Answer:
Force is 432.94 N along the rebound direction of ball.
Explanation:
Force is rate of change of momentum.
[tex]\texttt{Force}=\frac{\texttt{Final momentum-Initial momentum}}{\texttt{Time}}[/tex]
Final momentum = 0.38 x -1.70 = -0.646 kgm/s
Initial momentum = 0.38 x 2.20 = 0.836 kgm/s
Change in momentum = -0.646 - 0.836 = -1.472 kgm/s
Time = 3.40 x 10⁻³ s
[tex]\texttt{Force}=\frac{\texttt{Final momentum-Initial momentum}}{\texttt{Time}}=\frac{-1.472}{3.40\times 10^{-3}}\\\\\texttt{Force}=-432.94N[/tex]
Force is 432.94 N along the rebound direction of ball.
Final answer:
Using Newton's second law and the change in the ball's momentum, the average net force exerted on the billiard ball by the cushion is 436.47 N, directed away from the cushion.
Explanation:
The question relates to the concept of Newton's second law of motion, which states that the force on an object is equal to the mass of the object multiplied by the acceleration (F = ma). In this situation, a billiard ball strikes the cushion perpendicularly, changing its velocity and hence experiencing acceleration. To calculate the average net force exerted on the ball by the cushion, we can use the change in velocity and the time of impact in the following steps:
Calculate the change in momentum of the ball (Δp = m−v_f - m−v_i), where m is the mass, v_f is the final velocity, and v_i is the initial velocity.
Divide the change in momentum by the time of impact (Δt) to get the average force (F_avg). Use the formula F_avg = Δp / Δt.
Now let's apply these steps to the given values:
The change in momentum is Δp = 0.38 kg * (-1.70 m/s) - 0.38 kg * (+2.20 m/s) = -0.38 kg * (-1.70 - 2.20) m/s = -0.38 kg * (-3.90) m/s = 1.482 kg−m/s.
The average force is F_avg = 1.482 kg−m/s / 3.40 x 10^-3 s = 436.47 N.
The average net force exerted on the ball by the cushion is 436.47 N, directed away from the cushion since the ball rebounded after the collision.
What is the distance between adjacent crests of ocean waves that have a frequency of 0.20 Hz if the waves have a speed of 2.4 m/s?
Answer:
12 m
Explanation:
f = 0.2 Hz, v = 2.4 m/s
v = f x λ
Where, λ is the wavelength
λ = v / f = 2.4 / 0.2 = 12 m
The wavelength is defined as the distance between two consecutive crests or troughs.
So, the distance between two consecutive crests is 12 m.
An electron is to be accelerated in a uniform electric field having a strength of 2.00 × 106 V/m. (a) What energy in keV is given to the electron if it is accelerated through 0.400 m? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?
Answer:
a) 800 keV
b) 25 km
Explanation:
[tex]Strength\ of\ Electric\ field=2\times 10^6\ V/m\\a)\ Potential\ Difference=Strength\ of\ Electric\ field\times Distance\\\Rightarrow Potential\ Difference=Kinetic\ Energy\ =2\times 10^6\times 0.4\\\therefore Energy=0.8\times 10^6\ eV=800\ keV\\[/tex]
[tex]b)\ Potential\ difference=50\ GeV=50\times 10^9\ eV\\Distance=\frac{Potential\ difference}{electric\ field}\\\Rightarrow Distance=\frac{50\times 10^9}{2\times 10^6}\\\Rightarrow Distance=25\times 10^3\ m\\\therefore Distance=25\ km[/tex]
Answer:
a) 800 keV
b) 24.996 km.
Explanation:
(a) we have
[tex]\large \Delta K.E=q\Delta V[/tex] .............(1)
where,
[tex]\large \Delta K.E[/tex] = Change in kinetic energy
[tex]q[/tex] = charge of an electron
[tex]\Delta V[/tex] = Potential difference
also
[tex]\large E=\frac{V}{d}[/tex] .......(2)
E = electric field
d = distance traveled
Now from (1) and (2) we have,
[tex]\large \Delta K. E=qV=qEd[/tex]
substituting the values in the above equation, we get
[tex]\large \Delta K. E=(1.6\times 10^{-19}C)(2\times 10^6V/m)(0.400m)(\frac{1eV}{1.6\times 10^{-19}J})(\frac{1keV}{1000eV})[/tex]
[tex]\large \Delta K. E=800keV[/tex]
Thus, the energy gained by the electron is 800 keV if it is accelerated over a distance of 0.400 m.
(b) Using the equation (1), we have
[tex]\large d=\frac{\Delta K.E}{qE}[/tex]
[tex]\large d=\frac{(50\times 10^9eV)}{(1.6\times 10^{-19C})(2\times 10^6V/m)}(\frac{1.6\times 10^{-19}J}{1eV})[/tex]
or
[tex]\large d=2.4996\times 10^4m[/tex]
or
[tex]\large d=24.996\times 10^3m=24.996km[/tex]
Thus, to gain 50.0 GeV of energy the electron must be accelerated over a distance of 24.996 km.
The pressure of a monatomic ideal gas is doubled, while the volume is cut in half. By what factor is the internal energy of the gas multiplied? (a) 1/4 (b) 1/2 (c) 1 (d) 2 (e) 4
Answer:
option C
multiplication factor n = 1 when volume change to half and pressure become double
Explanation:
we know by Ideal Gas law:
P1V1 = nRT1
P2V2 = nRT2
according to the question
pressure is doubled and volume is reduced to half
so we have
new pressure = 2*P1
new volume = V1/2
hence,
(2P1) * (V1/2) = nRT2
P1V1 = nRT2
we have now
nRT1 = nRT2
we get
T1 = T2
thus no change in temperature
we know that internal energy is given as
internal energy = nCvT,
since temperature is directly proportional to internal energy and since temperature remains constant therefore internal energy remains constant
So there is no change in internal energy
thus, multiplication factor n = 1 when volume change to half and pressure change to double
A flowerpot falls off a windowsill and passes the win- dow of the story below. Ignore air resistance. It takes the pot 0.380 s to pass from the top to the bottom of this window, which is 1.90 m high. How far is the top of the window below the window- sill from which the flowerpot fell?
