Answer: Its b, The only problem with this is is there supposed to be a picture?
Explanation: NASA has used there fancy gadgets to figure this out but if there was a picture, this answer could be different.
Part A The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. 3.0 m, 5.0 m 4.0 m, 5.0 m 2.0 m, 3.0 m 1.0 m
Answer:
The location are [tex]x_1 = 2 \ and \ x_2 = 3[/tex]
Explanation:
From the question we are told that
The potential energy is [tex]U(x) = (2.0 \ J/m^3) * x^3 - (15 \ J/m) * x^2 + (36 \ J/m) * x - 23 \ J[/tex]
The force on the mass can be mathematically evaluated as
[tex]F = - \frac{d U(x)}{d x } = -( 6 x^2 - 30x +36)[/tex]
The negative sign shows that the force is moving in the opposite direction of the potential energy
[tex]F = - 6 x^2 + 30x - 36[/tex]
At critical point
[tex]\frac{d U(x)}{dx} = 0[/tex]
So
[tex]- 6 x^2 + 30x - 36 = 0[/tex]
[tex]- x^2 + 5x - 6 = 0[/tex]
Using quadratic equation formula to solve this we have that
[tex]x_1 = 2 \ and \ x_2 = 3[/tex]
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, counting the ground level as the first,
A. What is the energy E of the emitted photon in electron volts?、
B. What is the wavelength in nanometers of the emitted photon?
C. What is the radius of the hydrogen atom in nanometers in its initial 5th energy level?
Answer:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
[tex]E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})[/tex] (1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):
[tex]E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV[/tex]
B. The energy of the emitted photon is given by the following formula:
[tex]E=h\frac{c}{\lambda}[/tex] (2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:
[tex]2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J[/tex]
Next, you use the equation (2) and solve for λ:
[tex]\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm[/tex]
C. The radius of the orbit is given by:
[tex]r_n=n^2a_o[/tex] (3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:
[tex]r_5=5^2(2.380)=59.5[/tex]
hence, the radius of the atom in its 5-th state is 59.5 anstrongs
A) The energy E of the emitted photon in electron volts is; E = 2.856 eV
B) The wavelength in nanometers of the emitted photon is; λ = 434.4nm
C) The radius of the hydrogen atom in nanometers in its initial 5th energy level is; rₙ = 1.323 nm
A) Formula for the energy E of the emitted photons is;
E = -13.6([tex]\frac{1}{n_{2}^2} - \frac{1}{n_{1}^2}[/tex])
We are given;
n₂ = 5
n₁ = 2
Thus;
E = -13.6([tex]\frac{1}{5^2} - \frac{1}{2^2}[/tex])
E = 2.856 eV
B) The formula for the wavelength is;
λ = hc/E
where;
h is Planck's constant = 6.626 × 10⁻³⁴ m².kg/s
c is speed of light = 3 × 10⁸ m/s
E is energy of photon
λ is wavelength of the photon
Earlier we saw that E = 2.856 eV. Converting to Joules gives;
E = 4.5758 × 10⁻¹⁹ J
Thus;
λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(4.5758 × 10⁻¹⁹)
λ = 4.344 × 10⁻⁷ m
Converting to nm gives;
λ = 434.4nm
C) Formula for the radius of the hydrogen atom is;
rₙ = n²a₀
where;
a₀ is bohr's radius = 5.292 × 10⁻¹¹ m
n = 5
Thus;
rₙ = 5² × 5.292 × 10⁻¹¹
rₙ = 1.323 × 10⁻⁹
rₙ = 1.323 nm
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Unit conversion
The choices are in units
A,GA,MA,uA,kA,mA,nA,pA. Pick one the units
Answer:
1.234567 kA
Explanation:
The prefix k stands for kilo-, or 10³. The prefix m stands for milli-, or 10⁻³. The sum shown is ...
1.234 kA + 0.000567 kA = 1.234567 kA
A 9.0-V battery (with nonzero resistance) and switch are connected in series across the primary coil of a transformer. The secondary coil is connected to a light bulb that operates on 120 V. Determine the ratio of the secondary to primary turns needed for the bells transformer. Determine the ratio of the secondary to primary turns needed for the bells transformer. Ns/Np=?
