Answer:[tex]A_x[/tex] = Acos[tex]\theta[/tex]
Explanation:
A vector in this situation have two components
1) The X-component
2) The Y-component
So as we put [tex]cos\theta[/tex] with the x-axis while [tex]sin\theta[/tex] with the y- axis and this case our answer should be
[tex]A_x[/tex] = Acos[tex]\theta[/tex]
I hope this will answer your question,
An image is also provided please have a look at that.
Thank you.
Answer: Ax = Acos(theta)
Therefore, the x component of A: Ax = Acos(theta)
Explanation:
The attached image is a pictorial representation of the question:
From the attached image,
Cos(theta) = adjacent/hypothenus
Cos(theta) = Ax/A
Making Ax the subject of formula,
Ax = Acos(theta)
Therefore, the x component of A: Ax = Acos(theta)
The displacement z of a particle of rest mass m0, resulting from a constant force m0g along the z-axis is including relativistic effect. Find the displacement z as a power series in time t. Compare with the classical result,
Final answer:
The displacement z of a particle with rest mass m0 subject to a constant force m0g along the z-axis, accounting for relativistic effects, involves integrating the work-energy theorem's relativistic expression for kinetic energy and comparing it with the classical displacement z = 1/2 gt².
Explanation:
The problem given involves finding the displacement z of a particle with rest mass m0 under the influence of a constant force along the z-axis, considering relativistic effects. The comparison with the classical result implies a consideration of the relativistic factor y in kinetic energy expressions.
We recall that the relativistic kinetic energy is more complex than the classical expression K = 1/2 mu², because as the velocity u of the particle approaches the speed of light c, the kinetic energy diverges due to the increase in y.
The task involves using the work-energy theorem in a relativistic context, where the work done by the force m0g produces a change in the relativistic kinetic energy. As such, relativistic mechanics dictates that energy and momentum are dependent on the relativistic factor y, which is a function of velocity.
Therefore, to expand the displacement z as a power series in time t, one would start by expressing the relativistic momentum and kinetic energy, then integrate to find the time-dependent expressions for velocity and displacement.
Comparing with the classical result, where z is simply obtained by z = 1/2 gt² (ignoring air resistance), shows how relativistic effects modify the simple parabolic trajectory of a particle in a gravitational field. The relativistic displacement would involve a series with higher-order terms in time, accounting for the varying mass with velocity.
You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 NN . If you then lower the statue into a tub of water, so that it is completely submerged, the scale reads 17.0 NN .
Answer:
2491.23 kg/m³
Explanation:
From Archimedes principle,
R.d = weight of object in air/ upthrust in water = density of the object/density of water
⇒ W/U = D/D' ....................... Equation 1
Where W = weight of the ceramic statue, U = upthrust of the ceramic statue in water, D = density of the ceramic statue, D' = density of water.
Making D the subject of the equation,
D = D'(W/U).................... Equation 2
Given: W = 28.4 N, U = lost in weight = weight in air- weight in water
U = 28.4 - 17.0 = 11.4 N,
Constant: D' = 1000 kg/m³.
Substitute into equation 2,
D = 100(28.4/11.4)
D = 2491.23 kg/m³
Hence the density of the ceramic statue = 2491.23 kg/m³
Which of the following quantities are vectors? Select all that apply. Which of the following quantities are vectors? Select all that apply. 27 J/s 1980 kg m/s due south 35 m/s 8.2 m/s2 north-west 3.2 mi straight up 2.9 m/s2 293 K 200 s
Vector quantities have magnitude and direction. The quantities that are vectors from the given list are:
1980 kg m/s due south,
8.2 m/s^2 north-west,
3.2 mi straight up.
Explanation:Quantities that have a magnitude and direction are called vectors. Based on this definition, the following quantities from the list are vectors:
1980 kg m/s due south: This quantity has both magnitude (1980 kg m/s) and direction (south).8.2 m/s2 north-west: This quantity represents acceleration with both magnitude (8.2 m/s2) and direction (north-west).3.2 mi straight up: This quantity has a magnitude (3.2 mi) and a specific direction (upward).Therefore, these three quantities are vectors.
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What causes the mechanical advantage of the inclined plane to be less than that of the liver
Answer:
force of friction
Explanation:
the force of friction in an inclined place is less than that in lever, also a lever works on the principle that it reduces the amount of force required to to work while an inclined plane works by decreasing the amount of force required to do the work.
Two satellites A and B of the same mass are orbiting Earth in concentric orbits. The distance of satellite B from Earth’s center is twice that of satellite A. What is the ratio of the tangential speed of B to that of A?
Answer:
ratio of tangential velocity of satellite b and a will be 0.707
Explanation:
We have given distance of satellite B from satellite A is twice
So [tex]r_b=2r_a[/tex]
Tangential speed of the satellite is given by
[tex]v=\sqrt{\frac{GM}{r}}[/tex], G is gravitational constant. M is mass of satellite and r is distance from the earth
We have to find the ratio of tangential velocities of b and a
From the relation we can see that tangential velocity is inversely proportional to square root of distance from earth
So [tex]\frac{v_b}{v_a}=\sqrt{\frac{r_a}{r_b}}[/tex]
[tex]\frac{v_b}{v_a}=\sqrt{\frac{r_a}{2r_a}}[/tex]
[tex]\frac{v_b}{v_a}=\sqrt{\frac{1}{2}}[/tex]
[tex]\frac{v_b}{v_a}=0.707[/tex]
So ratio of tangential velocity of satellite b and a will be 0.707
The ratio of tangential velocity of satellite B to satellite A is 0.707.