Answer:
d = 0.50 m
Explanation:
Let say the speed at the top and bottom of the window is
[tex]v_1 \: and \: v_2[/tex] respectively
now we have
[tex]d = \frac{v_1 + v_2}{2}t[/tex]
[tex]1.90 = \frac{v_1 + v_2}{2} (0.380)[/tex]
[tex]v_1 + v_2 = 10 [/tex]
also we know that
[tex]v_2 - v_1 = 9.8(0.380)[/tex]
[tex]v_2 - v_1 = 3.72[/tex]
now we have from above equations
[tex]v_2 = 6.86 m/s[/tex]
[tex]v_1 = 3.14 m/s[/tex]
now the distance from which it fall down is given as
[tex]v_f^2 - v_i^2 = 2ad[/tex]
[tex]3.14^2 - 0^2 = 2(9.8)d[/tex]
[tex]d = 0.50 m[/tex]
A negative charge -0.550 HC exerts an upward 0.700 N force on an unknown charge 0.220 m directly below it. (a) What is the unknown charge (magnitude and sign)? (b) What are the magnitude and direction of the force that the unknown charge exerts on the -0.550 HC charge? downward upward
Answer:
Part a)
[tex]q_2 = -6.8 \mu C[/tex]
Part b)
[tex]F = 0.700 N[/tex]
direction = downwards
Explanation:
As we know that the negative charge will experience the force due to some other charge below it
the force is given as
[tex]F = 0.700 N[/tex]
now we know that
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
now plug in all data
[tex]0.700 = \frac{(9 \times 10^9)(0.550\mu C)q_2}{0.220^2}[/tex]
[tex]0.700 = 1.022\times 10^5 q_2[/tex]
[tex]q_2 = -6.8 \mu C[/tex]
since this is a repulsion force so it must be a negative charge
Part b)
As per Newton's III law it will exert equal and opposite force on it
So here the force on the charge below it will be same in magnitude but opposite in direction
so here we have
[tex]F = 0.700 N[/tex]
direction = downwards
Water is flowing in a straight horizontal pipe of variable cross section. Where the cross-sectional area of the pipe is 3.70·10-2 m2, the pressure is 6.10·105 Pa and the velocity is 0.260 m/s. In a constricted region where the area is 9.50·10-4 m2, what is the velocity?
Answer:
v = 10.1 m/s
Explanation:
As we know that by the law of conservation of volume the rate of volume flowing through the pipe will remain conserved
so here we have flow rate given as
[tex]Q = Area\times velocity[/tex]
now we have
[tex]A_1 v_1 = A_2 v_2[/tex]
now we have
[tex]A_1 = 3.70 \times 10^{-2} m^2[/tex]
[tex]v_1 = 0.260 m/s[/tex]
[tex]A_2 = 9.50 \times 10^{-4} m^2[/tex]
now from above equation we have
[tex]v_2 = \frac{A_1}{A_2} v_1[/tex]
[tex]v_2 = \frac{3.70\times 10^{-2}}{9.50\times 10^{-4}}(0.260)[/tex]
[tex]v_2 = 10.1 m/s[/tex]
A velocity selector has an electric field of magnitude 2170 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 5.45 × 103 m/s, enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of +4.10 × 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.54 × 10-9 N, pointing directly upward. What is the speed of this particle?
Answer:
[tex]v = 4.51 \times 10^3 m/s[/tex]
Explanation:
electric field = 2170 N/C
now the speed of the charge particle is given as
[tex]v = 5.45 \times 10^3 m/s[/tex]
here we know that charge particle moves without any deviation
so we will have
[tex]qvB = qE[/tex]
now magnetic field in this region is given as
[tex]B = \frac{E}{v}[/tex]
[tex]B = \frac{2170}{5.45 \times 10^3}[/tex]
[tex]B = 0.398 T[/tex]
Now another charge particle enters the region with different speed and experience the force upwards
[tex]F = qE - qvB[/tex]
[tex]1.54 \times 10^{-9} = (4.10\times 10^{-12})[2170 - v(0.398)][/tex]
[tex]375.6 = 2170 - v(0.398)[/tex]
[tex]v = 4.51 \times 10^3 m/s[/tex]
The answer is: [tex]9.43 \times 10^5 \, \text{m/s}[/tex].
To determine the speed of the different particle that enters the velocity selector, we need to consider the forces acting on it due to the electric and magnetic fields. The net force acting on a charged particle in an electric field [tex]\( E \)[/tex] and a magnetic field [tex]\( B \)[/tex] is given by the Lorentz force equation:
[tex]\[ \vec{F} = q\vec{E} + q(\vec{v} \times \vec{B}) \][/tex]
where [tex]\( q \)[/tex] is the charge of the particle, [tex]\( \vec{v} \)[/tex] is the velocity of the particle, [tex]\( \vec{E} \)[/tex] is the electric field, and [tex]\( \vec{B} \)[/tex] is the magnetic field.
For the particles that pass through the velocity selector without being deflected, the forces due to the electric and magnetic fields must cancel each other out. This means that:
[tex]\[ qE = qvB \][/tex]
where [tex]\( v \)[/tex] is the speed of the particles that pass through undeflected.