Answer:
N₂ / N₁ = 13.3
Explanation:
A transformer is a system that induces a voltage in the secondary due to the variation of voltage in the primary, the ratio of voltages is determined by the expression
ΔV₂ = N₂ /N₁ ΔV₁
where ΔV₂ and ΔV₁ are the voltage in the secondary and primary respectively and N is the number of windings on each side.
In this case, they indicate that the primary voltage is 9.0 V and the secondary voltage is 120 V
therefore we calculate the winding ratio
ΔV₂ /ΔV₁ = N₂ / N₁
N₂ / N₁ = 120/9
N₂ / N₁ = 13.3
s good clarify that in transformers the voltage must be alternating (AC)
At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to help people avoid collisions. The magnitude of the mirror's radius of curvature is 0.560 m.
A) Locate the image of a patient10.6m from the mirror. B) Indicate whether the image is upright or inverted.C) Determine the magnification of the image.
Answer:
Explanation:
For a convex mirror, the value of its image distance and its focal length are negative.
using the mirror formula 1/f = 1/u+1/v
f is the focal length = Radius of curvature/2 = 0.560/2
f= 0.28m
u is the object distance = 10.6m
v is the position of the image = ?
On substitution;
1/0.28 = 1/10.6 + 1/-v
3.57 = 0.094 - 1/v
3.57 - 0.094 = -1/v
3.476 = -1/v
v = -1/3.476
v = -0.2877m
B) Since the image distance is negative, this means that the image is an upright and a virtual image. All Upright images has their image distance to be negative.
C) Magnification = Image distance/object distance
Magnification = 0.2877/10.6
Magnification = 0.0271
In this problem you will consider the balance of thermal energy radiated and absorbed by a person.Assume that the person is wearing only a skimpy bathing suit of negligible area. As a rough approximation, the area of a human body may be considered to be that of the sides of a cylinder of length L=2.0m and circumference C=0.8m.For the Stefan-Boltzmann constant use σ=5.67×10−8W/m2/K4.Part aIf the surface temperature of the skin is taken to be Tbody=30∘C, how much thermal power Prb does the body described in the introduction radiate?Take the emissivity to be e=0.6.Express the power radiated into the room by the body numerically, rounded to the nearest 10 W.part bFind Pnet, the net power radiated by the person when in a room with temperature Troom=20∘C
Answer:
The thermal power emitted by the body is [tex]P_t = 286.8 \ Wm^{-2}[/tex]
The net power radiated is [tex]P_{net} = 460 \ W[/tex]
Explanation:
From the question we are told that
The length of the assumed hum[tex]T_{room} = 20 ^oC[/tex]an body is L = 2.0 m
The circumference of the assumed human body is [tex]C = 0.8 \ m[/tex]
The Stefan-Boltzmann constant is [tex]\sigma = 5.67 * 10^{-8 } \ W\cdot m^{-2} \cdot K^{-4}.[/tex]
The temperature of skin [tex]T_{body} = 30^oC[/tex]
The temperature of the room is
The emissivity is e=0.6
The thermal power radiated by the body is mathematically represented as
[tex]P_t = e * \sigma * T_{body}^4[/tex]
substituting value
[tex]P_t = 0.6 * 5.67*10^{-8} * (303)^4[/tex]
[tex]P_t = 286.8 \ Wm^{-2}[/tex]
The net power radiated by the body is mathematically evaluated as
[tex]P_{net} = P_t * A[/tex]
Where A is the surface area of the body which is mathematically evaluated as
[tex]A = C* L[/tex]
substituting values
[tex]A = 0.8 * 2[/tex]
[tex]A = 1.6 m^2[/tex]
=> [tex]P_{net} = 286.8 * 1.6[/tex]
=> [tex]P_{net} = 460 \ W[/tex]
Which symbol in a chemical equation separates the reactants from the products?
Answer:
the arrow symbol ⇒ in irreversible reactions and doble arrow symbol in reversible reactios⇔
Explanation:
i hope this will help you
28 points!! please help
n astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a time interval 11.2 s . What are the average acceleration in each interval? Assume that the positive direction is to the right.
Answer:-
-1 m/s^2
Explanation:
The average acceleration is given by dividing the change in velocity by change in time;
[tex]a_f=\frac{v_f-v_i}{t_f-t_i}[/tex]
[tex]=\frac{(0-11.2)}{(11.2-0)}=-1 m/s^2[/tex]
the point to be noted here is if the velocity is to the left we substitute it with a negative sign and if it is to the right we substitute it with a positive sign.
