Answer:
Explanation:
a ) If x be the position of n the bright fringe on the screen , following formula holds .
x = n (λD / d) ; λ is wave length , D is screen distance and d is slit separation .
If we increase the value of λ or wave length, x will increase so central fringe along with all the fringes will shift away from the centre .
If we increase the value of D or screen distance , it will also increase x , so fringes along with central fringe will shift away from the center.
b ) Fringes ,whether bright or dark , both shift together , either towards or away from the center .
So to move the dark spots towards the center , we need to do the opposite to what we did in the first case , ie decrease the wavelength or decrease the screen distance .
Which of the following do not make their own energy through nuclear fusion? Select all that apply.
A. giant star
B. protostar
C. dwarf star
D. black hole
E. neutron star
F. main sequence star
Tamsen and Vera imagine visiting another planet, planet X, whose gravitational acceleration, gX, is different from that of Earth's. They envision a pendulum, whose period on Earth is 2.243 s, that is set in motion on planet X, and the period is measured to be 2.000 s. What is the ratio of gX/gEarth? Neglect any effects caused by air resistance.
Answer:
1.27
Explanation:
Period of a pendulum T is
T = 2¶(l/g)^0.5
Where g is acceleration due to gravity
l is lenght of pendulum
For earth, T = 2.243 s
2.243 = 2 x 3.142 x (l/9.81)^0.5
0.36 = (l^0.5)/3.13
1.13 = l^0.5
l = 1.28 m
For planet X of the same lenght
Period T = 2.00 s
2 = 2 x 3.142 (1.28/gx)^0.5
0.32 = 1.13 / gx^0.5
3.53 = gx^0.5
gx^0.5 = 3.53
gx = 12.46 m/s^2
gx/gearth = 12.46/9.81 = 1.27
The ratio of the gravitational acceleration on planet X to that on Earth is 1.26, calculated using the periods of a pendulum on both planets and the relationship between period and gravitational acceleration.
Explanation:The period of a simple pendulum is given by the formula T = 2*pi*sqrt(L/g) where T is the period of the pendulum, L is the length of the pendulum, and g is the acceleration due to gravity. To find the ratio gX/gEarth, we can use the formula T_earth/T_x = sqrt(g_x/g_earth) (let's denote T_earth as the period on earth and T_x as the period on planet X).
Since we know the periods, we can substitute these values into the formula: 2.243s/2.000s = sqrt(g_x/g_earth), which gives us 1.1215 = sqrt(g_x/g_earth). To find the ratio g_x/g_earth, square both sides, giving (1.1215)² = g_x/g_earth. Thus, g_x/g_earth = 1.26.
Learn more about Gravitational Acceleration here:https://brainly.com/question/30429868
#SPJ11
A 94.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 44.0 m/s. If both are initially at rest and if the ice is frictionless,
how far does the player recoil in the time it takes the puck to reach the goal 16.5 m away?
Answer:
0.026 meter
Explanation:
using distance time equation to determine the time for the puck to move 16.5 meters.
distance 'd' = velocity'v' x time't'
16.5 = 44 x t
t =0.375 second
Here momentum is conserved. Since both objects are initially at rest, the initial momentum is 0.
Next is to determine the puck’s momentum.
Momemtum 'p' = m x v => 0.15 x 44 = 6.6kg⋅m/s
The player momentum is -6.6kg⋅m/s .
In order to determine the player’s velocity, we'll use p=mv
-6.6 = 94v
v= -0.0702 m/s
The above negative sign represents that the player is moving in the opposite direction of the puck.
Lastly, how far does the player recoil in the time it takes the puck to reach the goal 16.5 m away?
d = v x t = 0.0702 x 0.375= 0.026 meter
Final answer:
The recoil velocity of the ice hockey player is 0.07021 m/s, and the time it takes for the puck to reach the goal is 0.375 s. Therefore, the player recoils a distance of 2.633 cm in the time the puck takes to reach the goal 16.5 m away.
Explanation:
When a hockey player hits a puck on frictionless ice, this scenario is an excellent example of the conservation of momentum where the total momentum before and after the event must be equal. Since the player and the puck are initially at rest, their combined momentum is zero. Thus, when the player imparts a velocity to the puck, the player must recoil with a momentum equal in magnitude and opposite in direction to preserve the momentum balance.
To find the recoil speed of the player, we set the momentum of the puck equal to the momentum of the recoiling player:
Momentum of puck = mass of puck \\( imes\\) velocity of puckMomentum of player = mass of player \\( imes\\) recoil velocity of playerSo, \\(0.150 kg\\) \\( imes\\) 44.0 m/s = \\(94.0 kg\\) \\( imes\\) recoil velocity of player
Recoil velocity of player = \\((0.150 kg \\( imes\\) 44.0 m/s) / 94.0 kg)
Recoil velocity of player = 0.07021 m/s
To find the distance the player recoils, you first calculate the time it takes for the puck to reach the goal. Since velocity = distance / time, we rearrange to find time = distance / velocity.