Given that the electric field [tex]\( E \)[/tex] is 2170 N/C directed vertically upward, and the particles are traveling east, the magnetic field [tex]\( B \)[/tex] must be directed south to produce a force that cancels the electric force. The speed [tex]\( v \)[/tex] of the undeflected particles is given as [tex]\( 5.45 \times 10^3 \)[/tex] m/s.
Now, for the different particle with a charge of [tex]\( +4.10 \times 10^{-12} \)[/tex] C, the net force [tex]\( F \)[/tex] acting on it is [tex]\( 1.54 \times 10^{-9} \)[/tex] N, pointing directly upward. This means that the magnetic force is not sufficient to cancel the electric force, and the net force is due to the electric force only:
[tex]\[ F = qE \][/tex]
We can use this equation to find the speed [tex]\( v' \)[/tex] of the different particle. Since the net force is equal to the electric force, the magnetic force must be zero. This implies that the velocity of the particle is such that the magnetic force component is equal and opposite to the electric force component, but since the net force is not zero, the particle is not moving at the correct speed to pass through undeflected.
Let's solve for the speed [tex]\( v' \)[/tex] of the different particle:
[tex]\[ F = qE \][/tex]
[tex]\[ 1.54 \times 10^{-9} \, \text{N} = (4.10 \times 10^{-12} \, \text{C})(2170 \, \text{N/C}) \][/tex]
[tex]\[ v' = \frac{F}{qB} \][/tex]
We know [tex]\( F \), \( q \), and \( E \)[/tex], but we need to find [tex]\( B \)[/tex] from the information given for the undeflected particles:
[tex]\[ qvB = qE \][/tex]
[tex]\[ vB = E \][/tex]
[tex]\[ B = \frac{E}{v} \][/tex]
[tex]\[ B = \frac{2170 \, \text{N/C}}{5.45 \times 10^3 \, \text{m/s}} \][/tex]
[tex]\[ B = 3.98 \times 10^{-4} \, \text{T} \][/tex]
Now we can find [tex]\( v' \)[/tex]:
[tex]\[ v' = \frac{1.54 \times 10^{-9} \, \text{N}}{(4.10 \times 10^{-12} \, \text{C})(3.98 \times 10^{-4} \, \text{T})} \][/tex]
[tex]\[ v' = \frac{1.54 \times 10^{-9}}{1.6322 \times 10^{-15}} \][/tex]
[tex]\[ v' = 9.43 \times 10^5 \, \text{m/s} \][/tex]
Therefore, the speed of the different particle is [tex]\( 9.43 \times 10^5 \) m/s[/tex].
An object moves on a trajectory given by Bold r left parenthesis t right parenthesis equals left angle 10 cosine 6 t comma 10 sine 6 t right angle for 0 less than or equals t less than or equals pi. How far does it travel?
The object is moving along a circular path. Given the time restriction, it completes a half revolution, which equates to half the circumference of the circle. Therefore, the object travels approximately 31.4 units.
Explanation:The object is moving on a circular trajectory defined by r(t) = <10 cos 6t, 10 sin 6t>. This equation describes a circular path with a radius of 10 units. To find out how far the object travels, we can use the formula for the circumference of a circle, which is 2πr. However, since the time t varies from 0 to π, the object only makes a half revolution along the circle. So, the total distance the object travels would be equal to half the circumference of the circle, which is πr.
Substituting r = 10, we get, distance traveled = π*10 = 31.4 units approx.
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Which of the following is the kinetic rate equation for the addition-elimination mechanism of nucleophilic aromatic substitution? rate = k[aryl halide] rate = k[nucleophile] rate = k[aryl halide][nucleophile] rate = k[aryl halide][nucleophile]2
Answer:
Rate = k[aryl halide][nucleophile]
Explanation:
The simple aryl halides are almost inert to usual nucleophilic reagents but considerable activation on the ring can be produced by the addition of strongly electron-attracting substituents on either the ortho or para positions, or both. These groups deactivate the ring to allow the attack of the nucleophille on the ring.
Thus, these reactions can occur by following addition-elimination mechanism in which the nucleophille first attacks the aryl halide and then the elimination of the leaving group takes place.
Kinetic studies of this type of mechanism demonstrate that the reactions are of second-order kinetics– first order w.r.t. nucleophile and also, first-order w.r.t. aromatic substrate. The rate determining step (r.d.s.) is the formation of the addition intermediate.
