Answer:
[tex]\theta=44.03^{o}[/tex]
Explanation:
Here we have an inelastic collision problem. We can use the momentum (p = mv) conservation law in each component of the displacement.
So, [tex]p_{i}=p_{f}[/tex]
X-component:
[tex]m_{1}v_{i1x}+m_{2}v_{i2x}=m_{1}v_{f1x}+m_{2}v_{f2x}[/tex] (1)
Now,
v(i1x) is 0, because the first car just moving in y-directionv(i2x) is 164 km/hv(f1x)=v(f2x), because both cars stick together after the collision, so they have the same x-component velocity.Then, using this information we can rewrite the equation (1).
[tex]m_{2}v_{i2x}=v_{fx}(m_{1}+m_{2})[/tex]
[tex]v_{fx}=\frac{m_{2}v_{i2x}}{m_{1}+m_{2}}=\frac{2597.2*164}{1720+2597.2}[/tex]
[tex]v_{fx}=98.66 km/h[/tex]
Y-component:
[tex]m_{1}v_{i1y}+m_{2}v_{i2y}=m_{1}v_{f1y}+m_{2}v_{f2y}[/tex] (2)
We can do the same but with the next conditions:
v(i1y) is 239.44 km/hv(i2y) is 0, because the second car just moving at the x-directionv(f1y)=v(f2y), because both cars stick together after the collision, so they have the same y-component velocity.Then, using this information we can rewrite the equation (2).
[tex]m_{1}v_{i1y}=v_{fy}(m_{1}+m_{2})[/tex]
[tex]v_{fy}=\frac{m_{1}v_{i1y}}{m_{1}+m_{2}}=\frac{1720*239.44}{1720+2597.2}[/tex]
[tex]v_{fy}=95.39 km/h[/tex]
Now, as we have both components of the final velocity, we can find the angle East of North. Using trigonometric functions, we have:
[tex]tan(\theta)=\frac{v_{y}}{v_{x}}[/tex]
[tex]\theta=arctan(\frac{v_{y}}{v_{x}})=arctan(\frac{95.39}{98.66})[/tex]
[tex]\theta=44.03^{o}[/tex]
I hope it helps you!
What is the average useful power output (in W) of a person who does 6.60 ✕ 106 J of useful work in 8.40 h?
Answer:
218.25 W.
Explanation:
Power: This is defined as the rate at which work is done. The S.I unit is Watt (W). Mathematically, it can be expressed as
P = W/t ............................... Equation 1.
Where P = useful power output, W = useful work, t = time taken to do the work.
Given: W = 6.6×10⁶ J, t = 8.4 h = 8.4×60×60 = 30240 s.
Substitute into equation 1
P = 6.6×10⁶/30240
P = 218.25 W.
Hence the useful power output = 218.25 W.
What do we expect to happen to Earth’s air temperature if the amount of energy from the sun that reaches the Earth’s surface is reduced by half?
Explanation:
The Natural Greenhouse Effect However, the amount that directly escapes to space is only about 12 percent of incoming solar energy. The remaining fraction—a net 5-6 percent of incoming solar energy—is transferred to the atmosphere when greenhouse gas molecules absorb thermal infrared energy radiated by the surface.
An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. At what angle (in degrees) must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s?
Answer:
The answer to your question is angle = 18.46°
Explanation:
Data
d = 75 m
v₀ = 35 m/s
α = ?
Formula
[tex]d = \frac{vo^{2}sin2\alpha}{g}[/tex]
solve for sin2α
sin 2α = [tex]\frac{dg}{vo^{2}}[/tex]
Substitution
sin 2α = [tex]\frac{75(9.81)}{35^{2}}[/tex]
Simplify
sin 2α = [tex]\frac{735.75}{1225}[/tex]
Divide
sin 2α = 0.600
Get sin⁻¹
2α = 36.9°
Divide by 2
α = 18.5°
melvin pulls a sled across level snow with a force of 317 N along a rope that is 33 degrees above the horizont?