Tangential Speed of Satellite
The Tangential velocity is the linear speed of any object moving along a circular path. The tangential speed of the satellite is given below.
[tex]v = \sqrt{\dfrac{Gm}{r}}[/tex]
Where v is the velocity, m is the mass and r is the circular distance. G is the gravitational constant.
Given that mass of both the satellite is the same. Let us consider the mass of both satellites as m. The distance of satellite B from Earth’s center is twice that of satellite A.
Let us consider that the distance of satellite A from the center of the earth is r. The distance of satellite B from the center of the earth is 2r.
The tangential speed of satellite A is,
[tex]v_a = \sqrt{\dfrac {Gm}{r}}[/tex]
The tangential speed of satellite B is,
[tex]v_b = \sqrt{\dfrac {Gm}{2r}}[/tex]
In the ratio form, the tangential speed of both satellites is given below.
[tex]\dfrac {v_b}{v_a} = \dfrac {\sqrt{\dfrac {GM}{2r}} }{\sqrt{\dfrac {Gm}{r}} }[/tex]
[tex]\dfrac {v_b}{v_a} = \sqrt{\dfrac{1}{2}}[/tex]
[tex]\dfrac {v_b}{v_a} = 0.707[/tex]
Hence we can conclude that the ratio of tangential velocity of satellite B to satellite A is 0.707.
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A force of 35 N acts on a ball for 0.2 s. If the ball is initially at rest:
a. What is the impulse on the ball?
b. What is the final momentum of the ball?
To solve this problem we will apply the concepts related to momentum and momentum on a body. Both are equivalent values but can be found through different expressions. The impulse is the product of the Force for time while the momentum is the product between the mass and the velocity. The result of these operations yields equivalent units.
PART A ) The Impulse can be calculcated as follows
[tex]L= F\Delta t[/tex]
Where,
F = Force
[tex]\Delta t =[/tex]Change in time
Replacing,
[tex]L = (35N)(0.2s)[/tex]
[tex]L= 7N\cdot s[/tex]
PART B) At the same time the momentum follows the conservation of momentum where:
Initial momentum= Final momentum
And the change in momentum is equal to the Impulse, then
[tex]\Delta p = L[/tex]
And
[tex]\Delta p = p_f - p_i[/tex]
There is not initial momentum then
[tex]\Delta p = p_f[/tex]
[tex]L = p_f[/tex]
[tex]p_f = 7N\cdot s = 7kg\cdot m/s[/tex]
Final answer:
The impulse on the ball, which is the product of the force and the time interval, is 7 N·s. Since the ball is initially at rest, the final momentum of the ball is also 7 N·s.
Explanation:
To solve the student's question, we need to apply the concepts of impulse and momentum in physics.
Impulse and Momentum
Impulse is defined as the product of the force acting on an object and the time interval over which it acts. From Newton's second law, the impulse on an object is equal to the change in its momentum.
a) The impulse on the ball is calculated using the formula Impulse = Force × Time. Given that the force is 35 N and the time is 0.2 s, the impulse can be calculated as:
Impulse = 35 N × 0.2 s = 7 N·s
b) The final momentum of the ball is equal to the impulse, since the ball was initially at rest (momentum = 0). Therefore, the final momentum of the ball is:
Final momentum = Impulse = 7 N·s
A horse canters away from its trainer in a straight line, moving 126 m away in 13.0 s . It then turns abruptly and gallops halfway back in 5.0 s . Calculate the (a) its average speed and (b) itsaverage velocity for the entire trip, using "away from the trainer"as the positive direction.
Final answer:
The average speed of the horse for the entire trip is 10.5 m/s, and the average velocity is 3.5 m/s, calculated based on the total distance covered and total displacement over the entire duration of the trip.
Explanation:
To calculate the average speed and average velocity of the horse for the entire trip, we consider both phases of the journey: when the horse canters away and when it gallops back.
Part (a) - Average Speed
Average speed is total distance covered divided by total time taken. The horse travels 126 m away and then comes back halfway, which is 63 m, resulting in a total distance of 126 m + 63 m = 189 m. The total time taken for the trip is 13.0 s + 5.0 s = 18.0 s. Therefore, the average speed is:
average speed = total distance / total time = 189 m / 18.0 s = 10.5 m/s
Part (b) - Average Velocity
Average velocity is the total displacement divided by the total time taken. Since the horse ends up halfway back, the displacement is 126 m - 63 m = 63 m, in the initially defined positive direction (away from the trainer). Using a total time of 18.0 s:
average velocity = total displacement / total time = 63 m / 18.0 s = 3.5 m/s
In a particular crash test, an automobile of mass 1271 kg collides with a wall and bounces back off the wall. The x components of the initial and final speeds of the automobile are 19 m/s and 3.3 m/s, respectively. If the collision lasts for 0.16 s, find the magnitude of the impulse due to the collision. Calculate the magnitude of the average force exerted on the automobile during the collision.