Time for puck to reach the goal = Distance to goal / Speed of puck
Time for puck to reach the goal = 16.5 m / 44.0 m/s = 0.375 s
Then, we find the distance the player recoils by using the recoil velocity we found earlier:
Distance player recoils = Recoil velocity of player \\( imes\\) Time for puck to reach the goal
Distance player recoils = 0.07021 m/s \\( imes\\) 0.375 s
Distance player recoils = 0.02633 m, or 2.633 cm
Which part of the electromagnetic Spectrum is nearest to X-rays?
microwaves
infrared light
gamma rays
radio waves
Answer:gamma rays
Explanation:
Out of all these spectrum listed in The question , gamma rays is closest to x-ray in The electromagnetic spectrum
A refrigerator removes heat from the freezing compartment at the rate of20 kJ per cycle and ejects 24 kJ into the room each cycle. How much energy is used in each cycle?
Answer:
Energy used = 4KJ
Explanation:
Second law of thermodynamics states that as energy is transferred or transformed, more and more of it is wasted. The Second Law also states that there is a natural tendency of any isolated system to degenerate into a more disordered state.
Now when we apply that to heat engines, we'll see that;
Heat expelled = Heat removed + Work Done
We can write it as;
Q_h = Q_c + W
We are given that;
Heat removed; Q_c = 20KJ
Heat expelled into the room in each cycle; Q_h = 24KJ
Thus; plugging these 2 values into the equation, we obtain;
24 = 20 + W
W = 24 - 20
W = 4 KJ
Work done is energy used.
Thus, energy used = 4 KJ
Final answer:
The refrigerator uses 4 kJ of energy each cycle, calculated by the work done W, which is the difference between the heat ejected Qh (24 kJ) and the heat removed Qc (20 kJ).
Explanation:
The question is calculating the energy used by a refrigerator in each cycle. The refrigerator absorbs heat Qc from the inside and ejects a larger amount of heat Qh to the room. The extra energy being ejected comes from the work W done by the refrigerator, which is the energy used by it in each cycle.
If the refrigerator removes 20 kJ (Qc) from the freezing compartment and ejects 24 kJ (Qh) into the room each cycle, the work done (energy used) W can be found using the first law of thermodynamics which states that the conservation of energy principle - the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
W = Qh - Qc
Substituting in the given values:
W = 24 kJ - 20 kJ
W = 4 kJ
Thus, the work done (energy used) by the refrigerator in each cycle is 4 kJ.
As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.15 mm apart and position your screen 3.53 m from the slits. Although Young had to struggle to achieve a monochromatic light beam of sufficient intensity, you simply turn on a laser with a wavelength of 639 nm . How far on the screen are the first bright fringe and the second dark fringe from the central bright fringe
Answer:[tex]y_1=1.9\ mm[/tex]
Explanation:
Given
slit width [tex]d=1.15\ mm[/tex]
Distance of screen [tex]D=3.53\ m[/tex]
wavelength [tex]\lambda =639\ nm[/tex]
Position of any bright fringe is given by
[tex]y_n=\dfrac{n\lambda D}{d}[/tex]
[tex]y_1=\frac{1\times 639\times 10^{-9}\times 3.53}{1.15\times 10^{-3}}[/tex]
[tex]y_1=0.0019\ m[/tex]
[tex]y_1=1.9\ mm[/tex]
Position of dark fringe is given by
[tex]y_D=\dfrac{(2n+1)\lambda D}{2d}[/tex]
for second dark fringe [tex]n=1[/tex]
[tex]y_D=\dfrac{1.5\times 639\times 10^{-9}\times 3.53}{1.15\times 10^{-3}}[/tex]
[tex]y_D=0.00294\ m\approx 2.94\ mm[/tex]
PART ONE
A steel railroad track has a length of 28 m
when the temperature is 2◦C.
What is the increase in the length of the
rail on a hot day when the temperature is
35 ◦C? The linear expansion coefficient of
steel is 11 × 10^−6(◦C)^−1
.
Answer in units of m
PART TWO
Suppose the ends of the rail are rigidly
clamped at 2◦C to prevent expansion.
Calculate the thermal stress in the rail if
its temperature is raised to 35 ◦C. Young’s
modulus for steel is 20 × 10^10 N/m^2
Answer in units of N/m^2
Answer:[tex]\Delta L=0.0101\ m[/tex]
Explanation:
Given
Length of track [tex]L_o=28\ m[/tex] when
[tex]T_o=2^{\circ}C[/tex]
Coefficient of linear expansion [tex]\alpha =11\times 10^{-6}\ ^{\circ}C^{-1}[/tex]
When Temperature rises to [tex]T=35^{\circ}C[/tex]
[tex]\Delta T=35-2=33^{\circ}C[/tex]
and we know length expand on increasing temperature
[tex]L=L_o[1+\alpha \Delta T][/tex]
[tex]L-L_o=L_o\alpha \Delta T[/tex]
[tex]\Delta L=28\times 11\times 10^{-6}\times (33)[/tex]
[tex]\Delta L=0.0101=10.164\ mm[/tex]
(b)When rails are clamped thermal stress induced
we know [tex]E=\frac{stress}{strain}[/tex]
[tex]Stress=E\times strain[/tex]
[tex]Stress=20\times 10^{10}\times \frac{\Delta L}{L_o}[/tex]
[tex]Stress=20\times 10^{10}\times \frac{0.0101}{28}[/tex]
[tex]Stress=72.14\ MPa[/tex]
[tex]Stress=72.14\times 10^{6}\ N/m^2[/tex]
A 0.468 g sample of pentane, C5H12, was burned in a bomb calorimeter. The temperature of the calorimeter and 1.00 kg of water in it rose from 20.45 °C to 23.65 °C. The heat capacity of the calorimeter by itself is 2.21 kJ/°C and the specific heat capacity of water is 4.184 J/g.°C What is the heat of combustion per mole of pentane?
