Answer:
c) 12 eV
Explanation:
The electron crosses a potential difference of 12 V in total, This means that its initial electric potential energy as it leaves the negative plate is equal to
[tex]E=q\Delta V=e (12 V) = 12 eV[/tex]
due to the law of conservation of energy, when the electron moves towards the positive plate this electric energy is all converted into kinetic energy of the electron (in fact, the speed of the electron increases).
When the electron reaches the positive plate, all the electric potential energy has been converted into kinetic energy, which is therefore exactly 12 eV.
The kinetic energy of the electron when it hits the positively-charged plate of a capacitor, having been accelerated across a 12 V potential difference, will be 12 eV. This is determined by the formula KE = eV, where e is the electron charge and V is the potential difference.
The question asks about the kinetic energy of an electron when it hits the positively-charged plate of a capacitor after being accelerated from the negatively-charged plate by a 12 V potential difference. The concept here revolves around the relationship between electric potential difference and kinetic energy. When an electron is accelerated through a potential difference, it gains kinetic energy equal to the charge of the electron multiplied by the potential difference, with the unit of energy being electron volts (eV).
Since an electron has a charge of approximately 1e (or one elementary charge), the kinetic energy gained by an electron when accelerated by a potential difference of 1 V is 1 eV. Therefore, if an electron is accelerated by a 12 V potential difference, it will gain a kinetic energy of 12 eV. This is a straightforward application of the formula KE = eV, where e represents the charge of the electron and V the potential difference.
So, the correct answer to the student's question is (c) 12 eV. The electron, upon hitting the positively-charged plate, will have 12 eV of kinetic energy due to the 12 V potential difference it was accelerated across.
A leaky 10-kg bucket is lifted from the ground to a height of 16 m at a constant speed with a rope that weighs 0.7 kg/m. initially the bucket contains 48 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 16-m level. find the work done. (use 9.8 m/s2 for g.) show how to approximate the required work by a riemann sum. (let x be the height in meters above the ground. enter xi* as xi.)
[tex]\displaystyle W =\lim_{n\to\infty}{\sum_{i=0}^{n}{(678.16 - 36.26\;x_i)\cdot\dfrac{16}{n}}}[/tex].
ExplanationThe mass comes in three parts:
the mass of the rope,the mass of the water in the leaky bucket, andthe mass of the bucket.Both the mass of the rope [tex]m_\text{rope}[/tex] and the mass of the water in the bucket [tex]m_\text{water}[/tex] varies with the height [tex]x[/tex] of the bucket. Express the two masses as a function of [tex]x[/tex]:
[tex]m_\text{rope} = 0.7\;(16 - x)[/tex],The water in the bucket behaves like yet another rope of density [tex]48 \;\text{kg}/ 16 \;\text{m}= 3.0\;\text{kg}\cdot\text{m}^{-1}[/tex]. The mass of the water left in the bucket at height [tex]x[/tex] will be
[tex]m_\text{water} = 3.0\;(16 - x)[/tex].The mass of the bucket is [tex]10\;\text{kg}[/tex]. Combining the three:
[tex]m(x) = m_\text{rope}(x) + m_\text{water}(x) + m_\text{bucket} \\\phantom{m(x)} = 0.7\;(16-x) + 3.0\;(16 - x) + 10\\\phantom{m(x)} = 69.2 - 3.7\;x[/tex].Weight of the bucket at height [tex]x[/tex]:
[tex]F_\text{weight}(x) = m(x)\cdot g = 9.8\;(69.2 - 3.7\;x) = 678.16 - 36.26\;x[/tex].The bucket moves upward at a constant speed. As a result,
[tex]F(x) = W(x) = 678.16 - 36.26\;x[/tex].Express the work required as a definite integral:
[tex]\displaystyle W = \int_{0}^{16}{F(x) \cdot dx} = \int_{0}^{16}{(678.16 - 36.26\;x)\cdot dx}[/tex].Rewrite the definite integral as a Riemann Sum:
[tex]\displaystyle W = \int_{0}^{16}{(678.16 - 36.26\;x)\cdot dx} \\\phantom{W}= \lim_{n\to\infty}{\sum_{i=0}^{n}{(678.16 - 36.26\;x_i)\cdot\frac{16-0}{n}}}\\\phantom{W} = \lim_{n\to\infty}{\sum_{i=0}^{n}{(678.16 - 36.26\;x_i)\cdot\frac{16}{n}}}[/tex].The work done lifting a bucket of water draining at a constant rate can be calculated using the physics Work formula and approximating the varying force with a Riemann sum. The total work is the sum of the work done across several small intervals, calculated by multiplying the force required to lift the remaining water plus the bucket and the rope by the small altitude.
Explanation:The work done in lifting an object can be calculated using the formula Work = Force x Distance = (mass x gravity) x height. However, this situation is a bit more complex since the mass of both the bucket and the water it contains decrease continuously as the bucket is lifted.