An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associated with this wave points vertically up. The direction of the electric field vector of this wave is:___________
Answer:
either +z direction or -z direction.
Explanation:
The direction of the electric field, in an electromagnetic wave always is perpendicular to the direction of the magnetic field and the direction of propagation of the wave.
You assume a system of coordinates with the negative x axis as the west direction, and the y axis as the up direction
In this case, the wave is propagating toward the west (- x direction), and the magnetic field vector points up (+ y direction), then, it is mandatory that the electric field vector points either +z direction or -z direction.
I really need help with this question someone plz help !
Answer:
The answer is option 2.
Explanation:
Both sides are pulling the rope with equal force where the rope doesn't move. So they have a balanced forces.
A woman weighs 129 lb. If she is standing on a spring scale in an elevator that is traveling downward, but slowing up, the scale will read:___________.
A) more than 129 lb
B) 129 lb
C) less than 129 lb
D) It is impossible to answer this question without knowing the acceleration of the elevator.
Answer:
C) less than 129 lb.
Explanation:
Let the elevator be slowing up with magnitude of a . That means it is accelerating downwards with magnitude a .
If R be the reaction force
For the elevator is going downwards with acceleration a
mg - R = ma
R = mg - ma
R measures its apparent weight . Spring scale will measure his apparent weight.
So its apparent weight is less than 129 lb .
Of one of the planets becomes a black hole , what would the escape speed be?
Answer:
If, instead, that rocket was on a planet with the same mass as Earth but half the diameter, the escape velocity would be 15.8 km/s Any object that is smaller than its Schwarzschild radius is a black hole – in other words, anything with an escape velocity greater than the speed of light is a black hole.
Explanation:
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A uniform disk with a 25 cm radius swings without friction about a nail through the rim. If it is released from rest from a position with the center level with the nail, then what is its angular velocity as it swings through the point where the center is below the na
Answer:
Explanation:
During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy
mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .
mgr = 1/2 x ( 1/2 m r²+ mr²) x ω²
gr = 1/2 x 1/2 r² x ω² + 1/2r² x ω²
g = 1 / 4 x ω² r + 1 / 2 x ω² r
g = 3 x ω² r/ 4
ω² = 4g /3 r
= 4 x 9.8 / 3 x .25
= 52.26
ω = 7.23 rad / s .
Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistivities of these two wires compare?
Explanation:
The relation between resistance and resistivity is given by :
[tex]R=\rho \dfrac{l}{A}[/tex]
[tex]\rho[/tex] is resistivity of material
l is length of wire
A is area of cross section of wire
Resistivity of a material is the hidden property. If one wire has 3 times the length of the other, then it doesn't affect its resistivity. Hence, the resistivity of two wires is
A 2-kilogram toy car is traveling forward at 1 meter per second when it is hit in the rear by a 3-kilogram toy truck that was traveling at 3 meters per second just before impact. If the two toys stick together, their speed immediately after the collision is
Answer:
v = 1.4 m/s
Explanation:
This problem is about an inelastic collision. The total momentum before the collision is equal to total momentum after (because of the conservation of momentum law):
[tex]m_1v_1-m_2v_2=(m_1+m_2)v[/tex] (1)
m1: mass of the toy car = 2 kg
m2: mass of the toy truck = 3 kg
v1: speed of the toy car = 1 m/s
v2: speed of the truck car = 3 m/s
v: speed of both car and truck after the collision = ?
In the equation (1) the negative sign of m2v2 is because of the opposite direction of the toy truck respect to the toy car.
You solve the equation (1) for v, and you replace the values of all variables involved:
[tex]v=\frac{m_1v_1-m_2v_2}{m_1+m_2}\\\\v=\frac{(2kg)(1m/s)-(3kg)(3m/s)}{2kg+3kg}=-1.4\frac{m}{s}[/tex]
this velocity is negative, then, the direction of motion of both car and truck is in the direction of the truck
Hence, the speed of both car and truck toys is 1.4 m/s
If a metal rod is moved through magnetic field, the charged particles will feel a force, and if there is a complete circuit, a current will flow. We talk about the induced emf of the rod. The rod essentially acts like a battery, and the induced emf is the voltage of the battery. A magnetic field with a strength of 0.732 T is pointing into the page and a metal rod L=0.362 m in length is moved to the right at a speed v of 15.1m/s.