Answer:
E = 3,154.37 KJ
The heat of combustion per mole of pentane is 3,154.37 KJ
Explanation:
Given;
Change in temperature of system ∆T = 23.65-20.45 = 3.2 °C
Mass of water m1 = 1 kg
Specific heat capacity of water C1 = 4.184J/g°C = 4184J/kg °C
Heat capacity of calorimeter mC2 = 2.21 kJ/°C = 2210J/°C
Heat gained by both calorimeter and water is;
H = (m1C1 + mC2)∆T
Substituting the values;
H = (1×4184 + 2210)×3.2
H = 20460.8 J
Mass of pentane burned = 0.468 g
Molecular mass of pentane = 72.15g
If 0.468g of pentane releases 20460.8 J of heat,
72.15g of pentane will release;
E = (72.15/0.468) × 20460.8 J
E = 3154373.333333J
E = 3,154.37 KJ
The heat of combustion per mole of pentane is 3,154.37 KJ
This question checks that you can use the formula of the electric field due to a long, thin wire with charge on it. The field due to an infinitely long, thin wire with linear charge density LaTeX: \lambda λ is LaTeX: \vec{E} = \frac{1}{4\pi\epsilon_0}\frac{2\lambda}{r}\hat{r}E → = 1 4 π ϵ 0 2 λ r r ^. Imagine a long, thin wire with a constant charge per unit length of -6.6LaTeX: \times10^{-7}× 10 − 7 C/m. What is the magnitude of the electric field at a point 10 cm from the wire (assuming that the point is much closer to the wire's nearest point than to either of its ends)? Give your answer in units of kN/C.
Answer:
E = -118 KN / C
Explanation:
In this exercise we are given the expression of the electric field for a wire
E = 1 /4πε₀ 2λ / r
they also indicate the linear density of charge
λ = 6,6 10⁻⁷ C / m
ask to calculate the electric field at a position at r = 10 cm from the wire
k = 1 /4πε₀ = 8,988 10⁹ N m² / C²
E = 8,988 10⁹ 2 (-6.6 10⁻⁷ / 0.10)
E = -1.18 105 N / C
we reduce to KN / C
1 KN / C = 10³ N / C
E = -1.18 10² KN / C
E = -118 KN / C
the negative sign indicates that the field is directed to the charged
At the surface of Venus the average temperature is a balmy 460∘C due to the greenhouse effect (global warming!), the pressure is 92 Earth-atmospheres, and the acceleration due to gravity is 0.894 gEarthgEarth. The atmosphere is nearly all CO2 (molar mass 44.0g/m) and the temperature remains remarkably constant. We shall assume that the temperature does not change at all with altitude.
a. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Earth-atmospheres.
b. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Venus-atmospheres.
c. What is the root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km?
Answer:
a) 86 atm
b) 86 atm
c) 645 m/s
Explanation:
See attachment for calculations on how i arrived at the answer
The atmospheric pressure of 2.00 km above the surface of Venus is 86 atm and the root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km is 645 m/sec.
Given :
At the surface of Venus, the average temperature is a balmy 460∘C due to the greenhouse effect (global warming!), the pressure is 92 Earth-atmospheres, and the acceleration due to gravity is 0.894 g.The atmosphere is nearly all CO2 (molar mass 44.0g/m) and the temperature remains remarkably constant.a) and b) In order to determine the atmospheric pressure of 2.00 km above the surface of Venus using the formula given below:
[tex]\rm P = P_0\times L^{\frac{Mg}{\rho T}}[/tex] --- (1)
The value of the expression [tex]\rm Mg/\rho T[/tex] is given below:
[tex]\rm \dfrac{Mg}{\rho T} = \dfrac{44\times 10^{-3}\times 9.8\times 10^3}{8.314\times 733}[/tex]
[tex]\rm \dfrac{Mg}{\rho T}=0.02076[/tex]
Now, substitute the values of the known terms in the expression (1).
[tex]\rm P=92\times L^{0.02076}[/tex]
P = 86 atm
c) The root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km can be calculated as given below:
[tex]\rm V_{rms}=\sqrt{\dfrac{3RT}{M}}[/tex]
Now, substitute the values of the known terms in the above formula.
[tex]\rm V_{rms}=\sqrt{\dfrac{3\times 8.314\times 733}{44\times 10^{-3}}}[/tex]
Simplify the above expression.
