Answer:
towards the west
One can conclude that the magnetic field must have a component towards the east. The correct option is c.
What is magnetic field?A magnetic field is a vector field that describes the magnetic influence on moving charges, currents, and magnetic materials.
A moving charge in a magnetic field is subjected to a force that is perpendicular to both its own velocity and the magnetic field.
Every magnet possesses a north as well as a south pole. The opposite poles attract each other, whereas the same poles repel each other.
When you rub a piece of iron against a magnet, the atoms' north-seeking poles line up in the same direction.
Moving north, a proton enters a magnetic field. The proton curves downward as a result of this field. We can conclude that the magnetic field must have an eastward component.
Thus, the correct option is c.
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Which best identifies the requirements for work to be performed? an object that has a force acting on it an object that is moving and has no net force a force acting on a motionless object a force that moves an object
Answer:
a force that moves an object
Explanation:
the formula for work is force * distance
This question involves the concepts of work, force, and displacement.
The statement that best identifies the requirements for work to be performed is "a force that moves an object".
Work is defined as the product of force applied on an object and the distance moved by the object. Mathematically,
Work = (Force)(Displacement)
Hence, both the applied force and the displacement of the object as a result of the application of the force is necessary for the work to be done. If any one of these values becomes zero, the work automatically becomes zero, which means no work is performed.
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6. If you wanted to develop a telescope, what kind of lenses would you use for the objective lens (the lens that collects the light) and the eyepiece? Explain your reasoning. Draw a picture with ray tracing of your setup.
Answer:
objetive: a converging lens for large diameter lenses
eyepiece you must select a lens with a small focal length and the diameter is not important
The selected lenses should decrease chromatic aberration.
Explanation:
A telescope is an instrument that collects light from very distant objects, therefore very weak.
Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.
For the eyepiece you must select a lens with a small focal length and the diameter is not important
the telescope magnification is
m = f_objective / F_ocular
The selected lenses should decrease chromatic aberration.
In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.
Answer: this the real answer try it objetive: a converging lens for large diameter lenseseyepiece you must select a lens with a small focal length and the diameter is not importantThe selected lenses should decrease chromatic aberration.Explanation:A telescope is an instrument that collects light from very distant objects, therefore very weak.Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.For the eyepiece you must select a lens with a small focal length and the diameter is not importantthe telescope magnification is m = f_objective / F_ocularThe selected lenses should decrease chromatic aberration.In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.
Explanation:
3. What color of laser light shines through a diffraction grating with a line density of 500 lines/mm if the third maxima from the central maxima (m=3) is at an angle of 45°?
Answer:
Wavelength is 471 nm
Explanation:
Given that,
Lines per unit length of diffraction grating is 500 lines/mm.
The third maxima from the central maxima (m=3) is at an angle of 45°
We need to find the color of laser light shines through a diffraction grating.
The condition for maxima is :
[tex]d\sin\theta=m\lambda[/tex]
d = 1/N, N = number of lines per mm
[tex]\lambda=\dfrac{1}{Nm}\sin\theta\\\\\lambda=\dfrac{10^{-3}}{500\times 3}\sin(45)\\\\\lambda=4.31\times 10^{-7}\\\\\text{or}\\\\\lambda=471\ nm[/tex]
The temperature difference between the inside and the outside of a house on a cold winter day is 33°F. (a) Express this difference on the Celsius scale. 0.55 Incorrect: Your answer is incorrect. °C (b) Express this difference on the Kelvin scale. 273.7 Incorrect: Your answer is incorrect. K
Answer:
a) 0.56°C
b) 273.56 K
Explanation:
If we want to convert from Fahrenheit scale to Celcius scale we use the formula;
T(°C) = (T(°F) - 32) × 5/9
Where T(°F) = 33°F
Hence;
T(°C) = (33°F - 32) × 5/9
T°C = 0.56°C
b)
T(K) = T°C + 273
T(K) = 0.56 + 273
T(K) = 273.56 K
A mass m = 0.3 kg is released from rest at the origin point 0. The mass falls under the influence of gravity. When the mass reaches point A, it has a velocity of v downward and when the mass reaches point B its velocity is 5v. What is the distance between points A & B divided by the distance between points 0 & A?