Answer:
28343.3 kgm/s
177145.625 N
Explanation:
u = Initial velocity = 19 m/s
v = Final velocity = -3.3 m/s (opposite direction)
m = Mass of car = 1271 kg
t = Time taken = 0.16 s
Impulse is given by
[tex]J=m(v-u)\\\Rightarrow J=1271(-3.3-19)\\\Rightarrow J=-28343.3\ kgm/s[/tex]
The magnitude of the impulse due to the collision is 28343.3 kgm/s
Force given by
[tex]F=\dfrac{J}{t}\\\Rightarrow F=\dfrac{-28343.3}{0.16}\\\Rightarrow F=-177145.625\ N[/tex]
The magnitude of the average force exerted on the automobile is 177145.625 N
Final answer:
The magnitude of the impulse during the collision is 28345.3 kg·m/s, and the magnitude of the average force exerted on the automobile is 177158 N.
Explanation:
In the crash test scenario described, the impulse experienced by the automobile can be determined by using the change in momentum due to the collision. Momentum (p) is the product of mass (m) and velocity (v), and impulse is the change in momentum which can be calculated using the initial (vi) and final velocities (vf) of the car:
Impulse = Change in momentum
= m × (vf - vi)
For the automobile:
Mass (m) = 1271 kgInitial velocity (vi) = 19 m/s (positive in the direction of motion)Final velocity (vf) = -3.3 m/s (negative since the car bounces back)Impulse = 1271 kg × (-3.3 m/s - 19 m/s)
= 1271 kg × (-22.3 m/s)
= -28345.3 kg·m/s (to one decimal placing)
The magnitude of the impulse is the absolute value:
28345.3 kg·m/s.
To calculate the average force exerted on the automobile, we use the fact that impulse also equals the average force (Favg) multiplied by the time of impact (t):
Impulse = Favg × t
Thus, the average force can be found by dividing the impulse by the collision time.
Favg = Impulse / t
= 28345.3 kg·m/s / 0.16 s
= 177158.1 N (to one decimal placing)
The magnitude of the average force is the absolute value: 177158 N.
Imagine that two charged objects are the system of interest. When the objects are infinitely far from each other, the electric potential energy of the system is zero. When the objects are close to each other, the electric potential energy is positive. Which of the following statements is(are) incorrect
(a) Both objects are positively charged.
(b) Both objects are negatively charged.
(c) One object is negatively charged and the other one is positively charged.
Final answer:
Statement (c) is incorrect because if one object is negatively charged and the other is positively charged, they would attract each other, resulting in a negative potential energy as they come closer, not a positive one.
Explanation:
The question is addressing the concept of electric potential energy between two charged objects. When the electric potential energy of the system is positive as the two objects come close, we can infer that the objects have like charges, either both positive or both negative. This is due to the fact that work needs to be done against the electrical repulsion to bring like charges together, increasing the system's potential energy.
This makes statements (a) Both objects are positively charged and (b) Both objects are negatively charged possibly correct scenarios, as they would lead to a positive potential energy when the objects are brought together. Statement (c) One object is negatively charged and the other one is positively charged would be incorrect in this context, because a positive and a negative charge would attract each other, and the system would do work on the surroundings as they come closer to each other making the potential energy negative. Therefore, the incorrect statement, given that the electric potential energy is positive when they are near each other, is (c).
A hot-air balloon is rising upward with a constant speed of 2.51 m/s. When the balloon is 3.16 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground
Answer:
t = 1.099 s
Explanation:
given,
constant speed = 2.51 m/s
height of balloon above ground = 3.16 m
time elapsed before it hit the ground = ?
Applying equation of motion to the compass
[tex]y = u t + \dfrac{1}{2}at^2[/tex]
[tex]-3.16 = 2.51 t + \dfrac{1}{2}\times (-9.8)t^2[/tex]
[tex]4.9 t^2 - 2.51 t - 3.16 = 0[/tex]
using quadratic formula to solve the equation
[tex]t = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]t = \dfrac{-(-2.51)\pm \sqrt{2.51^2-4(4.9)(-3.16)}}{2\times 4.9}[/tex]
t = 1.099 s, -0.586 s
hence, the time elapses before the compass hit the ground is equal to 1.099 s.
A vacuum chamber contains a uniform electric field directed downward. If a proton is shot horizontally into this region, it's acceleration is? Give direction and relative magnitude (Ex 45 degrees west of north, and is exponentially decreasing)
Answer:
It is directed downward. Its relative magnitude is a= qp*E/mp = 0.96*10⁸*E m/s².
Explanation:
Assuming gravitational force to be negligible, the only force acting on the proton once within the vacuum chamber, is the one due to the electric field.
As the proton is a positive charge, the electric force on it due to the field has the same direction than the field, as the direction of the field by convention, is the one that would take a positive test charge.
According to electric field definition, and to Newton's 2nd Law, we can find the acceleration of the proton as follows:
F = m*a = q*E ⇒ a= q*E / m = 1.6*10⁻¹⁹ C * E (N/C) / 1.67*10⁻²⁷ kg
⇒ a = 0.96* 10⁸ * E m/s²
A proton shot horizontally into a vacuum chamber with a uniform electric field directed downward will experience an upward acceleration.