[tex]\rm V_{rms} = 645\;m/sec[/tex]
For more information, refer to the link given below:
https://brainly.com/question/7213287
Certain neutron stars (extremely dense stars) are believed to be rotating at about 0.83 rev/s. If such a star has a radius of 40 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation? Number Enter your answer in accordance to the question statement Units Choose the answer from the menu in accordance to the question statement
Answer:
The mass of the star would be [tex]M = 2.644*10^{24} \ kg[/tex]
Explanation:
From the question we are told that
The angular speed is [tex]w = 0.83\ rev/s = 0.83 * 2 \pi = 1.66 rad/s[/tex]
The radius of the star is [tex]r = 40km = 40 *1000 = 40 * 10^{3} m[/tex]
Generally the minimum mass of the start is mathematically evaluated as
[tex]M = \frac{r^3 w^2}{G}[/tex]
Where is the gravitational constant with a values of [tex]G = 6.67*10^{-11} N \cdot m^2 /kg[/tex]
[tex]M = \frac{(40*10^3)^3 * 1.66^2}{6.67*10^{-11}}[/tex]
[tex]M = 2.644*10^{24} \ kg[/tex]
9) A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk (M = 0.10 kg, R = 0.10 m) rotating in a horizontal plane around the vertical axis through its center. The disk is rotating at 10.0 rad/s. The bug crawls to the center of the disk. (a) What is the new angular velocity of the disk? (b) What is the change in the kinetic energy of the system? (c) What is the cause of the increase and decrease of kinetic energy?
The bug moving toward the center of the disk changes the moment of inertia and the angular velocity, resulting in a change in the rotational kinetic energy of the system. The principles of conservation of angular momentum and calculations of kinetic energy apply here.
Explanation:This problem involves the principle of conservation of angular momentum. Initially, when the bug is at the edge of the disk, the system's angular momentum (L) is conserved. The initial angular momentum (L_initial) is the sum of the angular momenta of the bug and the disk. It can be calculated as L_initial = I_bug(ω) + I_disk(ω) where I_bug = m*R², I_disk = (1/2)*M*R², m is the mass of the bug, M is the mass of the disk, R is the radius of the disk, and ω is the initial angular velocity.
When the bug moves to the center of the disk, the angular velocity will change but the total angular momentum will still remain conserved as L_final = I_bug(ω') + I_disk(ω') where ω' is the final angular velocity and I_bug is now 0 because the bug is at the center. From the conservation of angular momentum, we can solve for ω'.
The change in kinetic energy can be calculated from the difference between the initial kinetic energy and the final kinetic energy of the system. An increase or decrease in kinetic energy in this scenario is due to the displacement of the bug from the edge to the center of the disk, which changes the moment of inertia of the system and thus, the rotational kinetic energy.
Learn more about Conservation of Angular Momentum here:https://brainly.com/question/1597483
#SPJ11
Final answer:
The explanation covers the new angular velocity of the disk, the change in kinetic energy, and the causes for the increase and decrease in kinetic energy.
Explanation:
A: The new angular velocity of the disk can be calculated using the principle of conservation of angular momentum. The bug crawling towards the center decreases the moment of inertia, resulting in an increase in angular velocity.
B: The change in kinetic energy of the system can be determined by calculating the initial and final kinetic energies and finding their difference.
C: The increase in kinetic energy is due to the bug moving towards the center, reducing the moment of inertia. The decrease in kinetic energy occurs due to the redistribution of mass towards the center of rotation.
You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength of the laser by doing a double slit experiment. Shining the laser through a double slit with a slit separation of 0.329 mm on the wall 2.20 m away the first bright fringe is 2.90 mm from the center of the pattern. What is the wavelength
Explanation:
We have,
Slit separation, d = 0.329 mm
Distance between slit and screen, D = 2.20 m
The first bright fringe is formed at a distance off 2.9 mm from the center of the pattern. We need to find the wavelength.
For double slit experiment, the fringe width is given by :
[tex]\beta=\dfrac{D\lambda}{d}[/tex]
[tex]\lambda[/tex] is wavelength
[tex]\lambda=\dfrac{\beta d}{D}\\\\\lambda=\dfrac{2.9\times 10^{-3}\times 0.329\times 10^{-3}}{2.2}\\\\\lambda=4.33\times 10^{-7}\ m\\\\\lambda=433\ nm[/tex]
So, wavelength is 433 nm.
A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]. Being more practical-minded, you measure the rope to have a length of 1.35 mm and a mass of 3.38 grams. Assume that the ends of the rope are held fixed and that there is both this traveling wave and the reflected wave traveling in the opposite direction.
What is the wave function y(x,t) for the standing wave that is produced?
Answer:
Explanation:
y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]
The angular velocity ω = 742 rad /s
wave function k = 6.98 rad /m
Amplitude A = 2.3 mm
y(x,t) = A cos( ωt + kx )
equation of wave reflected wave
y(x,t) = A cos( ωt - kx )
resultant standing wave
= y₁ +y₂
= A cos( ωt + kx ) +A cos( ωt - kx )
y(x,t) = 2 A cosωt cos kx
= 2 x 2.3 cos 742t .cos6.98x
= 4.6 mm . cos 742t .cos6.98x
Determine the magnetic flux density at a point on the axis of a solenoid with radius b and length L, and with a current I in its N turns of closely wound coil. Show that the result reduces to that given in equation 6-14 when L approaches infinity.