Answer:
24
Explanation:
The mass = 3 kg
at point O all the mechanical energy of the system is due to its potential energy PE. The body is at rest.
PE = mgh
but ME = PE + KE = constant (law of energy conservation)
KE is the kinetic energy
since KE is zero at this point, then,
ME = mgh
where m is the mass
g is acceleration due to gravity = 9.81 m/s^2
h is the height = O
ME = 3 x 9.81 x O
ME = 29.43-O
At point A the total ME is due to its PE and its kE
PE at this point = mgh = 3 x 9.81 x A = 29.43-A
KE = [tex]\frac{1}{2}mv^{2}[/tex]
velocity = v at this point, therefore,
KE = [tex]\frac{1}{2}*3*v^{2}[/tex] = [tex]\frac{3}{2} }v^{2}[/tex]
therefore,
ME = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]
Equating ME for the points O and A, we have
29.43-O = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]
29.43-O - 29.43-A = [tex]\frac{3}{2} }v^{2}[/tex]
(O - A)29.43 = [tex]\frac{3}{2} }v^{2}[/tex]
O - A = 0.051[tex]v^{2}[/tex] this is the distance between point O and A
For point B
PE = 29.43-B
KE = [tex]\frac{25}{2}*3*v^{2}[/tex] = 37.5[tex]v^{2}[/tex] (velocity is equal to [tex]5v[/tex] at this point)
therefore,
ME = 29.43-B + 37.5[tex]v^{2}[/tex]
Equating the ME for points A and B, we have
29.43-A + [tex]\frac{3}{2} }v^{2}[/tex] = 29.43-B + 37.5[tex]v^{2}[/tex]
29.43-A - 29.43-B = 37.5[tex]v^{2}[/tex] - [tex]\frac{3}{2} }v^{2}[/tex]
(A - B)29.43 = 36[tex]v^{2}[/tex]
A - B = 1.22[tex]v^{2}[/tex] this is the distance between points A and B
The distance between points A & B divided by the distance between points 0 & A will be
1.22[tex]v^{2}[/tex]/0.051 = 23.9 ≅ 24
Each proton-proton cycle generates 26.7 MeV of energy. If 9.9 Watts are generated via by the proton-proton cycle, how many billion neutrinos are produced
Answer:
4.635 *10^12 Neutrinos
Explanation:
Here in this question, we are to determine the number of neutrinos in billions produced, given the power generated by the proton-proton cycle.
We proceed as follows;
In proton-proton cycle generates 26.7 MeV of energy and in this cycle two neutrinos are produced.
From the question, we are given that
Power P = 9.9 watts = 9.9 J/s
Watts is same as J/s
The number of proton-proton cycles required to generate E energy is N = E / E '
Where E ' = Energy generated in proton-proton cycle which is given as 26.7 Mev in the question
Converting Mev to J, we have
= 26.7 x1.6 x10 -13 J
To get the number N which is the number of proton-proton cycle required, we have;
N = 9.9 /(26.7 x1.6 x10^-13) = 2.32 * 10^12
Since we have two proton cycles( proton-proton), it automatically means 2 neutrinos will be produced.
Therefore number of neutrions produced = 2 x Number of proton-proton cycles = 2 * 2.32 * 10^12 = 4.635 * 10^12 neutrinos
Meaning of power in physics
Answer:
The rate of doing work is called power.
Answer:
The amount of energy transported or transformed per unit time is referred to as power in physics. The watt, which is equal to one joule per second in the International System of Units, is the unit of power.
OAmalOHopeO
What did the results of photoelectric-effect experiments establish?
Answer:
Option A
Electrons are emitted if low intensity, high-frequency light hits a metal surface.
Explanation:
From the experiments conducted to study the photoelectric effect, conclusions were made that the key factor that contributes to the emission of electrons from the surface of the metal is the frequency of the beam of light. This frequency has to be beyond a minimum threshold, if not, there will be no emission of electrons from the metal surface no matter the intensity of the beam of light or the length of time it is incident upon the metal surface.