Answer: work Melvin did=9000J
Explanation:
Given to complete the question: If the sled moved 33.9m,how much work did Melvin do? Answer in unit of J and round to the nearest thousandth.
W = F ×S
W = 317 × cos 33°×33.9
W=9012.6055J
W=9000J to the nearest thousandth
Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms folded in and legs together. Upon ending, their arms extend outward, proclaiming their finish. Not quite as noticeably, one leg goes out as well.Suppose that the moment of inertia of a skater with arms out and one leg extended is 2.9 kgm2 and for arms and legs in is 0.90 kgm2 . If she starts out spinning at 4.5rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward?
Answer:
Her angular speed (in rev/s) when her arms and one leg open outward is 1.4 rev/s
Explanation:
given information:
moment inertia of arm and leg when in, I₁ = 0.9 kgm²
moment inertia of arm and leg when extended, I₂ = 2.9 kgm²
angular speed when in, ω₁ = 4.5 rev/s
so, her angular speed (in rev/s) when her arms and one leg open outward is
L₁ = L₂
I₁ω₁ = I₂ω₂
ω₂ = I₁ω₁/I₂
= 0.9 x 4.5/2,9
= 1.4 rev/s
If an insulated (power side) wire rubbed through a part of the insulation and the wire conductor touched the steel body of a vehicle, the type of failure would be called a(an)
Answer:
a short-to-ground
Explanation:
Many failures that occur in vehicles can be diagnosed as electrical problems. The reason for these usually relate mainly to the electricity generated in the battery or alternator, in the ignition system as well as those produced by blown wires or fuses.
Cables or fuses: a blown fuse or a damaged electrical wire or one that is making ground or earth effect can be the cause of an electrical problem. Keep in mind that everyone is connected to each other through the wires of the electrical system. Fuses protect car components from power surges. If you have a problem with one, check it to see if you will have to replace it.
A failure where an insulated wire rubs through and touches the steel body of a vehicle is known as a short circuit.
Explanation:If an insulated (power side) wire rubs through a part of the insulation and the wire conductor touches the steel body of a vehicle, this type of failure is known as a short circuit. This happens when a wire carrying current touches the vehicle's body, which is grounded. As the body acts like a path of least resistance, the current prefers flowing through it rather than its intended path, creating a short circuit. This can result in various problems like damage to components, blowing a fuse, or even triggering a fire.
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A cyclist maintains a constant velocity of 4.1 m/s headed away from point A. At some initial time, the cyclist is 244 m from point A. What will be his displacement from his starting position after 60 s?
Answer:
[tex]d=490\ m[/tex] is his final displacement from the point A after 60 seconds.
Explanation:
Given:
Cyclist is moving away from A.
velocity of cyclist, [tex]v=4.1\ m.s^{-1}[/tex]displacement of the cyclist from point A at the time of observation, [tex]d_i=244\ m[/tex]time after which the next observation is to be recorded, [tex]t=60\ s[/tex]Now as the cyclist is moving away from point A his change in displacement after the mentioned time:
[tex]\Delta d=v.t[/tex]
[tex]\Delta d = 4.1 \times 60[/tex]
[tex]\Delta d=246\ m[/tex]
Now the the final displacement from point A after the mentioned time:
[tex]d=d_i+\Delta d[/tex]
[tex]d=244+246[/tex]
[tex]d=490\ m[/tex]
Which conditions will result in the smallest change in momentum? a. a large force over a long time period b. a large force over a short time period c. a small force over a long time period d. a small force over a short time period
Answer:
A small force over a long time period.
Explanation:
The second law of motion gives the relationship between the external force and the change in momentum of the object. The rate of change of linear momentum is equal to the external force applied. It is given by :
[tex]F=\dfrac{\Delta p}{\Delta t}[/tex]
[tex]\Delta p=F\times \Delta t[/tex]
It is clear that for the smallest change in momentum, a small force should be applied for a long period of time. Hence, the correct option is (c) "a small force over a long time period".
Answer:
D. a small force over a short time period
C is incorrect.