Answer:
See explaination
Explanation:
Magnetic flux density definition, a vector quantity used as a measure of a magnetic field.
It can be defined as the amount of magnetic flux in an area taken perpendicular to the magnetic flux's direction.
See attached file for detailed solution of the given problem.
The magnetic flux density inside a solenoid can be derived using the formula B = μ₀(NI)/L. As L approaches infinity, this simplifies to B = μ₀nI, aligning with the standard magnetic field expression for a long solenoid.
To find the magnetic flux density at a point on the axis of a solenoid with radius b, length L, a current I, and N turns of closely wound coil, use the following steps:
Step 1 : Consider an infinitesimal element dz of the solenoid. The element is located at a distance z from the center of the solenoid. The magnetic field contribution ( dB ) from this element at point ( P ) is given by the Biot-Savart law as -
[tex]\[ dB = \frac{\mu_0 I d\ell}{4\pi} \frac{1}{r^2} \][/tex]
Step 2 : For a solenoid, considering a differential element, the distance r from the element to the point P on the axis is -
[tex]\( \sqrt{b^2 + (z - x)^2} \).[/tex]
Step 3 : For a small segment dz, the current element is -
[tex]\[ dB = \frac{\mu_0 (n I dz)}{4\pi} \frac{1}{(b^2 + (z - x)^2)^{3/2}} \][/tex]
Step 4 : Integrate dB from -L/2 to L/2 :
[tex]\[ B = \int_{-L/2}^{L/2} \frac{\mu_0 (n I dz)}{4\pi} \frac{1}{(b^2 + (z - x)^2)^{3/2}} \][/tex]
Substitute, [tex]\( n = \frac{N}{L}[/tex]
[tex]\[ B = \frac{\mu_0 I N}{4\pi L} \int_{-L/2}^{L/2} \frac{dz}{(b^2 + (z - x)^2)^{3/2}} \][/tex]
Solving the integral part -
[tex]\[ \int_{-L/2}^{L/2} \frac{dz}{(b^2 + (z - x)^2)^{3/2}} = \frac{z}{b^2 \sqrt{b^2 + (z - x)^2}} \Bigg|_{-L/2}^{L/2} \][/tex]
[tex]\[ \left[ \frac{L/2}{b^2 \sqrt{b^2 + (L/2 - x)^2}} - \frac{-L/2}{b^2 \sqrt{b^2 + (-L/2 - x)^2}} \right] \][/tex]
Step 5: When L is very large, the ends of the solenoid are far away, and we can approximate:
[tex]\[ \frac{L/2 - x}{b^2 \sqrt{b^2 + (L/2 - x)^2}} \approx \frac{1}{b^2} \quad \text{and} \quad \frac{-L/2 - x}{b^2 \sqrt{b^2 + (-L/2 - x)^2}} \approx -\frac{1}{b^2} \][/tex]
This simplifies the result to -
[tex]\[ \left[ \frac{L/2 - x}{b^2 (L/2 - x)} - \frac{-L/2 - x}{b^2 (L/2 + x)} \right] = \left[ \frac{1}{b^2} - \left(-\frac{1}{b^2}\right) \right] = \frac{2}{b^2} \][/tex]
Conclusion, For a very long solenoid:
[tex]\[B = \mu_0 \frac{N}{L} I\][/tex]
Which simplifies to,
[tex]\[B = \mu_0 n I\][/tex]
Thus, for a very long soleniod the magnetic flux density can be approximated to [tex]\[B = \mu_0 n I\][/tex].
A string is 1.6 m long. One side of the string is attached to a force sensor and the other side is attached to a ball with a mass of 200 g. The ball is lifted to a height of 1.5 m above the ground and then released from rest. The ball swings to its lowest point where the string breaks. The ball is then in free-fall until it hits the ground. How far would the ball travel in the horizontal direction between points B and C (i.e. what is the range)?