This makes option A correct because it highlights the contributions made by the threshold frequency to the photoelectric effect.
One of the two slits in a Young’s experiment is painted over so that it transmits only one-half the intensity of the other slit. As a result:
A. the fringe system disappears
B. the bright fringes get brighter and the dark ones get darker
C. the fringes just get dimmer
D. the dark fringes just get brighter
E. the dark fringes get brighter and the bright ones get darker
Answer:
B. the bright fringes get brighter and the dark ones get darker
Explanation:
Let us consider when the intensities are equal, we use the equation
[tex]I_{max} = I_{1} + I_{2} + 2\sqrt{I*I}[/tex] for light fringes and,
[tex]I_{min} = I_{1} + I_{2} - 2\sqrt{I*I}[/tex] for dark fringes
where [tex]I_{1}[/tex] and [tex]I_{1}[/tex] are the light intensities from the first and second slits respectively.
For the first case where the light from the two slits have the same intensities, we can say both have intensity [tex]I[/tex]
[tex]I_{max} = I + I + 2\sqrt{I*I}[/tex] = [tex]2I + 2I = 4I[/tex]
[tex]I_{min} = I + I - 2\sqrt{I*I} = 2I - 2I = 0[/tex]
For the case where one of the intensities has half the intensity of the other.
one has intensity [tex]I[/tex] and the other one has intensity [tex]\frac{I}{2}[/tex]
inserting, we have
[tex]I_{max} = I + \frac{I}{2} + 2\sqrt{I*\frac{I}{2}} = 2.932I[/tex]
[tex]I_{min} = I + \frac{I}{2} - 2\sqrt{I*\frac{I}{2}} = 0.068I[/tex]
this shows that the bright fringes get brighter and the dark ones get darker.
A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0 m/s (about 22 mi/h). Just as the motorist passes, a police officer on a motorcycle stopped at the comer starts off in pursuit with constant acceleration of 3.0 m/S2. (a) How much time elapses before the officer catches up with the motorist? (b) What is the officer's speed at that point? (c) What is the total distance each vehicle has traveled at that point? Please help me
Answer:
(a) 10 s
(b) 30 m/s
(c) 150 m
Explanation:
The motorist's position at time t is:
x = 15t
The officer's position at time t is:
x = ½ (3) t² = 1.5 t²
(a) When they have the same position, the time is:
15t = 1.5 t²
t = 0 or 10 s
(b) The officer's speed is:
v = 3t
v = 30 m/s
(c) The position is:
x = 15t = 150 m
It takes 85N of force to accelerate a ball at a rate of 15 m/s². What is the mass of the ball?
Explanation:
F = ma
85 N = m (15 m/s²)
m ≈ 5.7 kg
Describe at least two unique characteristics of the new space telescope.
Answer: Astro 1 will have a 10x larger field of view than the Hubble.
Explanation: The hubble will also be extremely light weight that way it can go further into space and the mission will be able to last a longer amount of time.
A bucket filled with water has a mass of 23 Kg and is attached to a rope, which in turn is wound around a 0.050 m radius cylinder at the top of a well. What torque does the weight of water and bucket produce on the cylinder if the cylinder is ont permitted to rotate? (g= 9.8 m/s2)
Answer:
The torque is 11.27 N m
Explanation:
Recall that torque is the vector product of the force times the distance to the pivoting point. So in our case, the distance to the pivoting point is the radius of the cylinder (0.05 m), and the force is given by the weight of the bucket full of water (W = 9.8 * 23 N = 225.4 N)
Then the torque is: 0.05 * 225.4 N m = 11.27 N m
You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball is kicked at 31° from the edge of the building with an initial velocity of 15 m/s and lands 63 meters away from the wall. How tall, in meters, is the building that the child is standing on?