A child has been diagnosed with an autoimmune disease that predominantly affects epithelial tissue. Histologic examination should include which body parts as potential sites of involvement? Select all that apply.a) Heartb) Bone marrowc) Bladderd) Epidermise) Blood vessel lining
Answer:C) BLADDER
D) EPIDERMIS (E) BLOOD VESSEL LINING
Explanation: Autoimmune diseases are diseases caused by the body fighting against its self. In autoimmune diseases the body's defense system tends to attack the body cells, tissues, Organs and the entire system mistaking it for a foreign body such as Viruses or Bacteria,Fungi etc.
Epithelial tissues are tissues which form the surface covering of most of the hollow Organs,all body Organs and are present inside the glands. The body parts that will be involved are the BLADDER, EPIDERMIS AND BLOOD VESSEL LINING.
When a .22-caliber rifle is fired, the expanding gas from the burning gunpowder creates a pressure behind the bullet. This pressure causes the force that pushes the bullet through the barrel. The barrel has a length of 0.61 m and an opening whose radius is 2.8 x 10²³ m. A bullet (mass = 2.6 x 10²³ kg) has a speed of 370 m/s after passing through this barrel. Ignore friction and determine the average pressure of the expanding gas.
Answer:
[tex]1.18454\times 10^{-19}\ Pa[/tex]
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
m = Mass of bullet = [tex]2.6\times 10^{23}\ kg[/tex]
r = Radius of barrel = [tex]2.8\times 10^{23}\ m[/tex]
[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{370^2-0^2}{2\times 0.61}\\\Rightarrow a=112213.11475\ m/s^2[/tex]
Pressure is given by
[tex]P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{ma}{\pi r^2}\\\Rightarrow P=\dfrac{2.6\times 10^{23}\times 112213.11475}{\pi (2.8\times 10^{23})^2}\\\Rightarrow P=1.18454\times 10^{-19}\ Pa[/tex]
The pressure of the expanding gas is [tex]1.18454\times 10^{-19}\ Pa[/tex]
A car is driving around a circular track at constant speed. At one instant, the car is driving northward and sometime later the car is driving westward. What is the direction of the car’s average change in velocity during this time interval?
Answer:
Southwest
Explanation:
when the car is moving around a circular track at constant speed, velocity changes due to change in the direction. The direction at any instant is given by tangent drawn at the point on the circular path.
At instant 1: direction is northwards
At instant 2: direction is westwards
Let the constant speed be v. Then, at instant 1, the velocity would be:
[tex]v_1=v \hat j[/tex]
At instant 2, the velocity would be:
[tex]v_2 = v (-\hat i)[/tex]
Change in velocity = [tex]v_2-v_1 = v(-\hat i - \hat j)[/tex]
The direction of car's average change would be southwest.
The car's average change in velocity has a northwest direction.
Explanation:The direction of the car's average change in velocity is determined by the direction of the velocity vectors at the two instances. In this case, when the car is driving northward, its velocity vector points towards the north. Later, when the car is driving westward, its velocity vector points towards the west. Therefore, the car's average change in velocity will have a direction that is a combination of north and west, which is the northwest direction.
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What is it called when a solid turns directly into a gas
Answer:
Sublimation
Explanation:
Sublimation is the process of conversion of solids directly into vapor on the supply of heat. This occurs because the vapor pressure of the sublimate becomes higher than the atmospheric pressure and the melting has not yet happened because of insufficient heat. In other words, the substance cannot exist in its liquid form at the given temperature and the atmospheric pressure.Example: Dry ice, camphor
Which are true?
1) In a uniform electric field, the field lines are straight, parallel, and uniformly spaced.
2) Electric field lines near positive point charges radiate outward.
3) The electric force acting on a point charge is proportional to the magnitude of the point charge.
4) Electric field lines near negative point charges circle clockwise.
5) The electric field created by a point charge is constant throughout space.
Answer:
Following are the correct options.
1. In a uniform electric field, the field lines are straight, parallel and uniformly spaced. ( As the field strength does not change in the uniform electric field that is why the field lines are parallel, equally spaced etc).