Thus,
Rate = k[aryl halide][nucleophile]
horizontal block–spring system with the block on a frictionless surface has total mechanical energy E 5 47.0 J and a maximum displacement from equilibrium of 0.240 m. (a) What is the spring constant? (b) What is the kinetic energy of the system at the equilibrium point? (c) If the maximum speed of the block is 3.45 m/s, what is its mass? (d) What is the speed of the block when its displacement is 0.160 m? (e) Find the kinetic energy of the block at x 5 0.160 m. (f) Find the potential energy stored in the spring when x 5 0.160 m. (g) Suppose the same system is released from rest at x 5 0.240 m on a rough surface so that it loses
Answer:
Part a)
[tex]k = 1632 J[/tex]
Part b)
[tex]KE = 47 J[/tex]
Part c)
[tex]m = 7.9 kg[/tex]
Part d)
[tex]v = 2.57 m/s[/tex]
Part e)
[tex]KE = 26.1 J[/tex]
Part f)
[tex]PE = 20.9 J[/tex]
Explanation:
Total Mechanical energy is given as
[tex]E = 47.0 J[/tex]
Its maximum displacement from mean position is given as
[tex]A = 0.240 m[/tex]
Part a)
Now from the formula of energy we know that
[tex]E = \frac{1}{2}kA^2[/tex]
[tex]47 = \frac{1}{2}k(0.240)^2[/tex]
[tex]k = 1632 J[/tex]
Part b)
At the mean position of SHM whole mechanical energy will convert into kinetic energy
so it is given as
[tex]KE = 47 J[/tex]
Part c)
As per the formula of kinetic energy we know that
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]47 = \frac{1}{2}m(3.45^2)[/tex]
[tex]m = 7.9 kg[/tex]
Part d)
As we know by the equation of the speed of SHM is given as
[tex]v = \sqrt{\frac{k}{m}(A^2 - x^2)}[/tex]
[tex]v = \sqrt{\frac{1632}{7.9}(0.24^2 - 0.16^2)}[/tex]
[tex]v = 2.57 m/s[/tex]
Part e)
As we know by the formula of kinetic energy
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]KE = \frac{1}{2}(7.9)(2.57^2)[/tex]
[tex]KE = 26.1 J[/tex]
Part f)
As per energy conservation we know
KE + PE = Total energy
[tex]26.1 + PE = 47[/tex]
[tex]PE = 20.9 J[/tex]
(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]
(b) The KE at equilibrium point =47j
(c) The mass =7.9kg
(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]
(e)The KE of block 0.160m will be 26.1
(f) The PE will be 20.9j
What will be the asked values of the spring-mass system in the question?(a) for finding spring constant
KE=47 j
A=0.240m
By using the formula
[tex]E=\dfrac{1}{2} kx^2[/tex]
[tex]47=\dfrac{1}{2} k(0.240)^2[/tex]
[tex]k=1632\frac{N}{m^2}[/tex]
(b) At the mean position the whole mechanical energy will be equal to KE so
KE=47j
(c) The mass of the system
[tex]KE =\dfrac{1}{2} mv^2[/tex]
[tex]47=\dfrac{1}{2} m(3.45^2)[/tex]
[tex]m=7.9kg[/tex]
(d)Now the speed of the block
[tex]v=\sqrt{\dfrac{k}{m} (A^2-x^2)}[/tex]
[tex]v=\sqrt{\dfrac{1632}{7.9} (0.24^2-0.16^2)}[/tex]
[tex]v=2.57\frac{m}{s}[/tex]
(e) The KE of the block
[tex]KE=\dfrac{1}{2} mv^2=\dfrac{1}{2} 7.9(2.57)^2=26.1J[/tex]
(f) The PE of the system
[tex]Total Energy = KE+PE[/tex]
[tex]PE= 47-26.1 =20.9J[/tex]
Thus
(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]
(b) The KE at equilibrium point =47j
(c) The mass =7.9kg
(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]
(e)The KE of block 0.160m will be 26.1
(f) The PE will be 20.9j
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A metal ring 4.20 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.240 T/s . (A) What is the magnitude of the electric field induced in the ring? (B) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?
Answer:
09m
Explanation:
yes the current flow in clockwise direction
Explanation:
Given that,
Diameter of the metal ring, d = 4.2 cm
Radius, r = 2.1 cm
Initial magnetic field, B = 1.12 T
The magnetic field is decreasing at the rate of, [tex]\dfrac{dB}{dt}=0.24\ T/s[/tex]
Due to change in magnetic field, an emf is induced in it. And hence, electric field is induced. It is given by :
[tex]\int\limits {E.dl}=\dfrac{d}{dt}(\int\limits{B.dA)}[/tex]
[tex]E.(2\pi r)=\dfrac{dB}{dt}(\pi r^2)[/tex]
[tex]E=\dfrac{dB}{dt}\times \dfrac{r}{2}[/tex]
[tex]E=0.24\times \dfrac{2.1\times 10^{-2}}{2}[/tex]
[tex]E=0.00252\ N/C[/tex]
So, the magnitude of the electric field induced in the ring has a magnitude of 0.00252 N/C.
The direction of electric field will be counter clock wise direction as viewed by someone on the south pole of the magnet.
A wire placed on the plane of this screen carries a current toward the top of the screen. The wire feels a magnetic force toward the right. The direction of the magnetic field causing this force is which of the following? (a) outward out of the screen (b) inward into the screen (c) in the plane of the screen and toward the left edge (d) in the plane of the screen and toward the bottom edge.
Answer:
option (a)
Explanation:
The direction of magnetic field is given by Fleming's left hand rule.
If we spread the fore finger, middle finger and the thumb of our right hand such that they are mutually perpendicular to each other, then the direction of force is in the direction of thumb, direction of magnetic field is given by the direction of fore finger and the middle finger indicates the direction of current .
Here, current is upwards, magnetic force is rightwards, so the direction of magnetic field is given in outwards from the screen.
Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B carries a charge of -q. Sphere C carries no net charge. Spheres A and B are touched together and then separated. Sphere C is then touched to sphere A and separated from it. Lastly, sphere C is touched to sphere B and separated from it. (a) What is the ratio of the final charge on sphere C to q? What is the ratio of the final total charge on the three spheres to q (b) before they are allowed to touch each other and (c) after they have touched?
Answer:
Part a)
Final charge on C : q = 1.875
Part b)
Ratio for A = 6 : 1.25
Ratio for B = -1 : 1.875
Ratio for C = 0
Explanation:
When two identical metal sphere are connected together then the charge on them will get equally divided on both after connecting them by conducting wire
So here we have
[tex]q_A = + 6q[/tex]
[tex]q_B = -q[/tex]
[tex]q_c = 0[/tex]
Step 1: We connected A and B and then separate them
so we have
[tex]q_A' = q_B' = 2.5q[/tex]
Step 2: We connected A and C and then separate them
so we have
[tex]q_A'' = q_c' = 1.25q[/tex]
Step 3: We connected B and C and then separate them
so we have
[tex]q_c'' = q_b'' = 1.875q[/tex]
The final charge on sphere C is 2.5q. Before touching, the total charge is +5q, which remains the same after all interactions, demonstrating conservation of charge.