Answer:
69.58 m tall
Explanation:
Pls see attached file
When water freezes, it expands about nine percent. What would be the pressure increase inside your automobile engine block if the water in there froze? The bulk modulus of ice is 2.0 × 109 N/m2, and 1 ATM = 1.01 × 105 N/m2.
Answer:
The pressure increase inside the automobile engine block is 1782.18 ATM
Explanation:
Given;
the change in volume of water, ΔV = 9%
the bulk modulus of ice, K = 2 x 10⁹ N/m²
Bulk modulus is given by;
[tex]K = -V\frac{dP}{dV}[/tex]
for pressure increase in the automobile engine block, when the water in there froze;
[tex]dP = K(\frac{dV}{V} )\\\\dP = K(\frac{0.09V}{V} )\\\\dP = 0.09K\\\\dP = 0.09 (2*10^9)\\\\dP = 1.8 *10^{8} \ N/m^2\\\\dP = 1782.18 \ ATM[/tex]
Therefore, the pressure increase inside the automobile engine block is 1782.18 ATM
The pressure increase inside your automobile engine block will be 1782.18 atm. The force involved vertical to the surface of an object per unit area is pressure.
What is pressure?The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.
It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.
The given data in the problem is;
ΔV is the change in volume of water = 9%
K is the bulk modulus of ice = 2 x 10⁹ N/m²
dP is the change in the presure=?
The bulk modulus is found as;
[tex]\rm K=-V \frac{dp}{dv} \\\\[/tex]
The change in the presure is obtained as;
[tex]\rm dP = K\frac{dV}{V} \\\\ \rm dP = K\frac{0.09V}{V} \\\\ \rm dP = 0.09 K \\\\ \rm dP = 0.09 \times 2 \times 10^9 \\\\ \rm dP = 1.8 \times 10^8 \\\\ \rm dP =1782.18 \ atm[/tex]
Hence the pressure increase inside your automobile engine block will be 1782.18 atm.
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The orbital motion of Earth around the Sun leads to an observable parallax effect on the nearest stars. For each star listed, calculate the distance in parsecs before converting that distance to astronomical units. A. Sirius (0.38") B. Alpha Centauri A (0.75") C. Procyon (0.28") D. Wolf 359 (0.42") E. Epsilon Eridani (0.31") D(pc) = 1/parallax(arcsecs), D(a.u.) = D(pc) * 206265 (arcsecs per radian)
Answer:
Following are the answer to this question:
Explanation:
Formula:
[tex]D(PC) =\frac{1}{parallax}\\\\D(av)=D(PC) \times 20.626\ J[/tex]
Calculating point A:
when the value is [tex]0.38[/tex]
[tex]\to 0.38 \toD(PC)= \frac{1}{0.38}\\\\[/tex]
[tex]=2.632[/tex]
[tex]\to D(a.v) = \frac{1}{0.38} \times 206265\\[/tex]
[tex]=542,802.6[/tex]
Calculating point B:
when the value is [tex]0.75[/tex]
[tex]\to D(PC)=\frac{1}{0.75}[/tex]
[tex]=1.33[/tex]
[tex]\to D(a.v) = \frac{1}{0.75} \times 206265\\[/tex]
[tex]=275,020[/tex]
Calculating point C:
when the value is [tex]0.28[/tex]
[tex]\to D(PC)=\frac{1}{0.28}[/tex]
[tex]=3.571[/tex]
[tex]\to D(a.v) = \frac{1}{0.28} \times 206265\\[/tex]
[tex]=736660.7[/tex]
Calculating point D:
when the value is [tex]0.42[/tex]
[tex]\to D(PC)=\frac{1}{0.42}[/tex]
[tex]=2.38[/tex]
[tex]\to D(a.v) = \frac{1}{0.42} \times 206265\\[/tex]
[tex]=490910.7[/tex]
Calculating point E:
when the value is [tex]0.31[/tex]
[tex]\to D(PC)=\frac{1}{0.31}[/tex]
[tex]=3.226[/tex]
[tex]\to D(a.v) = \frac{1}{0.31} \times 206265\\[/tex]
[tex]=665370.97[/tex]
A front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the front tires
Answer:
static friction acting opposite to the direction of travel
Explanation:
Because the Frictional force of the front wheels act to oppose the spinning, so, For the front wheels to roll without slipping, the friction must be static friction pointing in the direction of travel of the car.