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The distance traveled in horizontal direction is [tex]D = 1.38 m[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]L = 1.6 \ m[/tex]
The mass of the ball is [tex]m = 200 g = \frac{200}{1000} = 0.2 \ kg[/tex]
The height of ball is [tex]h = 1.5 \ m[/tex]
Generally the work energy theorem can be mathematically represented as
[tex]PE = KE[/tex]
Where PE is the loss in potential energy which is mathematically represented as
[tex]PE =mgh[/tex]
Where h is the difference height of ball at A and at B which is mathematically represented as
[tex]h = y_A - y_B[/tex]
So [tex]PE =mg(y_A - y_B)[/tex]
While KE is the gain in kinetic energy which is mathematically represented as
[tex]KE = \frac{1}{2 } (v_b ^2 - 0)[/tex]
Where [tex]v_b[/tex] is the velocity of the of the ball
Therefore we have from above that
[tex]PE =KE \equiv mg (y_A - y_B) = \frac{1}{2} m (v_b ^2 - 0)[/tex]
Making [tex]v_b[/tex] the subject we have
[tex]v_b = \sqrt{2g (y_A - y_B)}[/tex]
substituting values
[tex]v_b = \sqrt{2g (1.5 - 0.40)}[/tex]
[tex]v_b = 4.6 \ m/s[/tex]
Considering velocity of the ball when it hits the floor in terms of its vertical and horizontal component we have
[tex]v_x = 4.6 m/s \ while \ v_y = 0 m/s[/tex]
The time taken to travel vertically from the point the ball broke loose can be obtained using the equation of motion
[tex]s = v_y t - \frac{1}{2} g t^2[/tex]
Where s is distance traveled vertically which given in the diagram as [tex]s = -0.4[/tex]
The negative sign is because it is moving downward
Substituting values
[tex]-0.4 = 0 -\frac{1}{2} * 9.8 * t^2[/tex]
solving for t we have
[tex]t = 0.3 \ sec[/tex]
Now the distance traveled on the horizontal is mathematically evaluated as
[tex]D = v_b * t[/tex]
[tex]D = 4.6 * 0.3[/tex]
[tex]D = 1.38 m[/tex]
Answer:
bjknbjk;jln
Explanation:
Polarizers #1 and #3 are "crossed" such that their transmission axes are perpendicular to each other. Polarizer #2 is placed between the polarizers #1 and #3 with its transmission axis at 60◦ with respect to the transmission axis of the polarizer #1 (see the sketch). #1 #2 #3 60◦ After passing through polarizer #2 the intensity I2 (in terms of the intermediate intensity I1) is
The intensity I2 after light passes through polarizer #2, oriented at 60° to polarizer #1, is 25% of the intermediate intensity I1.
To determine the intensity [tex]I_2[/tex] after light passes through the second polarizer, we use Malus's Law. Malus's Law states that when polarized light passes through a polarizing filter, the intensity of the transmitted light is given by:
[tex]I = I_0 * cos^2(\theta),[/tex]
where I0 is the initial intensity and θ is the angle between the light’s polarization direction and the axis of the polarizer.
In this case:
The light initially passes through polarizer #1, aligning its polarization with polarizer #1.When the light reaches polarizer #2 set at 60° to polarizer #1, the intensity I2 can be found using Malus's Law:[tex]I_2 = I_1 * cos^2(60^o)[/tex]Since [tex]cos(60^o) = 0.5:[/tex]
[tex]I_2 = I_1 * (0.5)^2= I_1 * 0.25[/tex]
Hence, the intensity [tex]I_2[/tex] is 25% of [tex]I_1[/tex].
There are two possibilities for final stage of extremely massive stars. The first is a
neutron star and the second is a
Answer: Black hole.
Explanation:
As the massive star "compacts" under its own gravity, it triggers a massive supernova, after this point the remains of the star can become a neutron star, which is a very compact star made primarily, as the name says, of neutrons. The other possibility is a black hole, which is a finite region of space wherein it's interior there is a big concentration of mass, which creates a gravitational field strong enough that there is no particle that can escape it.
a cross section of three parallel wires each carrying a current of 20 A. The current in wire B is out of the paper, while that in wires A and C are into the paper. If the distance R = 15.0 mm,
what is the magnitude of the force on a 200 cm length of wire C?
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The magnitude of force is [tex]F_R} = 5.33 \ m N[/tex]
Explanation:
From the question we are told that
The current in wire A , B and C are [tex]I _a = I_b =I_c = I= 20 A[/tex]
The distance is [tex]R = 15.0mm = \frac{15}{1000} = 0.015m[/tex]
The length of wire C is [tex]L_c = 200cm = \frac{200}{100} = 2 m[/tex]
Generally the force exerted per unit length that is acting in between two current carrying conductors can be mathematically represented as
[tex]\frac{F}{L} = \frac{\mu_o }{2 \pi} \cdot \frac{I_1 I_2 }{R}[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a constant value of
[tex]\mu_o = 4 \pi * 10^{-7} N/A^2[/tex]
When the current in the wire are of the same direction the force is positive and when they are in opposite direction the force is negative
Considering force between A and C
[tex]F_{A -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_a I_c }{2R} * L[/tex]
Considering force between B and C
[tex]F_{B -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_b I_c }{R} * L[/tex]
The resultant force is
[tex]F_R = F_{B -C} - F_{A -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_a I_c }{2R} * L - \frac{\mu_o }{2 \pi} \cdot \frac{I_b I_c }{R} * L[/tex]
[tex]F_R} = \frac{\mu_o }{2 \pi} \cdot \frac{I * I }{R} * L * ({1 - \frac{1}{2} })[/tex]
Substituting values
[tex]F_R} = \frac{4 \pi * 10^{-7}}{2 * 3.142} \cdot \frac{ 20*20 }{0.015} * 2 ({\frac{1}{2} })[/tex]
[tex]F_R} = 5.33 *10^{-3} N[/tex]
[tex]F_R} = 5.33 \ m N[/tex]
What is heat?