Explanation:When identical conductive spheres come into contact, the charges redistribute evenly across both spheres. If Sphere A is initially charged with +6q and Sphere B has a -q charge, touching them together allows their total charge to be shared, resulting in each sphere having (6q - q)/2 = 2.5q. After separation, both spheres A and B would have a charge of 2.5q. By touching Sphere C, which is uncharged, to A and then B in sequence, C gains a fraction of the charge from each, ending up with (2.5q)/2 from A and (2.5q)/2 from B, which totals 2.5q, since touching B does not change the charge obtained from A.
Before contact, the total charge is +5q (+6q from A and -q from B). After all the interactions, the total charge remains the same, +5q, but redistributed: A and B with 2.5q each and C with 2.5q.
In general, the total charge before and after remains constant, demonstrating conservation of charge.
A nonconducting sphere contains positive charge distributed uniformly throughout its volume. Which statements about the potential due to this sphere are true? All potentials are measured relative to infinity. (There may be more than one correct choice)
a) The potential is highest at the center of the sphere. b) The potential at the center of the sphere is the same as the potential at the surface. c) The potential at the surface is higher than the potential at the center. d) The potential at the center is the same as the potential at infinity. e) The potential at the center of the sphere is zero.
Answer:
a). The potential is highest at the center of the sphere
Explanation:
We k ow the potential of a non conducting charged sphre of radius R at a point r < R is given by
[tex]E=\left [ \frac{K.Q}{2R} \right ]\left [ 3-(\frac{r}{R})^{2} \right ][/tex]
Therefore at the center of the sphere where r = 0
[tex]E=\left [ \frac{K.Q}{2R} \right ]\left [ 3-0 \right ][/tex]
[tex]E=\left [ \frac{3K.Q}{2R} \right ][/tex]
Now at the surface of the sphere where r = R
[tex]E=\left [ \frac{K.Q}{2R} \right ]\left ( 3-1 \right )[/tex]
[tex]E=\left [ \frac{2K.Q}{2R} \right ][/tex]
[tex]E=\left [ \frac{K.Q}{R} \right ][/tex]
Now outside the sphere where r > R, the potential is
[tex]E=\left [ \frac{K.Q}{r} \right ][/tex]
This gives the same result as the previous one.
As [tex]r\rightarrow \infty , E\rightarrow 0[/tex]
Thus, the potential of the sphere is highest at the center.
A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of refraction of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its normal)?
The light leaves the glass at about 42° relative to its normal
[tex]\texttt{ }[/tex]
Further explanationLet's recall Snell's Law about Refraction as follows:
[tex]\boxed{n_1 \sin \theta_1 = n_2 \sin \theta_2}[/tex]
where:
n₁ = incident index
θ₁ = incident angle
n₂ = refracted index
θ₂ = refracted angle
[tex]\texttt{ }[/tex]
Given:
incident angle = θ₁ = 30°
index of refraction of air = n₃ = 1.0
index of refraction of glass = n₂ = 1.5
index of refraction of water = n₁ = 1.33
Asked:
refracted angle = θ₃ = ?
Solution:
Surface between Glass - Water:
[tex]n_1 \sin \theta_1 = n_2 \sin \theta_2[/tex]
[tex]\sin \theta_2 = \frac{n_1}{n_2} \sin \theta_1[/tex] → Equation A
[tex]\texttt{ }[/tex]
Surface between Air - Glass:
[tex]n_2 \sin \theta_2 = n_3 \sin \theta_3[/tex]
[tex]\sin \theta_3 = \frac{n_2}{n_3} \sin \theta_2[/tex]
[tex]\sin \theta_3 = \frac{n_2}{n_3} (\frac{n_1}{n_2} \sin \theta_1)[/tex] ← Equation A
[tex]\sin \theta_3 = \frac{n_1}{n_3} \sin \theta_1[/tex]
[tex]\sin \theta_3 = \frac{1.33}{1.0} \times \sin 30^o[/tex]
[tex]\sin \theta_3 = \frac{1.33}{1.0} \times \frac{1}{2}[/tex]
[tex]\sin \theta_3 \approx 0.665[/tex]
[tex]\boxed{\theta_3 \approx 42^o}[/tex]
[tex]\texttt{ }[/tex]
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Answer detailsGrade: High School
Subject: Physics
Chapter: Light
The angle at which the light leaves the glass is approximately 19.5°.
Explanation:When a light ray crosses from one medium to another, such as from water to glass, it undergoes refraction, which causes the ray to bend. The angle between the incident ray and the normal to the surface of the glass is called the angle of incidence. To find the angle of incidence, we can use Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the indices of refraction of the two media.
In this case, the index of refraction of water (n1) is 1.33 and the index of refraction of glass (n2) is 1.5.
The angle of incidence (θ1) is given as 30°. We can use the formula: n1*sin(θ1) = n2*sin(θ2) to solve for θ2, the angle at which the light leaves the glass.
Plugging in the values, we get: 1.33*sin(30°) = 1.5*sin(θ2).
Solving for θ2 gives us θ2 = sin^(-1)((1.33*sin(30°))/1.5).
Evaluating this expression, we find that θ2 is approximately 19.5°.
The small piston of a hydraulic lift has a diameter of 8.0 cm, and its large piston has a diameter of 40 cm. The lift raises a load of 15,000 N. Assume the pistons each have negligible weight. (a) Determine the force that must be applied to the small piston. (b) Determine the pressure applied to the fluid in the lift.
Answer:
a)
600 N
b)
1.2 x 10⁵ Pa
Explanation:
(a)
d₁ = diameter of small piston = 8 cm = 0.08 m
d₂ = diameter of large piston = 40 cm = 0.40 m
F₂ = force applied to large piston = 15000 N
F₁ = force applied to small piston = ?
