tension in the cable and
(b) the horizontal and vertical force components acting on the bridge at the castle end.
Answer:
The Tension T is 42120N
The Horizontal force component is 18322.2N
The Vertical force component is - 4729N
Explanation:
First, you have to find the angle between the drawbridge and the cable using sine and cosine rule. This will result in angle 44.2°. Hence, the angle between the horizontal axis and the cable will be 64.2° (44.2° + 20°).
Having done that, you apply two conditions of equilibrium.
1. THE VECTOR SUM OF ALL FORCES EQUAL ZERO.
∑Fx = 0
∑Fx = Rx - Tcos64.2 = 0
Rx = 0.435T
∑Fy = 0
∑Fy = Ry + Tsin64.2 - W - w = 0
W = 2000kg × 9.8 = 19600N
w =1000kg × 9.8 = 9800N
Ry + 0.9T = 29400N
Ry = 29400 - 0.9T
2. THE SUM TOTAL OF TORQUES EQUALS ZERO
Rx: τ = 0
Ry: τ = 0
T: τ = 5 × Tsin44.2
= 3.49T m
W: τ = 4 × 19600sin90
= 78400Nm
w: τ = 7 × 9800sin9
= 68600Nm
Note:
Rx = x component of Reaction force
Ry = y component of Reaction force.
T = Tension
W = weight of bridge
w = weight of Sir Lance a Lost and his steed
τ = torque
Note: The torque of Tension is counter clockwise while that of the weights is clockwise.
Hence,
∑τccw = ∑τcw
3.49T = 78400 + 68600
3.49T = 14700Nm
T = 147000/3.49
T = 42120N
Rx = 0.435 × 42120
Rx = 18322.2N
Ry = 29400N - (0.9×42120)N
Ry = 29400 - 34129
Ry = -4729N
Note: Ry being negative means that the hinge of the drawbridge exerts a downward force.
In this exercise we have to use the knowledge of tension and force, so we can say that this will result in:
A)The Tension T is 42120N
B)The Horizontal force component is 18322.2N
C)The Vertical force component is - 4729N
Before starting the calculations we have to remember some concepts such as:
Rx = x component of Reaction forceRy = y component of Reaction force.T = TensionW = weight of bridgew = weight of Sir Lance a Lost and his steedτ = torqueFirst we will add all the force vectors so that it results in zero and that means that the body is in equilibrium, so:
[tex]\sum F_x = 0\\ \sum F_x = R_x - Tcos(64.2) = 0\\ R_x = 0.435T\\ \sum F_y = 0\\ \sum F_y = R_y + Tsin(64.2) - W - w = 0\\ W = 2000kg * 9.8 = 19600N\\ w =1000kg * 9.8 = 9800N\\ R_y + 0.9T = 29400N\\ R_y = 29400 - 0.9T[/tex]
Secondly, we will add up all the torques so that it results in zero and that means that the body is in equilibrium, so:
[tex]T: T = 5 * Tsin(44.2) = 3.49T m\\ W: T = 4 * 19600sin(90)= 78400Nm\\ w: T = 7 * 9800sin(90) = 68600Nm\\ \sum Tccw = \sum Tcw\\ 3.49T = 78400 + 68600\\ 3.49T = 14700Nm\\ T = 147000/3.49\\ T = 42120N R_x = 0.435 * 42120\\ R_x = 18322.2N\\ R_y = 29400N - (0.9*42120)N\\ R_y = 29400 - 34129\\ R_y = -4729N[/tex]
See more about torque at brainly.com/question/6855614