We will approximate this varying force using a Riemann sum. We can imagine breaking up the 16 m into n small intervals. For each interval i from 0 to n-1, we calculate the force required to lift the remaining water plus the bucket and rope to a height xi*, and then multiply this by the small change in height Δx. This gives us the small amount of work done ΔW_i = (m(xi*)g)Δx, where m(xi*) is the mass of the system at height xi*.
The total work then is approximately the sum of all these small amounts of work, W ≈ Σ ΔW_i for i from 0 to n-1. The more intervals we use (the bigger n is), the better this approximation will be.
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A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its equilibrium position, what is its instantaneous velocity? 1.Instantaneous velocity cannot be determined without additional information2.At equilibrium, its instantaneous velocity is less than its maximum but not zero.3.At equilibrium, its instantaneous velocity is at maximum4.At equilibrium, its instantaneous velocity is zero.
Answer:
3.At equilibrium, its instantaneous velocity is at maximum
Explanation:
The motion of a mass on the end of a spring is a simple harmonic motion. In a simple harmonic motion, the total mechanical energy of the system is constant, and it is sum of the elastic potential energy (U) and the kinetic energy of the mass (K):
[tex]E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 = const.[/tex]
where
k is the spring constant
x is the displacement of the spring from equilibrium
m is the mass
v is the speed
As we see from the formula, since the total energy E is constant, when the displacement (x) increases, the speed (v) increases, and viceversa. Therefore, when the mass is at its equilibrium position (which corresponds to x=0), the velocity of the mass will be maximum.
At the equilibrium position in simple harmonic motion, the instantaneous velocity of the mass is at its maximum. This is because all the potential energy of the spring is converted into kinetic energy of the mass at this point.
Explanation:In a simple harmonic motion such as a mass on the end of a spring, the equilibrium position is the point where the spring is neither stretched nor compressed. This equilibrium is often denoted as x = 0. When the mass passes through this point, its velocity is at maximum, because at this instance, the entire potential energy of the spring is converted into kinetic energy of the mass. So, the correct statement among the provided options is, 'At equilibrium, its instantaneous velocity is at maximum'. The instantaneous velocity varies according to the displacement of the mass from the equilibrium during its oscillation.
In other words, when the displacement is at its extremes (maximum or minimum), the velocity becomes zero as the mass momentarily pauses before changing its direction of motion. On the other hand, when the displacement is zero at the equilibrium, the velocity is at its maximum.
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Find the TRUE statement. Use the underlined word to determine whether the statement is true or false. Question 3 options: Electromagnetic waves are mechanical because they require a medium to travel. Electromagnetic waves are transverse because they have electric and magnetic fields that move perpendicular to the direction of energy transfer. The speed of an electromagnetic wave depends on the frequency of the wave. Gamma rays are faster than radio waves. As the frequency of an EM wave increases, so does the wavelength. They are directly related.
(sorry if I'm wrong) :(
Final answer:
The accurate statement is that electromagnetic waves are transverse because their electric and magnetic fields move perpendicular to the direction of energy transfer. Other claims are false since electromagnetic waves do not need a medium to travel, and gamma rays are not faster than radio waves; all electromagnetic waves move at light speed, and frequency and wavelength are inversely related.
Explanation:
The true statement among the options given is that electromagnetic waves are transverse because they have electric and magnetic fields that move perpendicular to the direction of energy transfer.
Electromagnetic waves are not mechanical because they do not require a medium to travel through; they can move through the vacuum of space. The speed of an electromagnetic wave is constant at the speed of light (approximately 3.0 × 108 m/s), regardless of its frequency. Moreover, gamma rays are not faster than radio waves because all electromagnetic waves travel at the same speed. And as the frequency of an electromagnetic wave increases, the wavelength decreases; they are inversely related, not directly related.
In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was 55 m/s (terminal speed), that his mass (including gear) was 68 kg, and that the magnitude of the force on him from the snow was at the survivable limit of 1.1 × 105 N. What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?
a) 0.94 m
The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:
[tex]W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2[/tex]
where:
[tex]F=1.1 \cdot 10^5 N[/tex] is the force applied by the snow
d is the displacement of the man in the snow, so it is the depth of the snow that stopped him
m = 68 kg is the man's mass
v = 0 is the final speed of the man
u = 55 m/s is the initial speed of the man (when it touches the ground)
and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)
Solving the equation for d, we find:
[tex]d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m[/tex]
b) -3740 kg m/s
The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:
[tex]I=\Delta p = m \Delta v[/tex]
where
m = 68 kg is the mass of the man
[tex]\Delta v = 0-55 m/s = -55 m/s[/tex] is the change in velocity of the man
Substituting,
[tex]I=(68 kg)(-55 m/s)=-3740 kg m/s[/tex]
(a) The minimum depth of the snow is 0.935 m.