Using pascal's law
[tex]\frac{F_{1}}{(0.25)\pi d_{1}^{2}} = \frac{F_{2}}{(0.25)\pi d_{2}^{2}}[/tex]
Inserting the values
[tex]\frac{F_{1}}{(0.08)^{2}} = \frac{15000}{(0.40)^{2}}[/tex]
F₁ = 600 N
b)
Pressure applied is given as
[tex]P = \frac{F_{1}}{(0.25)\pi d_{1}^{2}}[/tex]
[tex]P = \frac{(600)}{(0.25)(3.14) (0.08)^{2}}[/tex]
P = 1.2 x 10⁵ Pa
To determine the force needed on the small piston of a hydraulic lift, the ratios of the pistons' areas are used in combination with the load on the large piston, applying Pascal's Principle. The pressure in the hydraulic fluid is then found by dividing the force by the area of the piston where the force is applied.
Force applied to small piston = (Force on large piston) × (Area of small piston / Area of large piston)Pressure in the hydraulic fluid = Force / AreaGiven the diameters of the pistons, we can find the areas by using the formula for the area of a circle, A = πr², where r is the radius of the piston.
For the provided diameters of 8.0 cm and 40 cm, the areas of the pistons are:
Area of small piston = π × (4 cm)²Area of large piston = π × (20 cm)²With a load of 15,000 N on the large piston:
Force applied to small piston = 15,000 N × (Area of small piston / Area of large piston)Pressure = 15,000 N / (Area of large piston)Once we perform these calculations, we can determine the required force to apply to the small piston and the pressure applied to the fluid within the lift.
When 1.50 ✕ 10^5 J of heat transfer occurs into a meat pie initially at 20.0°C, its entropy increases by 465 J/K. What is its final temperature (in degrees)?
Answer:
The final temperature is 79.16°C.
Explanation:
Given that,
Heat [tex]Q=1.50\times10^{5}\ J[/tex]
Temperature = 20.0°C
Entropy = 465 J/k
We need to calculate the average temperature
Using relation between entropy and heat
[tex]\Delta S=\dfrac{\Delta Q}{T}[/tex]
[tex]T=\dfrac{\Delta Q}{\Delta S}[/tex]
Where, T = average temperature
[tex]\Delta Q[/tex]= transfer heat
[tex]\Delta S[/tex]= entropy
Put the value into the formula
[tex]T=\dfrac{1.50\times10^{5}}{465}[/tex]
[tex]T=322.58\ K[/tex]
We need to calculate the final temperature
Using formula of average temperature
[tex]T = \dfrac{T_{i}+T_{f}}{2}[/tex]
[tex]T_{f}=2T-T_{i}[/tex]....(I)
Put the value in the equation (I)
[tex]T_{f}=2\times322.58-293[/tex]
[tex]T_{f}=352.16\ K[/tex]
We convert the temperature K to degrees
[tex]T_{f}=352.16-273[/tex]
[tex]T_{f}=79.16^{\circ}\ C[/tex]
Hence, The final temperature is 79.16°C.
A 19 kg rock slides on a rough horizontal surface at 9.25 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.63. What average thermal power (in Watts) is produced as the rock stops?
Answer:
541.89 Watt
Explanation:
Consider the motion of rock on rough surface
μ = Coefficient of kinetic friction = 0.63
acceleration caused to kinetic friction is given as
a = - μg
a = - (0.63) (9.8)
a = - 6.2 m/s²
v₀ = initial velocity of the rock = 9.25 m/s
t = time taken to stop
v = final velocity = 0 m/s
Using the equation
v = v₀ + a t
0 = 9.25 + (- 6.2) t
t = 1.5 sec
m = mass of the rock = 19 kg
Energy lost due to friction is given as
E = (0.5) m (v₀² - v²)
E = (0.5) (19) (9.25² - 0²)
E = 812.84 J
Average thermal power is given as
[tex]P = \frac{E}{t}[/tex]
[tex]P = \frac{812.84}{1.5}[/tex]
P = 541.89 Watt
A book slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m. Calculate the coefficient of kinetic friction ????k between the book and the carpet. Assume the only forces acting on the book are friction, weight, and the normal force.
A book that slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m, has a coefficient of kinetic friction of 0.166.
A book slides across a level, carpeted floor with an initial speed (u) of 3.25 m/s. It comes to rest (final speed = v = 0m/s) after 3.25 m (s).
Since this is a uniformly decelerated rectilinear motion, we can calculate the acceleration (a) using the following kinematic expression.
[tex]v^{2} = u^{2} + 2as\\\\(0m/s)^{2} = (3.25m/s)^{2} + 2a(3.25m)\\\\a = -1.63 m/s^{2}[/tex]
The negative sign only explains that the friction opposes the movement. We can calculate the force exerted by the friction (F) using Newton's second law of motion.
[tex]F = m \times a[/tex] [1]
where,
m: mass of the bookWe can also calculate the force of friction using the following expression.
[tex]F = k \times N = k \times m \times g[/tex] [2]
where,
k: coefficient of kinetic frictionN: normal forceg: gravity (9.81 m/s²)Given [1] = [2], we get,
[tex]m \times a = k \times m \times g\\\\k = \frac{1.63 m/s^{2} }{9.81 m/s^{2}} = 0.166[/tex]
A book that slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m, has a coefficient of kinetic friction of 0.166.
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This question involves the concepts of the equations of motion, Newton's Second Law, and Frictional Force.
The coefficient of kinetic friction is found to be "0.15".