Explanation:A proton shot horizontally into a vacuum chamber with a uniform electric field directed downward will experience an acceleration due to the force exerted by the electric field. Since the electric field is directed downward, the proton will experience an upward acceleration.
The magnitude of the proton's acceleration can be calculated using the equation:
a = qE/m
where a is the acceleration, q is the charge of the proton, E is the magnitude of the electric field, and m is the mass of the proton.
The direction of the acceleration will be opposite to the direction of the electric field, which in this case is upward.
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A human hair has a thickness of about 60 um. What is this in millimeters
Answer:
0.06
Explanation:
The thickness of a human hair is about 60 micrometers. When converted to millimeters, this measures to 0.06 millimeters.
Explanation:The thickness of a human hair is about 60 micrometers (um). To convert this to millimeters (mm), you need to understand that 1 millimeter is equal to 1000 micrometers. Therefore, you can convert 60 um to mm by dividing 60 by 1000.
This gives an answer of 0.06 mm. So, the thickness of a human hair is 0.06 millimeters.
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A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of electrons that can be transferred by this charger in 1.0 hour
Answer:
The maximum no. of electrons- [tex]2.25\times 10^{22}[/tex]
Solution:
As per the question:
Maximum rate of transfer of charge, I = 1.0 C/s
Time, t = 1.0 h = 3600 s
Rate of transfer of charge is current, I
Also,
[tex]I = \frac{Q}{t}[/tex]
Q = ne
where
n = no. of electrons
Q = charge in coulomb
I = current
Thus
Q = It
Thus the charge flow in 1. 0 h:
[tex]Q = 1.0\times 3600 = 3600\ C[/tex]
Maximum number of electrons, n is given by:
[tex]n = \frac{Q}{e}[/tex]
where
e = charge on an electron = [tex]1.6\times 10^{- 19}\ C[/tex]
Thus
[tex]n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}[/tex]
polybius believed that rome's successes stemmed from select one: a. its constitution and mixed government. b. conservative roman values. c. its geographical diversity. d. roman worship of greek deities. e. all these answers are correct.
Answer:
a. its constitution and mixed government.
Explanation:
Polybius tell us about the Roman mixed constitution as a fundamental in the Roman victory and the Carthaginian defeat in the Punic war, conceiving the mixed constitution as the best, since this constitution was at its peak, which implies that among the elements of the constitution, the aristocratic component was the dominant one.
What is the potential difference between the terminals of an ordinary AA or AAA battery? (If you’re not sure, find one and look at the label.)
The potential difference between the terminals of an ordinary AA or AAA battery is usually 1.5 volts. Voltage is the work required to move a charge, while current is the charge flow rate. Batteries can provide considerable energy for their size, analogous to lifting a significant mass against gravity.
Explanation:The potential difference between the terminals of an ordinary AA or AAA battery is the electromotive force (emf) when the battery is not part of a complete circuit, and is typically about 1.5 volts (V). If the battery is in a complete circuit, the potential difference is known as the terminal potential difference, which is also measured in volts but may differ slightly from the emf due to internal resistance and load on the battery. Voltage is the measure of work required to move a charge between two points, while current is the rate of charge flow, measured in amperes.
When a battery is used in a circuit, the conventional current flows from the positive terminal to the negative terminal, and this flow of charge creates the potential difference that does work in the circuit. An ammeter, which must be connected in series, is used to measure current. The common AA battery not only has a voltage of 1.5 V but also a significant amount of stored energy, which could be likened to a substantial mass being lifted against gravity, highlighting the compact energy storage capability of such batteries.
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You break a piece of Styrofoam 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 21 mN when they’re 15 mm apart, what’s the magnitude of the charge on each?
Answer:
2.24×10⁻⁸ C
Explanation:
From coulomb's law,
F = kq1q2/r² ............................ Equation 1
Where F = Repulsive force between the two charges, q1 = charge on the first sphere, q2 = charge on the second sphere, r = distance between the sphere, k = proportionality constant.
Note: q1 = q2 = q,
Then, we can rewrite equation 1 as,
F = kq²/r²
making q the subject of the equation
q = √(Fr²/k)................................. Equation 2
Given: F = 21 mN = 0.021 N, r = 15 mm = 0.015 m
Constant: k = 9.0×10⁹ Nm²/C²
Substituting these values into equation 2
q = √(0.021×0.015²/9.0×10⁹)
q = √(5×10⁻¹⁶)
q = 2.24×10⁻⁸ C.
Hence the charge on each sphere = 2.24×10⁻⁸ C
The magnitude of the charge on each sphere is [tex]2.29*10^{-8}C[/tex]
The force between two charges is given by coulombs law,
[tex]F=k\frac{Q_{1}*Q_{2}}{r^{2} }[/tex]
Where r is distance between charges.
And k is constant, [tex]k=9*10^{9} Nm^{2}/C^{2}[/tex]
Given that, [tex]Q_{1}=Q_{2},r=15mm=15*10^{-3}m,F=21mN=21*10^{-3}N[/tex]
Substitute above values in above formula.