A. A measure of the movement of molecules inside an object
B. The transfer of energy from a hot object to a cold object
C. The force exerted on an area by an object
D. A measure of mass per unit volume of an object
Answer:
B. The transfer of energy from a hot object to a cold object
Explanation:
Answer:
The answer is option B
Explanation:
What should Jaime do to increase the number of energy storage molecules that producers can make? You may choose more than one answer.
Decrease the amount of carbon dioxide in the ecosystem. Increase the amount of carbon dioxide in the ecosystem. Decrease the amount of sunlight in the ecosystem. Increase the amount of sunlight in the ecosystem.
Answer: Increase the amount of carbon dioxide in the ecosystem.
Explanation:
The producers are the organisms which can make their own food by conducting the process of photosynthesis. The carbon dioxide, water are the reactants and carbohydrates and oxygen are the products of photosynthesis. The entire process takes place in the presence of sunlight energy.
The carbon dioxide is the chief component for this reaction and the components of carbon are produced in the form of carbohydrates. The carbohydrates are the source of energy that are utilized by the producers for cellular metabolism and other functions. These get stored in the form of storage molecules in producers. Thus to increase the no. of storage molecules that producers can make Jaime need to increase the concentration of carbon dioxide in the ecosystem.
Assume that a certain location on the Earth reflects 33.0% of the incident sunlight from its clouds and surface. (a) Given that the intensity of solar radiation at the top of the atmosphere is 1368 W/m2, find the radiation pressure on the Earth, in pascals, at the location where the Sun is straight overhead. Pa (b) State how this quantity compares with normal atmospheric pressure at the Earth's surface, which is 101 kPa. Patm Prad
Answer:
a) The radiation pressure on the Earth is 6.065x10⁻⁶Pa
b) The atmospheric pressure is 1.665x10¹⁰ greater than the pressure exerted by the radiation.
Explanation:
Given:
I = intensity of solar radiation = 1368 W/m²
Earth reflects 33%, therefore Earth absorbs 67%
P = pressure = 101 kPa = 1.01x10⁵Pa
c = speed of light = 3x10⁸m/s
Questions:
a) Find the radiation pressure on the Earth, in pascals, at the location where the Sun is straight overhead, P = ?
b) State how this quantity compares with normal atmospheric pressure at the Earth's surface
a) To solve, it is necessary to calculate the pressure exerted by both the reflected light and the light that is absorbed, in this way:
The pressure exerted by the reflected light:
[tex]P_{1} =\frac{2*0.33*I}{c} =\frac{2*0.33*1368}{3x10^{8} } =3.01x10^{-6} Pa[/tex]
The pressure exerted by the absorbed light:
[tex]P_{2} =\frac{0.67*I}{c} =\frac{0.67*1368}{3x10^{8} } =3.055x10^{-6} Pa[/tex]
The radiation pressure on the Earth:
Pt = P₁ + P₂ = 3.01x10⁻⁶ + 3.055x10⁻⁶ = 6.065x10⁻⁶Pa
b) Comparing with normal atmospheric pressure
[tex]Ratio=\frac{P_{atm} }{P_{t} } =\frac{1.01x10^{5} }{6.065x10^{-6} } =1.665x10^{10}[/tex]
According to this result, the atmospheric pressure is 1.665x10¹⁰ greater than the pressure exerted by the radiation.
the personality test that presents an ambiguous stimulus picture to which the person may respond as he or she wishes is a ___.
A ring is attached at the center of the underside of a trampoline. A sneaky teenager crawls under the trampoline and uses the ring to pull the trampoline slowly down while his 67-kg mother is sleeping on it. When he releases the trampoline, she is launched upward. As she passes through the position at which she was before her son stretched the trampoline, her speed is 3.0m/s.
How much elastic potential energy did the son add to the trampoline by pulling it down? Assume the interaction is nondissipative.
Answer:
E = 301.5 J
Explanation:
We have,
Mass of mother, m = 67 kg
Here, a sneaky teenager crawls under the trampoline and uses the ring to pull the trampoline slowly down. As she passes through the position at which she was before her son stretched the trampoline, her speed is 3 m/s.
It is required to find the elastic potential energy the son add to the trampoline by pulling it down. It is based on the conservation of energy.
The elastic potential energy of the mother = the elastic potential energy the son add to the trampoline.
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2[/tex]
So, the elastic potential energy is :
[tex]E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\times 67\times 3^2\\\\E=301.5\ J[/tex]
So, the elastic potential energy of 301.5 J the son add to the trampoline by pulling it down.
A large diameter closed top tank is filled with a depth of 3 meters of a fluid (density 1200 kg/m3). A small pipe leading from the bottom of the tank must carry the fluid to a height of 5 meters (2 meters above the top level of the tank fluid level). a) What pressure must be maintained in the space above the fill line in the tank to provide an exit speed of 5 m/sec for the fluid out of the pipe? (the pressure at the exhaust end of the pipe is 1 atmosphere or about 1 x 105 pascal) 15 pt If the pipe diameter is 2 cm, what will be the initial volume rate of flow (m3 /sec) 5pt
Answer:
1.57×10^-3 m^3/s
Explanation:
Please see the attached filw for a detailed and step by step solution of the given problem.
The attached file explicitly solves it.