2. Electric field lines near positive point charges radiate outward. (Due to repulsive forces field lines radiate in outward direction)
3.The electric force acting on a point charge is proportional to the magnitude of the point charge. ( As columb law states that the electrostatic forces between the point charges is proportional to the product of their magnitude and inversely proportional to square of distance between them).
Explanation:
In a uniform electric field, electric field lines are straight, parallel, and uniformly spaced while near positive point charge it emits outwards. Also, the electric force on a point charge is proportional to its magnitude. But, lines near negative point charges radiate inward, not circle clockwise. The electric field created by a point charge isn't constant across space, but diminishes with distance.
Explanation:Of the five statements you provided:
In a uniform electric field, the field lines are indeed straight, parallel, and uniformly spaced. Electric field lines near positive point charges do radiate outward. The electric force acting on a point charge is proportional to the magnitude of the point charge.However, electric field lines near negative point charges don't circle clockwise, they radiate inward towards the negative charge.Additionally, the electric field created by a point charge isn't constant throughout space; it diminishes with distance from the point charge according to Coulomb's Law.Therefore, statements 1, 2 and 3 are true, while 4 and 5 are false.
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To win a prize at the county fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object. Should you choose to throw a rubber ball or a beanbag of equal size and weight?
Answer:
Being an elastic object, rubber ball will be an ideal choice as it will bounce off the bowling pit and will experience a large change in momentum in comparison with the beanbag which will either slow down or come to a halt upon hitting a bowling pit. That is why rubber ball will experience a greater impulse and the bowling pin will experience the negative impulse of the rubber ball.
For Rubber Ball
Upon elastic collision it will reverses the direction and move with velocity equal or less then original
change in momentum = P
[tex]P = m(v_{f} -v_{i})\\v_{f}=-v_{i} \\ P = -2mv_{i}[/tex]
For Beanbag
value of impulse will large if velocity is zero.
[tex]v_{f}=0\\ P = -mv_{i}[/tex]
Explanation:
The rubber ball will impact a greater force on the heavy bowling pin because its kinetic energy will be conserved while the beanbag will impact lesser force due to loss of kinetic energy.
A collision between two objects can be elastic or inelastic.
In elastic collision both momentum and kinetic energy are conserved.In inelastic collision only momentum is conserved.The impulse experienced by each throwing object is equal to change in the momentum of the object.
[tex]J = \Delta P[/tex]
The rubber ball will impact a greater force on the heavy bowling pin because its kinetic energy will be conserved while the beanbag will impact lesser force due to loss of kinetic energy.
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A 0.500 kg mass is oscillating on a spring with k=330 N/m.The total energy of its oscillation is 3.24 J. What is speed of the mass when it is 0.100 m from the EP?
(Unit=m/s)
The speed is 2.5 m/s
Explanation:
The total (mechanical) energy of the mass-spring system at any point during the motion is the sum of the kinetic energy (KE) and the potential energy (PE):
[tex]E=KE+PE=3.24 J[/tex]
and it is constant.
The kinetic energy can be written as
[tex]KE=\frac{1}{2}mv^2[/tex]
where
m = 0.500 kg is the mass
v is the speed
While the potential energy is
[tex]PE=\frac{1}{2}kx^2[/tex]
where
k = 330 N/m is the spring constant
x is the elongation
So the first equation becomes
[tex]E=\frac{1}{2}mv^2 + \frac{1}{2}kx^2[/tex]
Therefore, if we substitute
x = 0.100 m
We can find the speed when the elongation is x = 0.100 m:
[tex]v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(3.24)-(330)(0.100)^2}{0.500}}=2.5 m/s[/tex]
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Correct Answer:
2.5 m/s
A helicopter, initially hovering 40 feet above the ground, begins to gain altitude at a rate of 21 feet per second. Which of the following functions represents the helicopter’s altitude above the ground y, in feet, t seconds after the helicopter begins to gain altitude?