Required:
a. What is the induced emf in the rod?
b. Suppose the rod is sliding on conducting rails, and a complete circuit is formed. If the load resistance is 5.74Ω , what is the magnitude and direction (clockwise or counterclockwise) of the current flowing in the circuit?
Answer:
a. 4 V
b. 0.697 A
Explanation:
Magnetic field strength B = 0.732 T
length of rod l = 0.362 m
velocity of rod v = 15.1 m/s
a. EMF can be calculated as
E = Blv = 0.732 x 0.362 x 15.1 = 4 V
b. If the rod is connected to a conducting rail, with resistance R = 5.74Ω
current I = V/R = 4/5.74 = 0.697 A
the current flows in a clockwise direction
A point charge is located at the center of a thin spherical conducting shell of inner and outer radii r1 and r2, respectively. A second thin spherical conducting shell of inner and outer radii R1 and R2, respectively, is concentric with the first shell. The flux is as follows for the different regions of this arrangement. Ф -10.3 103 N-m2/C for
0 for r<2 4:
-36.8 x 10נ N-m2/c
0 for r > R2
36.8 x 10נ N-m2/c
Determine the magnitude ond sign of the point chorge ond the charge on the surface of the two shels point charge inner shell outer shel.
Answer:
the magnitude is 7 and sign of the point charge on the surface shell is -13
Explanation:
Coherent light that contains two wavelengths, 660 nm and 470 nm , passes through two narrow slits with a separation of 0.280 mm and an interference pattern is observed on a screen which is a distance 5.50 m from the slits.
Required:
What is the disatnce on the screen between the first order bright fringe for each wavelength?
Answer:
λ1 = 0.0129m = 1.29cm
λ2 = 0.00923m = 0.92 cm
Explanation:
To find the distance between the first order bright fringe and the central peak, can be calculated by using the following formula:
[tex]y_m=\frac{m\lambda D}{d}[/tex] (1)
m: order of the bright fringe = 1
λ: wavelength of the light = 660 nm, 470 nm
D: distance from the screen = 5.50 m
d: distance between slits = 0.280mm = 0.280 *10^⁻3 m
ym: height of the m-th fringe
You replace the values of the variables in the equation (1) for each wavelength:
For λ = 660 nm = 660*10^-9 m
[tex]y_1=\frac{(1)(660*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.0129m=1.29cm[/tex]
For λ = 470 nm = 470*10^-9 m
[tex]y_1=\frac{(1)(470*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.00923m=0.92cm[/tex]
a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B suspended by a cord of length l =2.2m. The block then swings through a maximum angle of theta = 60. Determine (a) the initial speed of the bullet vo, (b) the impulse imparted by the bullet on the block, (c) the force on the cord immediately after the impact
Answer:
(a) v-bullet = 399.04 m/s
(b) I = 2.38 kg m/s
(c) T = 2.59 N
Explanation:
(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.
The potential energy is given by:
[tex]U=(M+m)gh[/tex] (1)
U: potential energy
M: mass of the wood block = 1 kg
m: mass of the bullet = 6g = 6.0*10^-3 kg
g: gravitational constant = 9.8m/s^2
h: distance to the ground
The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:
[tex]h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m[/tex]
The potential energy is:
[tex]U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J[/tex]
Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:
[tex]U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}[/tex]
Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:
[tex]Mv_1+mv_2=(M+m)v[/tex] (2)
v1: initial velocity of the wood block = 0m/s
v2: initial speed of the bullet
v: speed of bullet and block = 2.38m/s
You solve the equation (2) for v2:
[tex]M(0)+mv_2=(M+m)v[/tex]
[tex]v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}[/tex]
The speed of the bullet before the impact with the wood block is 399.04 m/s
(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:
[tex]I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}[/tex]
The impulse is 2.38 kgm/s
(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:
[tex]T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N[/tex]
The force on the cord after the impact is 2.59N
Answer:
The initial speed of the bullet [tex]V_o = 777.97m/s[/tex]The force on the cord immediately after the impact = [tex]19.71N[/tex]Explanation:
Apply the law of conversion of energy
[tex]V_f = \sqrt{2gh}[/tex]
where,
h = height of which the bullet and block rise after impact
[tex]h = L - Lcos\theta\\\\h = 2.2 - (2.2*cos60)\\\\h = 1.1m[/tex]
Therefore,
[tex]V_f = \sqrt{2gh}\\\\V_f = \sqrt{2*9.8*1.1}\\\\V_f = 4.64m/s[/tex]
From conservation of momentum principle, [tex]m_Bv_B = 0[/tex]
[tex]m_ov_o + m_Bv_B = (m_b+m_B)V_f\\\\0.006V_o = (0.006+1)*4.64\\\\V_o = 777.97m/s[/tex]
C) The force in the cable is due to the centrfugal force of the system, which is due to the motion of the system is a curved path and weight of the system
[tex]F = \frac{m_b+m_B}{L}V_f^2 + (m_b+m_B)g\\\\F = \frac{0.006+1}{2.2}*4.64^2 + (0.006+1)9.81\\\\F = 19.71N[/tex]
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The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor.