Explanation:
A car moving east at 45 km/h turns and travels west at 30 km/h. What is the
magnitude and direction of the change in velocity?
mahalle 1.11
Explanation:
Change in Velocity = final velocity - initial velocity
Change in velocity = 30km/h - (- 45km/h )
= 75 km/h due west
Why would someone choose a for-profit fitness center over a nonprofit center?
For-profits have important social missions that deserve support. For-profits have important social missions that deserve support. , ,
For-profits are more likely to be located in low-income communities.For-profits are more likely to be located in low-income communities. , ,
For-profits have services like homework support and job training. , For-profits have services like homework support and job training. , ,
For-profits usually have nicer equipment than nonprofit centers.For-profits usually have nicer equipment than nonprofit centers. , ,
Someone would choose a for-profit fitness center over a nonprofit center because for-profits usually have nicer equipment than nonprofit centers.
A for-profit institution is an institution that provides a service to society to obtain an economic profit from this service. On the other hand, non-profit institutions are those that provide a service to society and do not intend to make a profit, they are generally institutions that are part of the government, foundations, or non-governmental organizations.
Non-profit institutions can function thanks to donations, volunteers, public resources, and do not charge for providing the service they provide, while for-profit institutions charge for their service and a percentage of that charge is profit for the owner of the institution and another is reinvested in equipment, human resources or others to improve the service.
According to this, for-profits are likely to have better equipment, while non-profit are characterized by supporting missions and offering services to low-income communities.
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if a 1-m diameter sewer pipe is flowing at a depth of 0.4 m and has a flow rate of 0.15 m^3/s, what will be the flow rate when the pipe flows full?
Answer:
0.35 m³/s
Explanation:
When the pipe's depth is 0.4 m, the area of the circular segment is:
A = ½ R² (θ − sin θ)
The depth of the water is:
h = R (1 − cos(θ/2))
Solving for θ:
0.4 = 0.5 (1 − cos(θ/2))
0.8 = 1 − cos(θ/2)
cos(θ/2) = 0.2
θ/2 = acos(0.2)
θ = 2 acos(0.2)
θ ≈ 2.74 rad
The area is therefore:
A = ½ (0.5 m)² (2.74 − sin 2.74)
A = 0.338 m²
The cross-sectional area when the pipe is full is:
A = π (0.5 m)²
A = 0.785 m²
The flow velocity is constant:
v = v
Q / A = Q / A
(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)
Q = 0.35 m³/s
Test Bank, Question 18.83 Inside a room at a uniform comfortable temperature, metallic objects generally feel cooler to the touch than wooden objects do. This is because: a given mass of wood contains more heat than the same mass of metal the human body, being organic, resembles wood more closely than it resembles metal metal conducts heat better than wood heat tends to flow from metal to wood
Answer:
metal conducts heat better than wood.
Explanation:
Metals are generally good conductors of heat, and they usually conduct heat at a relatively rapid rate. Inside the room with a uniform temperature, a metal when touched will rapidly conduct the heat from your hand, leaving your hand with a cooler feeling. Wood on the other hand is a poor heat conductor, so the heat is not conducted from your hand fast enough to cool it up to the point that your hand feels cool.
15. Food chain always start with
a. Photosynthesis
Decay
b. Respiration
d. N2 Fixation
C.Photosynthesis
In a physics laboratory experiment, a coil with 250 turns enclosing an area of 14 cm2 is rotated in a time interval of 0.030 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is 5.0×10^−5 T.Required:a. What is the total magnetic flux through the coil before it is rotated? After it is rotated? b. What is the average emf induced in the coil?