Answer: y = 40 + 21t
Explanation:
Apply the equation of distance covered.
d = vt + C
Where d is the distance covered
v = velocity , t = time
C = constant = initial distance covered
For the case above....
d = y
y(t) = vt + C
But y(0) =40 = C
C = 40ft
velocity v = 21 ft/s
Therefore, the equation of the altitude is given by;
y(t) = 21t +40
y = 40 + 21t
The function representing the helicopter's altitude as it gains height is y = 21t + 40. The helicopter gains altitude at a rate of 21 feet per second and it starts at 40 feet above the ground.
Explanation:The function that represents this situation is a linear function because the change in the helicopter's altitude is constant over time.
The formula for a linear function is y = mx + b, where m is the slope (rate of change), b is the y-intercept (initial value), y is the dependent variable (altitude in this case), and x is the independent variable (time in this case).
In this scenario, we're given that the helicopter starts 40 feet above the ground (our y-intercept, b) and it's gaining altitude at 21 feet per second (our slope, m).
Therefore, the function to represent the helicopter’s altitude, y, after t seconds can be written as y = 21t + 40.
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Satellite 1 revolves around a planet at the altitude equal to one-half the radius of the planet. The period of revolution of satellite 1 is . What is the period of revolution of an identical satellite 2 that revolves around the same planet at the altitude equal to the radius of the planet?
Answer:
To calculate the period of satellite orbiting around a planet, we use Kepler's third law;
Square of T = [(4π)/(G*m)] * R^3.
Therefore,
T = sqrt{[(4π)/(G*m)]*R^3}.
T is the period, m is mass orbiting satellite, G is gravitational constant, R is the radius of of the planet, r is the radius of the orbiting satellite.
For Satellite 1, r is one-half of the planet, that is r = (3/2) * R
For satellite 2, r = R
Explanation:
A 92-kg fullback moving south with a speed of 5.8 m/s is tackled by a 110-kg lineman running north with a speed of 3.6 m/s. Assuming momentum conservation, determine the speed and direction of the two players immediately after the tackle.Give the direction as an angle, in degrees, south of west.
They travel at a speed of 0.68 m/s south ([tex]90^{\circ}[/tex] south of west)
Explanation:
We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two players must be conserved before and after they collide.
Mathematically:
[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v[/tex]
where, taking north as positive direction:
[tex]m_1 = 92 kg[/tex] is the mass of the first player
[tex]u_1 = -5.8 m/s[/tex] is the initial velocity of the first player (south, so negative)
[tex]m_2 = 110 kg[/tex] is the mass of the second player
[tex]u_2 = 3.6 m/s[/tex] is the initial velocity of the second player (north, so positive)
[tex]v[/tex] is the final combined velocity of the two players
Solving for v, we find the final velocity of the two players combined:
[tex]v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(92)(-5.8)+(110)(3.6)}{92+110}=-0.68 m/s[/tex]
where the negative sign indicates their final direction is south.
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To find the final speed and direction after the tackle, we use the conservation of momentum. The fullback and lineman's initial momenta are calculated and combined to find the total initial momentum, which equals the final momentum because momentum is conserved. The direction is either south (90 degrees) or north (270 degrees) depending upon the sign of the final velocity.
Explanation:To solve this physics problem, we will use the principle of conservation of momentum, which states that in the absence of external forces, the total momentum of a system remains constant. The formula for momentum (p) is the product of mass (m) and velocity (v), or p = mv.
Before the tackle, the fullback's momentum is southward and the lineman's is northward. We treat south as the positive direction. Thus, the initial momentums of the fullback and lineman are:
Fullback momentum: pFB = 92 kg × 5.8 m/sLineman momentum: pL = 110 kg × (-3.6 m/s) (negative due to the northward direction)The total initial momentum (pinitial) is the sum of these:
pinitial = pFB + pL
After the tackle, the players stick together and move as one mass (M = 92 kg + 110 kg), with a final velocity (V), so:
pfinal = MV
Since momentum is conserved, pinitial = pfinal, and we can solve for V:
V = pinitial / M
Once we calculate the final velocity, the direction is south if V > 0, north if V < 0. To convert the direction into an angle south of west, note that if they move southward, the angle is simply 90 degrees (directly south), while any northward movement would result in a 270-degree angle.