[tex]21*10^{-3}=9*10^{9}*\frac{Q^{2} }{(15*10^{-3} )^{2} } \\\\Q^{2}=\frac{21*10^{-3}*225*10^{-6} }{9*10^{9} } \\\\Q^{2}=5.25*10^{-16} \\\\Q=\sqrt{5.25*10^{-16}} =2.29*10^{-8}C[/tex]
Hence, the magnitude of the charge on each sphere is [tex]2.29*10^{-8}C[/tex]
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In fair weather, the ground may become charged such that there is an electric field just above the surface of the Earth, pointing down toward the ground. What is the electric charge of the ground in this situation?
The electric field is defined as the electric force per unit of charge. The direction of the field is taken as the direction of the force it would exert on a positive test load. An electric field is always directed from its positive charge to the negative. Under this condition if the terrestrial electric field is pointing towards the ground, the terminal that is pointing is the negative and that of its surface is the positive. This implies that the sign of the charge of the electric field of the soil will be negative.
A typical adult human lung contains about 330 million tiny cavities called alveoli. Estimate the average diameter of a single alveolus. Assume the alveoli are spherical and a typical human lung is about 1.9 liters.
Answer:
The average diameter of a single alveolus is 0.0222 cm.
Explanation:
Volume of the lung ,V= 1.9 L
[tex]1 L = 1000 cm^3[/tex]
[tex]1.9 L=1.9\times 1000 cm^3=1900 cm^3[/tex]
Number of alveoli in a human lung = [tex]330\times 10^6[/tex]
Volume of single alveoli =v
[tex]v\times 330\times 10^6=V[/tex]
[tex]v=\frac{1900 cm^3}{330\times 10^6}[/tex]
[tex]v=5.7575\times 10^{-6} cm^3[/tex]
The alveoli are spherical.
Radius of an alveolus = r
Volume of the sphere = [tex]\frac{4}{3}\pi r^3[/tex]
[tex]v=\frac{4}{3}\pi r^3[/tex]
[tex]5.7575\times 10^{-6} cm^3=\frac{4}{3}\times 3.14\times r^3[/tex]
[tex]r=0.0111 cm[/tex]
Diameter of the alveolus =d
d = 2r = 2 × 0.0111 cm = 0.0222 cm
The average diameter of a single alveolus is 0.0222 cm.
The question seeks to find the average diameter of a single alveolus in the human lung, using information about the total volume of the lung and the number of alveoli. By using the formula for the volume of a sphere, one can calculate the volume of a single alveolus and derive its radius and thereby its diameter.
Explanation:The problem is essentially asking to find the average diameter of a single alveolus, given we know the total volume of the lung and the number of alveoli. Using the formula for the volume of a sphere, V=4/3πr³, where V is the volume and r is the radius, we can find the volume of a single alveolus by dividing the total volume of the lungs (1.9 liters, which equals to 1.9 x 10^9 cubic millimeters) by the number of alveoli (approximately 330 million). We can then calculate the radius by rearranging the sphere volume formula to r = ((3*V)/4π)^1/3. The diameter would be double the radius.
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A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V.
Use ϵ0 = 8.85×10⁻¹² C²/N⋅m².
The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.
Answer:
U_eq = 1.99 * 10^(-10) J
Explanation:
Given:
Plate Area = 10 cm^2
d = 0.01 m
k_dielectric = 3
k_air = 1
V = 15 V
e_o = 8.85 * 10 ^-12 C^2 / N .m
Equations used:
U = 0.5 C*V^2 .... Eq 1
C = e_o * k*A /d .... Eq 2
U_i = 0.5 e_o * k_i*A_i*V^2 /d ... Eq 3
For plate to be half filled by di-electric and half filled by air A_1 = A_2 = 0.5 A:
U_electric = 0.5 e_o * k_1*A*V^2 /2*d
U_air = 0.5 e_o * k_2*A*V^2 /2*d
The total Energy is:
U_eq = U_electric + U_air
U_eq = 0.5 e_o * k_1*A*V^2 /2*d + 0.5 e_o * k_2*A*V^2 /2*d
U_eq = (k_1 + k_2) * e_o * A*V^2 / 4*d
Plug the given values:
U_eq = (3 + 1) * (8.82 * 10^ -12 )* (0.001)*15^2 / 4*0.01
U_eq = 1.99 * 10^(-10) J
The energy U2 of the partially dielectric-filled capacitor is found by calculating the total capacitance Ct when the capacitor is half-filled with the dielectric and then using the formula U2 = ½ CtV2 with the given voltage.
Explanation:To calculate the energy U2 of the partially dielectric-filled capacitor, we start by finding the capacitance when the capacitor is half-filled with the dielectric. With a dielectric constant k of 3.00 and a vacuum permittivity ε0 of 8.85×10−12 C2/N·m2, the initial capacitance Ci entirely filled with the dielectric is:
Ci = kε0A/d
When the capacitor is half-filled, the capacitance can be represented as two capacitors in parallel: one with dielectric (Cd) and one without (C0). These can be described by:
Cd = ε0kA/(2d) and C0 = ε0A/(2d)
The total capacitance Ct when the capacitor is half-filled is then:
Ct = Cd + C0
Finally, the energy stored in the capacitor U2 is given by:
U2 = ½ CtV2
This formula allows us to plug in the calculated capacitance Ct and the constant voltage V to find the desired energy U2.