Using evidence from the article, defend the concept that
Earth's magnetic poles have swapped places over time.
Answer: Scientists found evidence of Earth's magnetic field reversal in rocks on the ocean floor at plate boundaries. These rocks have alternating polarity due to magnetization that were during their cooling period. Using radiometric dating, scientists estimate that reversals occur approximately every several hundred thousand years.
Explanation: The earth’s magnetic field impacts into the alignment of elements like iron in rocks due to Ferro-magnetization. This causes the elements to align in a north-south direction. Accordingly scientists have studied this alignment in ancient rocks, such as in the layers of sedimentary rocks, and realized that the earth magnetic field has flipped around 200,000 to 300,000 in the last 20 million years.
Answer:
Scientists found evidence of Earth's magnetic field reversal in rocks on the ocean floor at plate boundaries. These rocks have alternating polarity due to magnetization that occurred during their cooling period. Using radiometric dating, scientists estimate that reversals occur approximately every several hundred thousand years.
Explanation:
A traffic controller at the airport watches a commercial plane circling above at a distance of 24.0 km as the pilot waits for clearance to land. If the Moon subtends an angle of 9.89 10-3 radians at the controller's location, what is the distance the jet travels as its nose moves across the diameter of the Moon?
Answer:
The distance travelled is [tex]S = 237.36 \ m[/tex]
Explanation:
From the question we are told that
The distance of the airplane from the ground is [tex]L = 24.0\ km[/tex]
The angle subtended by the moon is [tex]\theta = 9.89*10^{-3} \ radians[/tex]
The distance traveled can be mathematically represented as
[tex]S = R \theta[/tex]
Substituting values
[tex]S = 24 *10^3 * 9.89*10^{-3}[/tex]
[tex]S = 237.36 \ m[/tex]
A ski lift is used to transport people from the base of a hill to the top. If the lift leaves the
base at a velocity of 15.5 m/s and arrives at the top with a final velocity of 0 m/s, what is the height of the hill? Round the answer to the nearest tenth.
Answer:
12.3
Explanation:
Answer:
12.3
Explanation:
got it right
To apply Problem-Solving Strategy 21.1 Faraday's law. A closely wound rectangular coil of 80 turns has dimensions 25.0 cm by 40.0 cm. The plane of the coil is rotated from a position in which it makes an angle of 37.0 degrees with a magnetic field of 1.10 T to a position perpendicular to the field. The rotation takes 0.0600 s. What is the average emf E induced in the coil
Answer:
58.37 V
Explanation:
Given that
Number of winding on coil, N = 80 turns
Area of the coil, A = 25 cm * 40 cm = 0.25 m * 0.4 m
Magnitude of magnetic force, B = 1.1 T
Time of rotation, t = 0.06 s
See the attachment for calculations
Answer:
ℰ[tex]Ф_{av}[/tex]=58.37V
Explanation:
Given:
Number of winding on coil 'N' = 80 turns
Area 'A' = 25 cm x 40 cm = 0.25 m x 0.4 m =>0.10m²
Magnitude of magnetic force 'B' = 1.1 T
Time ' t' = 0.06 s
The average magnitude of the induced emf is given by:
ℰ[tex]Ф_{av}[/tex]=N|ΔФB/Δt| => N |Ф[tex]Ф_{B,f}[/tex] - Ф[tex]Ф_{B,i}[/tex]| /Δt
The flux through the coil is Ф[tex]Ф_{B[/tex] = BA cos∅
ℰ[tex]Ф_{av}[/tex]=NBA |cos(∅[tex]Ф_{f}[/tex]) - cos(∅[tex]Ф_{i}[/tex])| /Δt
As the initial angle is ∅= 97-37 =>53° and the final angle is ∅=0°
ℰ[tex]Ф_{av}[/tex]=[tex]\frac{(80)(1.1)(0.1)|cos(0)-cos(53)|}{0.06}[/tex]
ℰ[tex]Ф_{av}[/tex]= [tex]\frac{(8.8)|1-0.002|}{0.06}[/tex]
ℰ[tex]Ф_{av}[/tex]=58.37V
The instructions for an electric lawn mower suggest that an A gauge extension cord ( cross sectional area = 4.2 x 10-7 m2 ) should only be used for distances up to 35 m. The resistivity of copper (used in the extension cord) is 1.72 x 10-8 .m at 20oC and the temperature coefficient of resistivity of copper is 0.004041 (oC)-1. What is the resistance of a A type extension cord of length 35 m at 40oC
Answer:
1.43 Ω
Explanation:
Given that
Cross sectional area of the cord, A = 4.2*10^-7 m²
Distance meant to be used, L = 35 m
Resistivity of copper, p = 1.72*10^-8 Ωm
Temperature of copper, t = 20° C
Temperature coefficient of copper, t' = 0.004041 °C^-1
To solve this, we would use the resistance and resistivity formula
R = pL / A, on substituting
R = (1.72*10^-8 * 35) / 4.2*10^-7
R = 6.02*10^-7 / 4.2*10^-7
R = 1.43 ohm
Therefore, the resistance of the extension cord is 1.43 ohm