First, we will use the third equation of motion to find out the acceleration of the book:
[tex]2as = v_f^2-v_i^2\\[/tex]
where,
a = acceleration = ?
s = distance covered = 3.25 m
vf = final speed = 0 m/s
vi = initial speed = 3.25 m/s
Therefore,
[tex]2a(3.25\ m) = (0\ m/s)^2-(3.25\ m/s)^2\\\\a=\frac{-10.56\ m^2/s^2}{7\ m/s}[/tex]
a = - 1.51 m/s² (negative sign indicates deceleration)
Now the force can be calculated using Newton's Second Law of motion:
F = ma
This force is also equal to the frictional force:
F = μmg
comparing both forces, we get:
ma = μmg
a = μg
[tex]\mu = \frac{a}{g} = \frac{1.51\ m/s^2}{9.81\ m/s^2}[/tex]
μ = 0.15
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The attached picture shows the equations of motion.
A research-level Van de Graaff generator has a 2.15 m diameter metal sphere with a charge of 5.05 mC on it. What is the potential near its surface in MV? (Assume the potential is equal to zero far away from the surface.)
Answer:
42.3 MV
Explanation:
d = diameter of the metal sphere = 2.15 m
r = radius of the metal sphere
diameter of the metal sphere is given as
d = 2r
2.15 = 2 r
r = 1.075 m
Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C
Potential near the surface is given as
[tex]V = \frac{kQ}{r}[/tex]
[tex]V = \frac{(9\times 10^{9})(5.05\times 10^{-3})}{1.075}[/tex]
V = 4.23 x 10⁷ volts
V = 42.3 MV
When a comet comes close to the Sun, what happens to it that makes it bright and easier to see?
Answer:
Explanation:
Comet is made by dust particles, icy particles, gases etc.
A comet has a fixed time to complete a revolution around the sun.
As a comet comes nearer to the sun, due to the heat of the sun the vapour and the icy particles becomes gases and due to the radial pressure of energy od sun, we observe a tail of comet which has vapours mainly. SI the comet is visible easily.
A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0; (5.0 m, 5.0 m) at t = 2.0 s; and (14.0 m, 12.0 m) at t = 5.0 s. (a) Find the magnitude of the average velocity from t = 0 to t = 2 s. (b) Find the magnitude of the average velocity from t = 0 to t = 5 s.
Answer:
a) Magnitude of the average velocity from t = 0 to t = 2 s = 2.24 m/s
b) Magnitude of the average velocity from t = 0 to t = 5 s = 3.16 m/s
Explanation:
a) Velocity is rate of change of position.
A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0
A particle's position coordinates (x, y) are (5.0 m, 5.0 m) at t = 2.0 s
Displacement = (5-1)i + (5-3)j = 4i + 2j
Change in time = 2s
Velocity
[tex]v=\frac{4i+2j}{2}=2i+j[/tex]
Magnitude of velocity
[tex]v=\sqrt{2^2+1^2}=2.24m/s[/tex]
b) Velocity is rate of change of position.
A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0
A particle's position coordinates (x, y) are (14.0 m, 12.0 m) at t = 5.0 s
Displacement = (14-1)i + (12-3)j = 13i + 9j
Change in time = 5s
Velocity
[tex]v=\frac{13i+9j}{5}=2.6i+1.8j[/tex]
Magnitude of velocity
[tex]v=\sqrt{2.6^2+1.8^2}=3.16m/s[/tex]
A 15,000 N truck starts from rest and moves down a 15∘ hill with the engine providing a 8,000 N force in the direction of the motion. Assume the rolling friction force between the truck and the road is very small. If the hill is 50 m long, what will be the speed of the truck at the bottom of the hill?
Answer:
[tex]v_f = 27.9 m/s[/tex]
Explanation:
Component of the weight of the truck along the inclined plane is given as
[tex]F_1 = W sin\theta[/tex]
[tex]F_1 = 15000 sin15[/tex]
[tex]F_1 = 3882.3 N[/tex]
also the engine is providing the constant force to it as
[tex]F_2 = 8000 N[/tex]
now the net force along the the plane is given as
[tex]F_{net} = 8000 + 3882.3[/tex]
[tex]F = 11882.3 N[/tex]
mass of the truck is given as
[tex]m = \frac{w}{g} = 1529 kg[/tex]
now the acceleration is given as
[tex]a = \frac{F}{m}[/tex]
[tex]a = 7.77 m/s^2[/tex]
now the speed of the truck after travelling distance of d = 50 m is given as
[tex]v_f^2 = v_i^2 + 2 a d[/tex]
[tex]v_f^2 = 0 + 2(7.77)(50)[/tex]
[tex]v_f = 27.9 m/s[/tex]
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.503 X 104 T) at a distance of 15 cm from the wire, what is the maximum current the wire can carry?
Answer:
37.725 A
Explanation:
B = magnitude of the magnetic field produced by the electric wire = 0.503 x 10⁻⁴ T
r = distance from the wire where the magnetic field is noted = 15 cm = 0.15 m
i = magnitude of current flowing through the wire = ?
Magnetic field by a long wire is given as
[tex]B = \frac{\mu _{o}}{4\pi }\frac{2i}{r}[/tex]
Inserting the values
[tex]0.503\times 10^{-4} = (10^{-7})\frac{2i}{0.15}[/tex]
i = 37.725 A
Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lightweight material of density 600 kg/m3. (Assume that acceleration due to gravity is g = 9.8 m/s2. Round your answer to one decimal place.)
Answer:
Work done = 35467.278 J
Explanation:
Given:
Height of the cone = 4m
radius (r) of the cone = 1.2m
Density of the cone = 600kg/m³
Acceleration due to gravity, g = 9.8 m/s²
Now,
The total mass of the cone (m) = Density of the cone × volume of the cone
Volume of the cone = [tex]\frac{1}{3}\pi r^2 h[/tex]
thus,
volume of the cone = [tex]\frac{1}{3}\pi 1.2^2\times 4[/tex] = 6.03 m³
therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg
The center of mass for the cone lies at the [tex]\frac{1}{4}[/tex]times the total height
thus,
center of mass lies at, h' = [tex]\frac{1}{4}\times4=1m[/tex]
Now, the work gone (W) against gravity is given as:
W = mgh'
W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J
The work against gravity required to build a cone of height 4 m and base of radius 1.2 m out of a material of density 600 kg/m3 is 35,467.278 J.