Consider a cloudless day on which the sun shines down across the United States. If 2073 kJ of energy reaches a square meter ( m 2 ) of the United States in one hour, how much total solar energy reaches the entire United States per hour
The total amount of energy per hour is [tex]2.039\cdot 10^{16} kJ[/tex]
Explanation:
In this problem we are told that the amount of energy reaching a square meter in the United States per hour is
[tex]E_1 = 2073 kJ[/tex]
The total surface area of the United States is
[tex]A=9.834\cdot 10^6 km^2[/tex]
And converting into squared metres,
[tex]A=9.834\cdot 10^6 \cdot 10^6 = 9.834\cdot 10^{12} m^2[/tex]
Therefore, the total energy reaching the entire United States per hour is given by:
[tex]E=AE_1 = (9.834\cdot 10^{12})(2073)=2.039\cdot 10^{16} kJ[/tex]
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Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note that the latent heat is equal to the change in enthalpy, and that liquid nitrogen boils at 77 K. The system is kept at a constant pressure of 1 atm. 20) Assuming that you can treat the gas as ideal gas and that the volume of the liquid compute the binding energy of a nitrogen molecule in the liquid. (the binding energy is the difference in internal energy per molecule between the liquid and gas) approximately zero,
a. 9.4 x 10-21 J
b. 3.8 х 1027 J
c. 4.2 x 10-18 J
d. 10-20 J e. 2.1 x 10-19 J
Answer:
The energy is [tex]9.4\times10^{-21}\ J[/tex]
(a) is correct option
Explanation:
Given that,
Energy = 4480 j
Weight of nitrogen = 20 g
Boil temperature = 77 K
Pressure = 1 atm
We need to calculate the internal energy
Using first law of thermodynamics
[tex]Q=\Delta U+W[/tex]
[tex]Q=\Delta U+nRT[/tex]
Put the value into the formula
[tex] 4480=\Delta U+\dfrac{20}{28}\times8.314\times77[/tex]
[tex]\Delta U=4480-\dfrac{20}{28}\times8.314\times77[/tex]
[tex]\Delta U=4022.73\ J[/tex]
We need to calculate the number of molecules in 20 g N₂
Using formula of number of molecules
[tex]N=n\times \text{Avogadro number}[/tex]
Put the value into the formula
[tex]N=\dfrac{20}{28}\times6.02\times10^{23}[/tex]
[tex]N=4.3\times10^{23}[/tex]
We need to calculate the energy
Using formula of energy
[tex]E=\dfrac{\Delta U}{N}[/tex]
Put the value into the formula
[tex]E=\dfrac{4022.73}{4.3\times10^{23}}[/tex]
[tex]E=9.4\times10^{-21}\ J[/tex]
Hence, The energy is [tex]9.4\times10^{-21}\ J[/tex]
The binding energy of a nitrogen molecule in the liquid is calculated using the formula for latent heat and Avogadro's number. Once the latent heat and the number of molecules are determined, we can find the energy per molecule and hence, the binding energy. After performing the calculations, we find the binding energy of a nitrogen molecule to be approximately 5.2 x 10^-21 J.
Explanation:To calculate the binding energy of a nitrogen molecule in the liquid, we can use the formula for latent heat. This is given by:
Q = mL, where,
Q = Heat energy applied (color change or state change)
m = mass
L = Latent heat
After calculating the latent heat, we can determine the number of moles of nitrogen gas formed and subsequently the number of molecules using Avogadro's number. Then, the energy per molecule is calculated by dividing the total heat absorbed by the number of molecules. Therefore, the binding energy will be the difference in energy per molecule (latent heat per molecule) between the liquid and the gaseous states.
The amount of heat given (Q) is 4480 J and the mass (m) of liquid nitrogen boiled is 20 g. The molar mass (M) of nitrogen (N2) is approximately 28 g/mol. Using this, we can rearrange the formula to find L which is the energy/mass.
Substituting the values, we get L = Q/m = 4480J / 20g = 224 J/g
The number of moles of N2(n) = m/M = 20g / 28 g/mol ~ 0.714 mol
The number of N2 molecules(N) = n x Avogadro's number = 0.714 mol x 6.022 x 10^23 mol^-1 ~ 4.3 x 10^23
Therefore, the binding energy of a nitrogen molecule (E) = L/N = 224J/g / 4.3 x 10^23 = ~5.2 x 10^-21 J.
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A ball of mass 0.7 kg flies through the air at low speed, so that air resistance is negligible. (a) What is the net force acting on the ball while it is in motionWhich components of the ball's momentum will be changed by this force? What happens to the x component of the ball's momentum during its flight? What happens to the y component of the ball's momentum during its flight? It decreases. What happens to the z component of the ball's momentum during its flight?
With negligible air resistance and low speed, the only significant net force on a 0.7 kg ball is gravity, affecting the ball's y component of momentum. The x component remains constant, and z component changes are not discussed without additional forces.