(b) The magnitude of the impulse on him from the snow is -3,740 Ns.
How to calculate the minimum depth of the snow?(a) The minimum depth of the snow is calculated by applying the following formula as follows;
Apply the principle of conservation of energy;
Fd = ¹/₂mu²
d = mu²/2F
where;
m is the massF is the applied forceu is the initial velocityd = (68 x 55² ) / ( 2 x 1.1 x 10⁵)
d = 0.935 m
(b) The magnitude of the impulse on him from the snow is calculated as follows;
J = mΔv
where;
Δv is change in speedJ = m(v - u)
J = 68 (0 - 55)
J = -3,740 Ns
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What is the strongest force on Earth?
P.S. It is not LOVE
lol if not LOVE then its , strongest fundamental force is the strong nuclear force; it is 100 times stronger than the electromagnetic force.
hope this helps:)sorry if it doesnt
Use the flow rate equation to help explain how the motion of a whitewater raft changes as the river channel becomes narrower.
as the flow becomes narrower, more pressure is exerted because the pressure is greater
Final answer:
As the river channel narrows, the water velocity increases due to the same volume of water needing to pass through a smaller area, speeding up the motion of a whitewater raft. This effect is similar to narrowing the opening of a hose to increase water speed.
Explanation:
The flow rate equation helps explain how the motion of a whitewater raft changes as the river channel becomes narrower by considering the flow velocity, which is influenced by the hydraulic radius, channel slope, and channel roughness. In simpler terms, the flow rate (Q) is the volume of water that moves past a point in a given time and is calculated by Q = wdv, where w is the width of the channel, d is the depth of the channel, and v is the velocity of the water.
When the river channel narrows, the same volume of water (discharge) must pass through a smaller cross-sectional area. Consequently, the velocity of the water increases, which can be thought of like putting a thumb over the end of a hose to speed up the water. This increase in velocity will cause the whitewater raft to move faster downstream when the channel is narrower.
Moreover, factors such as rain or snow melt can lead to an increase in discharge, which would further contribute to an increase in the velocity of the water in the narrowed parts of the river. On the contrary, as the channel widens downstream, the current generally slows, which may result in increased sedimentation and a slower moving whitewater raft.
The reason an astronaut in an earth satellite feels weightless is that
Answer:
Gravity is the reason why you feel this way
Explanation:
The velocity of a projectile at launch has a horizontal component vh and a vertical component vv. When the projectile is at the highest point of its trajectory, identify the vertical and the horizontal components of its velocity and the vertical component of its acceleration. Consider air resistance to be negligible
Answer:
- horizontal component of the velocity: [tex]v_h[/tex] (because it is constant)
- vertical component of the velocity: 0
- vertical component of the acceleration: [tex]-g=-9.8 m/s^2[/tex] (downward)
Explanation:
The motion of a projectile consists of two independent motions:
- Along the horizontal direction, there are no forces acting on the projectile (if we neglect air resistance), therefore the horizontal acceleration is zero and the horizontal component of the velocity, vh, is constant
- Along the vertical direction, there is only one force acting on the projectile: the force of gravity, downward, which produces a constant downward acceleration of [tex]g=9.8 m/s^2[/tex]. As a consequence, the vertical component of the velocity changes according to
[tex]v_v(t) = v_v-gt[/tex]
where vv is the initial vertical velocity and t the time. According to this equation, the vertical component of the velocity decreases first, then becomes zero at the point of maximum height, then becomes negative (= changes direction and points downward)
So, in summary, at the highest point of the trajectory we have:
- horizontal component of the velocity: [tex]v_h[/tex] (because it is constant)
- vertical component of the velocity: 0
- vertical component of the acceleration: [tex]-g=-9.8 m/s^2[/tex] (downward)
At the highest point, the final vertical velocity is zero, and the final horizontal velocity is equal to the initial horizontal velocity.
The acceleration due to gravity acting on the object is always directed downwards.
In a projectile motion, the velocity of projected upwards decreases as the object moves upwards and eventually become zero at the maximum height.
The velocity of an object projected upwards is given as;
[tex]v_y = v_0 -gt[/tex]
At the highest point, the final vertical velocity is given as;
[tex]v_y_f= 0[/tex]
The horizontal velocity is always constant because it is not affected by gravity.
At the highest point, the final horizontal velocity is given as;
[tex]v_x_f = v_x_0[/tex]
The acceleration due to gravity acting on the object is always directed downwards.
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A coil lies flat on a level tabletop in a region where the magnetic field vector points straight up. The magnetic field suddenly grows stronger. When viewed from above, what is the direction of the induced current in this coil as the field increases?a. Counterclockwiseb. Clockwise initially, then counterclockwise before stoppingc. Clockwised. There is no induced current in this coil
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