Given to us
Height of the cone = 4m
The radius of the cone = 1.2 m
Density of the material = 600 kg/m³
We know that the work gone against the gravity is given as,
[tex]W = mgh'[/tex]
where W is the work, m is the mass, g is the acceleration due to gravity, and h' is the center of mass.
Also, the mass of an object is the product of its volume and density, therefore,
[tex]m = v \times \rho[/tex]
We know the value of the volume of a cone,
[tex]m = \dfrac{1}{3}\pi r^2 h \times \rho[/tex]
The center of mass of a cone lies at the center of its base at 1/4 of the total height from the base,
[tex]h' = \dfrac{h}{4}[/tex]
Substitute all the values we will get,
[tex]W = mgh'\\\\W = (v \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi r^2h \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi (1.2)^2 \times4 \times 600) \times 9.81 \times \dfrac{4}{4}\\\\W = 35467.278\rm\ J[/tex]
Hence, the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lightweight material of density 600 kg/m3 is 35,467.278 J.
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A curve ball is a type of pitch in which the baseball spins on its axis as it heads for home plate. If a curve ball is thrown at 34.5 m/s (77 mph ) with a spin rate of 26 rev/s , how many revolutions does it complete before reaching home plate? Assume that home plate is 18.3 m (60 ft) from the pitching mound and that the baseball travels at a constant velocity.
Answer:
13.79 revolutions
Explanation:
Velocity of ball = v = 34.5 m/s
Spin rate of ball = 26 revolutions/s
Distance between home plate and pitching mound = s = 18.3 m
Time taken by the ball to reach the home plate = t
[tex]t=\frac{18.3}{34.5}\ s[/tex]
Number of revolutions the ball completes is
[tex]26\times \frac{18.3}{34.5}=13.79[/tex]
∴Number of revolutions the ball completes before reaching home plate is 13.79
Answer:
The number of revolutions is 13.78
Explanation:
Given data:
v = speed = 34.5 m/s
d = distance = 18.3 m
spin rate = 26 rev/s
The time taken is equal to:
[tex]t=\frac{d}{v} =\frac{18.3}{34.5} =0.53s[/tex]
The number of revolutions is equal:
N = 0.53 * 26 = 13.78 rev
A baseball player wants to hit a home run over the wall of a stadium. He hits the ball 0.751 m above the ground with at an angle of 30 degrees above the horizontal and a speed of 37.5 m/s. What is the tallest wall the player can clear if the wall is 150 m away?
Answer:
The baseball does not reach the wall, because the ball falls at 125.33 meters and the wall is at 150 meters
Explanation:
V= 37.5 m/s
α= 30º
g= 9.8 m/s²
Vx= V * cos(30º) = 32.47 m/s
Vy= V * sin(30º) = 18.75 m/s
flytime of the baseball:
t= 2 * Vy/g
t= 3.86 sec
distance of baseball fall:
d= Vx * t
d= 125.33 m
Answer: 18.67 m wall
Explanation:this is projectile.
For Max height,
H = v²sin²©/2g
H= (37.5)²sin²(30)/(2*9.81)
H = 17.919 m
But he is standing 0.751 m above ground, so total height = 17.919+0.751= 18.67m
A pump is used to empty a 5200 L wading pool. The water exits the 3.0-cm-diameter hose at a speed of 2.2 m/s .How long will it take to empty the pool? Express your answer using two significant figures.
Answer:
55.79 minutes
Explanation:
Volume of water = 5200 L = 5.2 m^3
Diameter = 3 cm
radius, r = 1.5 cm = 0.015 m
v = 2.2 m/s
The volume of water coming out in 1 second = velocity x area
= 2.2 x 3.14 x 0.015 x 0.015
= 1.55 x 10^-3 m^3
1.55 x 10^-3 m^3 water flows = 1 second
5.2 m^3 water flows = 5.2 / (1.55 x 10^-3) = 3345.57 second
= 55.79 minutes
What is the tangential velocity of a 15 kg mass tied to a string and moving in a circle of radious 10 m, if the period of the orbit is 0.2 seconds?
Answer:
Tangential speed, v = 314.15 m/s
Explanation:
It is given that,
Mass of the object, m = 15 kg
It is moving in a circle of radius, r = 10 m
Time period, t = 0.2 seconds
We need to find the tangential velocity of the object. It is given by :
[tex]v=\dfrac{2\pi r}{t}[/tex]
Where
v = tangential speed
[tex]v=\dfrac{2\pi \times 10\ m}{0.2\ s}[/tex]
v = 314.15 m/s
So, the tangential speed of the object is 314.15 m/s. Hence, this is the required solution.
A plank 2.65 m long is supported by a cable, and a small ledge. The cable attaches 79.7 cm from the ledge, and makes an angle of 40.3 with the plank. It can hold 2648 N. How close (in meters) to the end of the plank can a 76.4 kg person walk before the cable breaks? Ignore the mass of the plank itself.
Answer:
0.51 m
Explanation:
Given :
Plank length = 2.65 m
Distance from the ledge from which the cable is attached = 79.7 cm = 0.797 m
Angle = 40.3 degree
Mass of the man, m = 76.4 kg
Now the net torque acting on the plank is zero.
Therefore, maximum torque applied by cable is
= 2648 x 0.797 x cos (40.3)
= 1609.5 N-m
We know that the man should apply the same torque,
Therefore, x [tex]\times[/tex]76.4 [tex]\times[/tex] g = 1609.5
x =2.14 m
Therefore, length to the end of the plank is = 2.65-2.14
=0.51 m