Explanation:When a ball of mass 0.7 kg flies through the air at low speed with air resistance negligible, the net force acting on the ball while it is in motion is primarily due to gravity, which will be impacting the y component of the ball's momentum. The x component of the ball's momentum remains unchanged because no horizontal force is applied, while the y component changes due to gravity, and the z component would only change if there were forces acting in a direction out of the horizontal plane, which are not mentioned in the scenario. As for the Earth-ball system, momentum is conserved in the vertical direction because the system experiences no net external vertical force.
What is the magnitude of the force on an electron at a distance of 1.70 angstrom from the plutonium nucleus?
Answer:
Force, [tex]F=7.04\times 10^{-5}\ N[/tex]
Explanation:
Given that,
Distance between 1.70 A from the plutonium nucleus, [tex]d=1.7\times 10^{-10}\ m[/tex]
The number of electron in plutonium is 94.
To find,
The magnitude of the force on an electron.
Solution,
Total charge in the plutonium nucleus is, [tex]q=94\times 1.6\times 10^{-19}=1.504\times 10^{-17}\ C[/tex]. The electric force between charges is given by :
[tex]F=\dfrac{kq^2}{d^2}[/tex]
[tex]F=\dfrac{9\times 10^9\times (1.504\times 10^{-17})^2}{(1.7\times 10^{-10})^2}[/tex]
[tex]F=7.04\times 10^{-5}\ N[/tex]
So, the magnitude of the force on an electron is [tex]7.04\times 10^{-5}\ N[/tex]. Hence, this is the required solution.
Calculate the energy, in electron volts, of a photon whose frequency is the following.
(a) 585 THz
(b) 3.50 GHz
(c) 40.0 MHz
Planck's equation for the energy of a photon is E = hf, where f is the frequency and h is Planck's constant. We use 1 eV = 1.60 ✕ 10−19 J for units of energy.
(a) For the energy of the photon at a frequency of 585 THz, we have E = hf
Answer:
(a) [tex]E=2.42eV[/tex]
(b) [tex]E=1.45*10^{-5}eV[/tex]
(c) [tex]E=1.66*10^{-7}eV[/tex]
Explanation:
The Planck-Einstein relation allows us to know the energy (E) of a photon, knowing its frequency (f). According to this relation, the energy of the photon is defined as:
[tex]E=hf[/tex]
Here h is the Planck constant.
(a)
[tex]E=(4.14*10^{-15}eV\cdot s)(585*10^{12}Hz)\\E=2.42eV[/tex]
(b)
[tex]E=(4.14*10^{-15}eV\cdot s)(3.50*10^{9}Hz)\\E=1.45*10^{-5}eV[/tex]
(c)
[tex]E=(4.14*10^{-15}eV\cdot s)(40.0*10^{6}Hz)\\E=1.66*10^{-7}eV[/tex]
When two point charges are 2.0 cm apart, each one experiences a 1.0-N electric force due to the
other charge. If they are moved to a new separation of 8.0 cm, the electric force on each of them
is closest to
A) 1.0 N.
B) 4.0 N.
C) 16 N.
D) 0.25 N.
E) 0.063 N.
Answer:
option (E)
Explanation:
F = 1 N
r = 2 cm
r' = 8 cm
F' = ?
According to the Coulomb's law
[tex]F \alpha \frac{1}{r^{2}}[/tex]
So, [tex]\frac{F'}{F}=\left (\frac{r}{r'} \right )^{2}[/tex]
[tex]\frac{F'}{1}=\left (\frac{2}{8} \right )^{2}[/tex]
F' = 0.063 N
When they are moved to a new separation of 8.0 cm, the electric force on each of them will be closest to 0.063 N. Option E is correct.
Given:
The point charges are initially at a distance of [tex]d=2[/tex] cm.
The initial force experienced by them is [tex]f=1[/tex] N.
The electrostatic force experienced by two charges q and Q is defined as,
[tex]f=K\dfrac{qQ}{r^2}[/tex]
where K is the constant and r is the distance between the charges.
The initial force experienced by them will be,
[tex]f=K\dfrac{qQ}{d^2}=1\rm\; N[/tex]
Now, the final distance between the charges is changed to 8 cm which is equal to 4d.
So, the new force on the charges will be,
[tex]F=K\dfrac{qQ}{(4d)^2}\\F=K\dfrac{qQ}{d^2}\times \dfrac{1}{16}\\F=f\times \dfrac{1}{16}\\F=0.0625\approx0.063\rm\;N[/tex]
Therefore, when they are moved to a new separation of 8.0 cm, the electric force on each of them will be closest to 0.063 N. Option E is correct.
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A certain car has a fuel efficiency of 39.2 miles per gallon (mi/gal). Express this efficiency in kilometers per liter (km/L).
Answer:
e = 16.67 km/L
Explanation:
given,
fuel efficiency = 39.2 mi/gal
we need to convert fuel efficiency in Km/l
1 Km = 0.621 mile
1 gal = 3.786 L
now, efficiency
[tex]e = 39.2 \dfrac{mi}{gal}\times \dfrac{1\ km}{0.621\ mile}\times\dfrac{1\ gal}{3.786\ L}[/tex]
e = 16.67 km/L
hence, the efficiency of the car in km/L is equal to 16.67 km/L
A torque of 4 Nm is required to rotate the intermediate cylinder at 30 radians/min. Calculate the viscosity of the oil. All cylinders are 450 mm long. Neglect end effects.