Using conversions and data, determine the number of hydrogen atoms required to obtain 8.0 kg of hydrogen. A hydrogen atom has a mass of 1.0 u.
Answer:
The answer to your question is 4.82 x 10²⁷ atoms of Hydrogen
Explanation:
Data
final mass = 8.0 kg
mass of hydrogen = 1.0 u
number of hydrogens = ?
Process
1.- Use proportions to solve this problem
1 u ------------------------- 1.66 x 10⁻²⁷ kg
x u ------------------------- 8 kg
x = (8 x 1) / 1.66 x 10⁻²⁷
x = 8 / 1.66 x 10⁻²⁷
x = 4.82 x 10 ²⁷ u or 4.82 x 10²⁷ atoms of Hydrogen
__________ brake systems release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and traction. Control systems apply the brakes momentarily to one of the drive wheels whenever the wheel speed sensors indicate a wheel is going faster than the others during acceleration.
"Anti-Lock" brake systems release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and traction.
Explanation:
The safety anti-skid braking system is known as "anti-lock braking system" having huge application on land vehicles like one, two and multiple wheeler vehicles and aircraft. During braking, it avoids wheels to get locked by building tractive contacts to the road's surface.
This seems to be an automated system work on the principles of techniques - threshold and cadence braking. The wheel velocity sensors are utilized by ABS to find whether one or more than one wheels chose to get lock while braking.
Answer:
Anti- Lock
Explanation:
Anti- lock brake systems(ABS) is a safety system used in hindering the wheels from locking up when the brake is applied.
Braking release the brakes momentarily when wheel speed sensors indicate a locked wheel during braking and traction.
Control systems apply the brakes momentarily to one of the drive wheels whenever the wheel speed sensors indicate a wheel is going faster than the others during acceleration.
How star properties affect star formation? Provided following are the spectral types of four different main-sequence stars. Rank the stars based on the strength of the radiation pressure that pushes outward as they are forming, from highest pressure to lowest pressure.A) O9.B) A5.C) G2.D) M6.
Answer:
-M6
-G2
-A5
-O9
Explanation:
What type of nuclear radiation is emitted when carbon-14 decays
Answer:
Beta radiation
Explanation:
Beta radiation is a radioactive phenomenon of nuclear decay in which an unstable atom or isotop, by transforming a neutron into a proton, or by transforming a proton into a neutron, becomes stable. For example, the decay of carbon 14 produces beta radiation.
A car traveling at 7 m/s accelerates uniformly at 2.5 m/s to reach a speed of 12 m/s. How long does it take for this acceleration to occur?
There is one mistake in the question as unit of acceleration is not written correctly.The correct question is here
A car traveling at 7 m/s accelerates uniformly at 2.5 m/s² to reach a speed of 12 m/s. How long does it take for this acceleration to occur?
Answer:
time taken =2 seconds
Explanation:
Given Data
Initial Speed Vi= 7 m/s
Final Speed Vf=12 m/s
Acceleration a= 2.5 m/s²
To find
Time taken for this acceleration
Solution
As we know that
Final velocity=Initial velocity + acceleration×time
[tex]V_{f}=V_{i}+at\\ t=\frac{V_{f}-V_{t}}{a}\\ t=\frac{12m/s-7m/s}{2.5m/s^{2} }\\ t=2 seconds[/tex]
So car takes 2 seconds for this acceleration to occur
A proton is released from rest inside a region of constant, uniform electric field E 1 pointing due north. 27.3 s after it is released, the electric field instantaneously changes to a constant, uniform electric field E 2 pointing due south. 3.03 s after the field changes, the proton has returned to its starting point. What is the ratio of the magnitude of E 2 to the magnitude of E 1 ? You may neglect the effects of gravity on the proton.