a. What is the electric field strength and direction between the plates?
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules
The magnitude of the magnetic field at a certain distance from a long, straight conductor is represented by B. What is the magnitude of the magnetic field at twice the distance from the conductor
Answer:
B/4
Explanation:
The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:
The field at twice the distance is B/4.
250cm3 of fres
er of density 1000kgm-3 is mixed with 100cm3 of sea water of density 1030kgm-3. Calculate the density of the mixture. *
Answer:
1008.57kg/m3
Explanation:
Now the mass of fresh water is 250×1000 /1000000 = 0.25kg
Now the mass of salt water is
100×1030 /1000000 = 0.103kg
Note Density = mass / volume
Mass = volume × density
Note that converting from cm3 to m3 we divide by 1000000
Total mass = 0.25kg +0.103kg= 0.353kg.
Total volume also is (250 +100 )/1000000= 35 × 10^{-5}m3
Hence the density of the mixture= total mass / total volume
0.353kg/35 × 10^{-5}m3=1008.57kg/m3
g it as been suggested that solar powered space ships could get a boost from a laser either on earth or in orbit around earth. the laser would have to be very powerful to give any measurable benefit to the ship. if the laser produces a 0.18-m diameter beam of 490-nm light, what is the minimum angular spread of the beam?
Answer:
The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians (or 3.32 μrad).
Explanation:
The minimum angular spread of a wave is the ratio of its narrowest diameter to its wavelength.
From Rayleigh's formula,
Angular spread = 1.22 (wavelength ÷ diameter)
= 1.22 (λ ÷ D)
Given that:
diameter, D = 0.18 m and wavelength, λ = 490 nm, then;
Angular spread of the laser beam = 1.22 (λ ÷ D)
= 1.22[tex](\frac{490*10^{-9} }{0.18})[/tex]
= 1.22× 2.7222 × [tex]10^{-6}[/tex]
= 3.3211 × [tex]10^{-6}[/tex] rad
The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians.
As your bus rounds a flat curve at constant speed, a package with mass 0.900 kg , suspended from the luggage compartment of the bus by a string 50.0 cm long, is found to hang at rest relative to the bus, with the string making an angle of 30.0 â with the vertical. In this position, the package is 55.0 m from the center of curvature of the curve.
Required:
a. What is the radial acceleration of the bus?
b. What is the radius of the curve?
Answer:
a.[tex]5.66ms^{-2}[/tex]
b.55 m
Explanation:
We are given that
Mass ,m=0.9 kg
Length of string,l=50 cm=[tex]\frac{50}{100}=0.50 m[/tex]
1 m=100 cm
[tex]\theta=30^{\circ}[/tex]
R=55 m
a.Centripetal acceleration
[tex]a_c=gtan\theta[/tex]
[tex]a_c=9.8tan30^{\circ}[/tex]
[tex]a_c=5.66 m/s^2[/tex]
Hence, the radial acceleration of the bus=[tex]5.66m/s^2[/tex]
b. Radius of curve,R=55 m
A block of mass 15.0 kg slides down a ramp inclined at 28.0∘ above the horizontal. As it slides, a kinetic friction force of 30.0 N parallel to the ramp acts on it. If the block slides for 5.50 m along the ramp, find the work done on the block by friction.
Answer:
Work is done by friction = -165 J
Explanation:
Given:
Mass of block (m) = 15 kg
Ramp inclined = 28°
Friction force (f) = 30 N
Distance (d) = 5.5 m
Find:
Work is done by friction.