To calculate the viscosity of the oil, you need to utilize the concept of shear stress in fluid dynamics. You take the known values of torque, cylinder length, and rotational speed and substitute them into the rearranged shear stress formula. Ensure you convert length from millimeters to meters to maintain SI unit consistency.
Explanation:The question can be solved using the principles of fluid dynamics, specifically the concept of viscous drag in a medium. In this case, the medium is oil and the object moving through it is a cylinder. The viscosity of the oil can be calculated using the formula for shear stress (τ), given as τ = η (du/dy), where η represents viscosity, du represents the difference in velocity, and dy represents the distance between layers of fluid. As we're given torque (τ) in Newton-metres (Nm), cylinder length (L) in millimeters, and rotational speed in rad/s (radians per second), we can derive η (viscosity) using the rearranged formula: η = τ /(du/dy).
In this case, du is equivalent to the speed of the cylinder let's represent this as 'u', dy is equivalent to the length of the cylinder 'L', hence du/dy becomes u/L. Substitute these values into the rearranged formula and solve to get the viscosity. It's critical to convert the length from millimeters to meters before performing the calculation to maintain consistency in the SI units.
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Water from a 100 ft x 200 ft lawn converges into a gully. During a 1-hour storm, discharge into the gully is 0.37 cfs (ft3/s). The interception by grass is 0.02 acre-ft. Overall, 70% of the rain is infiltrated. What is the amount of rainfall?
Answer:
7344 ft^3
Explanation:
rain flown = 0.37 ft^3/s * 3600 s/hr = 1332 ft^3
rain intercepted by grass = 0.02 acre-ft = 0.02 * 43559.9 = 871.2 ft^3
AS 1 acre-ft = 43559.9 ft^3
LET THE TOTAL RAINFALL BE 'X'
Thus the equations become
x = 0.7x + 871.2 +1332 (as 70 % of rain in infiltrated)
x = 7344 ft^3
When jumping, a flea accelerates at an astounding 1000 m/s2, but over only the very short distance of 0.50 mm. If a flea jumps straight up, and if air resistance is neglected (a rather poor approximation in this situation), how high does the flea go?
Answer:
The flea will move high to a height of 0.05 meters.
Explanation:
Given that,
Acceleration of the flea, [tex]a=1000\ m/s^2[/tex]
Distance, d = 0.5 mm = 0.0005 m
Let u and v are the initial and final velocity of the flea. Using equation of motion as :
[tex]v^2-u^2=2ad[/tex]
[tex]v^2-u^2=2\times 1000\times 0.0005[/tex]
[tex]v^2-u^2=1[/tex]..........(1)
Using conservation of energy, we get :
[tex]\dfrac{1}{2}mu^2=\dfrac{1}{2}mv^2+mgh[/tex]
[tex]\dfrac{1}{2}u^2=\dfrac{1}{2}v^2+(-g)h[/tex]
[tex]\dfrac{1}{2}u^2=\dfrac{1}{2}v^2-gh[/tex]
[tex]\dfrac{1}{2g}(u^2-v^2)=-h[/tex]
[tex]h=\dfrac{1}{2g}[/tex]
[tex]h=\dfrac{1}{2\times 9.8}[/tex]
h = 0.05 meters
So, the flea will move high to a height of 0.05 meters. Hence, this is the required solution.
Two charged particles are a distance of 1.62 m from each other. One of the particles has a charge of 7.10 nc, and the other has a charge of 4.42 nc. (a) What is the magnitude (in N) of the electric force that one particle exerts on the other? (b) is the force attractive or repulsive? O attractive O repulsive
Answer:
A. F=107.6nN
B. Repulsive
Explanation:
According to coulombs law, the force between two charges is express as
F=(Kq1q2) /r^2
If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.
Note the constant K has a value 9*10^9
Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m
If we substitute values we have
F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)
F=(282.4×10^-9)/2.6244
F=107.6×10^-9N
F=107.6nN
B. Since the charges are both positive, the force is repulsive
The magnitude of the electric force between the two particles is approximately 0.012 N, according to Coulomb's Law. The force could be either attractive or repulsive, depending on whether the charges are the same or opposite, respectively.
Explanation:The subject of this question is Physics, specifically about Coulomb’s Law, which deals with the electric force between two charges.
(a) According to Coulomb’s Law, the electric force (F) between two charges is directly proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (d) between them. This is mathematically represented as F = k*q1*q2/d^2, where k is Coulomb's constant (8.988 x 10^9 Nm^2/C^2). Substituting in the given values, we find F = (8.988 * 10^9 Nm^2/C^2) * (7.10 * 10^-9 C) * (4.42 * 10^-9 C) / (1.62 m)^2, resulting in an electric force of approximately 0.012 N.
(b) Whether the force is attractive or repulsive depends on the nature of the charges. Same charges repel each other, while opposite charges attract each other. As the charges are not specified, we cannot definitively answer this part of the question.
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