Answer:
E2/E1 =99.2
Explanation:
time after release of E1 (t) = 27.3 s
time after release of E2 (t') = 3.03 s
acceleration (a) = [tex]\frac{QE1}{M}[/tex]
where
Q is the protons charge M is the massafter 27.3 s
velocity (V) = a x t = [tex]\frac{QE1}{M}[/tex] x 27.3 = [tex]\frac{27.3QE1}{M}[/tex]
distance to turning point (s) = 0.5a[tex]t^{2}[/tex] = 0.5 x [tex]\frac{QE1}{M}[/tex]x [tex]27.3^{2}[/tex] = [tex]\frac{372.65QE1}{M}[/tex]
now for its return back to its starting point
acceleration (a') = [tex]-\frac{QE2}{M}[/tex]
total distance S' = distance to turning point + distance from turning point to starting point
S' = S + vt' + 0.5 a'[tex]t'^{2}[/tex]
S' = [tex]\frac{372.65QE1}{M}[/tex] + ([tex]\frac{27.3QE1}{M}[/tex] x 3.03) + (0.5 x [tex]-\frac{QE2}{M}[/tex]x [tex]3.03^{2}[/tex])
S' is the distance at the starting point and = 0
0 = [tex]\frac{372.65QE1}{M} + \frac{82.72QE1}{M}-\frac{4.59QE2}{M}[/tex]
[tex]\frac{4.59QE2}{M}=\frac{372.65QE1}{M} + \frac{82.72QE1}{M}[/tex]
multiplying both side by M/Q we have
4.59.E2 = 372.65E1 + 82.72E1
4.59.E2 = 455.37E1
E2/E1 = 455.37 / 4.59
E2/E1 =99.2
The ratio of the magnitude of electric field E2 to E1, when a proton returns to its initial point after the direction of a uniform electric field changes, is approximately 0.0123.
Explanation:A proton is initially at rest in an electric field that points due north. After 27.3 seconds, the electric field changes direction and points due south, and after 3.03 seconds the proton returns to its starting point. To find the ratio of the magnitudes of E2 to E1, we must consider the distances covered by the proton under the influence of both fields given that it starts and ends at the same position.
The distance covered under E1 can be calculated using the formula s = 0.5 × a × t^2, where 'a' is the acceleration and 't' is the time. Since the proton moves for 27.3 s under E1, the distance s1 is s1 = 0.5 × a1 × (27.3)^2. The proton then moves in the opposite direction under E2 for 3.03 s, covering the same distance in opposing direction, thus s2 = 0.5 × a2 × (3.03)^2. Since s1 = s2, we can equate them: 0.5 × a1 × (27.3)^2 = 0.5 × a2 × (3.03)^2.
Furthermore, the acceleration of the proton is directly proportional to the electric field (since a = F/m and F = qE), giving us a1 ∝ E1 and a2 ∝ E2. By simplifying the above equation, we find that a1/a2 = (3.03/27.3)^2, and therefore the ratio of the magnitudes of electric fields is also E2/E1 = (3.03/27.3)^2. Calculating this, we get the ratio E2/E1 approximately equal to 0.0123.
What do you call the gravitational attraction between you and the earth?
A.weight
B. mass
Answer:
A. Weight
Explanation:
Mass And Weight
Mass is a fundamental property of objects and is defined as a measure of the amount of matter in the object. The symbol for mass is m and its SI unit is the kilogram (kg). At normal speeds, i.e. much smaller than the speed of light, the mass is considered as a constant property.
The weight W is defined as the force of gravity on the object, that is, the gravitational attraction between the object and the Earth. Its SI unit is the Newton (Nw or N).
_____ can be thrown, caught, and used to provide resistance for a variety of movements, in a variety of planes of motion, and at a variety of velocities.
Answer:
b. Medicine balls
Explanation:
A medicine ball (also called an exercise ball, or a health ball) is a weighted ball about the shoulder diameter (approx. 13.7 inches), often used for recovery and strength training.... In 1705 similar large balls had been used in Persia.So you can use your medicine ball as a projectile to boost the strength of your throws.