Computation:
Work is done by friction = -Fd
Work is done by friction = -(30)(5.5)
Work is done by friction = -165 J
70 pointss yall !!! helpp
Answer:
A= The type of plant
B= How tall the plant is
Explanation:
A transformer has a primary coil with 375 turns of wire and a secondary coil with 1,875 turns. An AC voltage source connected across the primary coil has a voltage given by the function Δv = (130 V)sin(ωt). What rms voltage (in V) is measured across the secondary coil?
Answer:
The rms voltage (in V) measured across the secondary coil is 459.62 V
Explanation:
Given;
number of turns in the primary coil, Np = 375 turns
number of turns in the secondary coil, Ns = 1875 turns
peak voltage across the primary coil, Ep = 130 V
peak voltage across the secondary coil, Es = ?
[tex]\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V[/tex]
The rms voltage (in V) measured across the secondary coil is calculated as;
[tex]V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V[/tex]
Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V
A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic field that produces on it an upward force exactly equal in magnitude to the wire's weight, causing the wire to "levitate."
Required:
a. What is the field's magnitude?
b. What is the field's direction?
Answer:
The classification of that same issue in question is characterized below.
Explanation:
The given values are:
Current, I = 50.0 A
Diameter, d = 0.10 cm
(a)...
As we know,
⇒ Magnetic force = Copper wire's weight
So,
⇒ [tex]B\times I\times L=M\times g[/tex]
On putting the estimated values, we get
⇒ [tex]B\times 50\times 1=7.037\times 10^{-3}\times 9.81[/tex]
⇒ [tex]50B=69.03297\times 10^{-3}[/tex]
⇒ [tex]B=1.38\times 10^{-3} \ T[/tex]
(b)...
As we know,
⇒ [tex]m=\delta\times L\times \frac{\pi \ d^2}{4}[/tex]
⇒ [tex]=8960\times 1\times \frac{\pi \ (0.001)^2}{4}[/tex]
⇒ [tex]=2240\times \pi \ 0.000001[/tex]
⇒ [tex]=7.037\times 10^{-3} \ kg[/tex]
Use the position function s(t) = -16t + v_0t + s_0 for free falling objects. A ball is thrown straight down from the top of a 600-foot building with an initial velocity of -30 feet per second. (a) Determine the position and velocity functions for the ball. (b) Determine the average velocity on the interval [1, 3]. (c) Find the instantaneous velocities when t=1 and t=3. (d) Find the time required for the ball to reach ground level. (e) Find the velocity of the ball at impact.
Answer:
a) v = -30 - 32 t , s (t) = 600 - 30 t -16 t² , b) v = -32 ft / s
c) v (1) = -62 ft / s, v (3) = -126 ft / s , d) t = 7.13 s , e) v = -258.16 ft / s
Explanation:
a) For this exercise they give us the function of the position of the ball
s (t) = s (o) + v_o t - 16 t²
notice that you forgot to write the super index
indicate the initial position of the ball
s (o) = 600 ft
also indicates initial speed
v_o = - 30 ft / s
let's substitute in the equation
s (t) = 600 - 30 t -16 t²
to find the speed we use
v = ds / dt
v = v_o - 32 t
v = -30 - 32 t
b) To find the average speed, look for the speed at the beginning and end of the time interval
t = 1 s
v (1) = -30 -32 1
v (1) = - 62 ft / s
t = 3 s
v (3) = -30 -32 3
v (3) = -126 ft / s
the average speed is
v = (v (3) -v (1)) / (3-1)
v = (-126 +62) / 2
v = -32 ft / s
c) instantaneous speeds, we already calculated them
v (1) = -62 ft / s
v (3) = -126 ft / s
d) the time to reach the ground
in this case s = 0
0 = 600 - 30 t -16 t²
t² + 1,875 t - 37.5 = 0
we solve the quadratic equation
t = [-1,875 ±√ (1,875² + 4 37.5)] / 2
t = [1,875 ± 12.39] / 2
t₁ = 7.13 s
t₂ = negative
Since the time must be positive, the correct answer is t = 7.13 s
e) the speed of the ball on reaching the ground
v = -30 - 32 t
v = -30 - 32 7.13
v = -258.16 ft / s