Carefully consider how the accelerations a1 and a2 are related. Solve for the magnitude of the acceleration, a1, of the block of mass m1, in meters per square second.
Answer:
a1 = 3.68m/s²
Explanation:
Given values:
Mass of the block placed on the table, m1 = 12.25 kg
Mass of the block hanging vertically, m2 = 7.5 kg
Acceleration due to gravity, g = 9.8 m/s2
Tension in the string is T
Let the acceleration of mass 1 and mass 2 be a1 and a2
a1 and a2 are equal in magnitude but different in direction. This because the string does not stretch. Hence the two bodies must move equal distances in equal times, and so their speechless at any instant must be equal. When the speeds change , they change by equal amounts in a given time, so the acceleration of the two bodies must have the same magnitude a,
a = m2*g/(m1 + m2)
a = 7.5 x 9.8 / (12.5 + 7.5)
a = 3.68 m/s²
a1 = a2 = 3.68m/s²
a1 is directed to the right and a2 is directed downwards
Below is a diamonds to show the geometrical arrangements of both masses
A rock is thrown upward with a velocity of 26 meters per second from the top of a 43 meter high cliff, and it misses the cliff on the way back down. When will the rock be 6 meters
To solve the problem, we use the kinematic equation for distance considering the initial velocity, acceleration due to gravity, and the distance from the top of the cliff to 6m above the ground. The equation gives two possible times: one when the rock is on the way up, and one when it's on the way down.
Explanation:In this physics question, we are dealing with kinematic equations which describe the motion of the rock. The equation we will use for this question is d = vit + 0.5gt^2, where 'd' represents total distance, 'vi' is initial velocity, 'g' is acceleration due to gravity and 't' is time.
Here, d = 43m - 6m = 37m (distance from the top of the cliff to 6m above the ground), vi = 26m/s (initial velocity of the rock when thrown upwards), and g = 9.8m/s^2 (acceleration due to gravity, but since the rock is thrown downwards, we take it as positive).
Now plug these values into the equation, you have two possible times- one for the rock on its way up and one for on its way down.
Learn more about Kinematic Equations here:https://brainly.com/question/24458315
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Compared with cool air, warm air is:________
a. more dense, and tends
b. to rise less dense, and tends
c. to rise more dense, and tends
d. to sink less dense, and tends to sink
Answer:
option (b)
Explanation:
The density of cool air is more so always falls downwards.
Compared to the cool air, the density of air is less and it tends to rise.
Thus, option (b) is correct.
An elevator packed with people has a mass of 1800 kg.
A) The elevator accelerates upward (in the positive direction) from rest at a rate of 1.5 m/s2 for 2.2 s. Calculate the tension in the cable supporting the elevator in newtons.
B) The elevator continues upward at constant velocity for 9 s. What is the tension in the cable during this time in newtons?
C) The elevator experiences a negative acceleration at a rate of 0.65 m/s2 for 2 s. What is the tension in the cable, in Newtons, during this period of negative accleration?
D) How far has the elevator moved above its original starting point in meters?
Answer:
a) 20.34 N
b)17.64 N
c)16.47 N
d)38.63m
Explanation:
a) F=m*a. Earth gravity is 9.8m/s2. If the elevator goes upwards, tension on the rope will be higher. To find the tension we need to all accelerates. Total accelerates effects the elevator is a=9.8+1.5=11.3 m/s2. Then;
[tex]F=1.8*11.3=20.34[/tex]
the tension is 20.34N
b) There is no accelerate in this situation therefore the tension is:
F=1.8*9.8=17.64 N
c) In this situation elevator goes down. We need to subtract gravity from elevator acceleration. a=9.8-0.65=9.15m/s2
F=1.8*9.15=16.47 N
d) Total distance is:
[tex]x=0.5*a*t^2+v*t+v*t-0.5*a*t^2\\=0.5*1.5*2.2^2+9*(1.5*2.2)+2*(1.5*2.2)-0.5*0.65*2^2\\=38.63 m[/tex]