Explanation:
Consider a loop of wire, which has an area of [tex]A=14 \mathrm{cm}^{2}[/tex] and [tex]N=250[/tex] turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in [tex]\Delta t=0.030[/tex] s. Given that the earth's magnetic field at the position of the loop is [tex]B=5.0 \times 10^{-5} \mathrm{T}[/tex], the flux through the loop before it is rotated is,
[tex]\Phi_{B, i} &=B A \cos \left(\phi_{i}\right)=B A \cos \left(0^{\circ}\right[/tex]
[tex]=\left(5.0 \times 10^{-5} \mathrm{T}\right)\left(14 \times 10^{-4} \mathrm{m}^{2}\right)(1)[/tex]
[tex]=7.0 \times 10^{-8} \mathrm{Wb}[/tex]
[tex]\quad\left[\Phi_{B, i}=7.0 \times 10^{-8} \mathrm{Wb}\right[/tex]
after it is rotated, the angle between the area and the magnetic field is [tex]\phi=90^{\circ}[/tex] thus,
[tex]\Phi_{B, f}=B A \cos \left(\phi_{f}\right)=B A \cos \left(90^{\circ}\right)=0[/tex]
[tex]\qquad \Phi_{B, f}=0[/tex]
(b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns, that is,
[tex]{\left|\mathcal{E}_{\mathrm{av}}\right|=N\left|\frac{\Phi_{B, f}-\Phi_{B, i}}{\Delta t}\right|}{=} & \frac{1.40 \times 10^{-5} \mathrm{Wb}}{0.030 \mathrm{s}}[/tex]
[tex]& 3.6 \times 10^{-4} \mathrm{V}=0.36 \mathrm{mV}[/tex]
[tex]\mathbb{E}=0.36 \mathrm{mV}[/tex]
(a) The initial and final flux through the coil is 1.75 × 10⁻⁵ Wb and 0 Wb
(b) The induced EMF in the coil is 0.583 mV
Flux and induced EMF:Given that the coil has N = 250 turns
and an area of A = 14cm² = 1.4×10⁻³m².
It is rotated for a time period of Δt = 0.030s such that it is parallel with the earth's magnetic field that is B = 5×10⁻⁵T
(a) The flux passing through the coil is given by:
Ф = NBAcosθ
where θ is the angle between area vector and the magnetic field
The area vector is perpendicular to the plane of the coil.
So, initially, θ = 0°, as area vector and earth's magnetic field both are perpendicular to the plane of the coil
So the initial flux is:
Φ = NABcos0° = NAB
Ф = 250×1.4×10⁻³×5×10⁻⁵ Wb
Ф = 1.75 × 10⁻⁵ Wb
Finally, θ = 90°, and since cos90°, the final flux through the coil is 0
(b) The EMF induced is given by:
E = -ΔФ/Δt
E = -(0 - 1.75 × 10⁻⁵)/0.030
E = 0.583 × 10⁻³ V
E = 0.583 mV
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A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the circle from which light escapes from the liquid into the air above the surface
Answer:
The radius is [tex]r = 3.1905 \ m[/tex]
Explanation:
From the question we are told that
The distance beneath the liquid is [tex]d = 2.70 \ m[/tex]
The refractive index of the liquid is [tex]n_i = 1.31[/tex]
Now the critical value is mathematically represented as
[tex]\theta = sin ^{-1} [\frac{1}{n_i} ][/tex]
substituting values
[tex]\theta = sin ^{-1} [\frac{1}{131} ][/tex]
[tex]\theta = 49.76^o[/tex]
Using SOHCAHTOA rule we have that
[tex]tan \theta = \frac{ r}{d}[/tex]
=> [tex]r = d * tan \theta[/tex]
substituting values
[tex]r = 2.7 * tan (49.76)[/tex]
[tex]r = 3.1905 \ m[/tex]
There is a hydraulic system that by means of a 5 cm diameter plunger to which a 5 N force is applied and that force is transmitted by means of a fluid to a 1 meter diameter plunger. Determine how much force can be lifted by the 1 m diameter plunger,
1) - 234 N
2) - 800 N
3) - 636 N
4) - 600 N
Explanation:
Pressure is the same for both plungers.
P = P
F / A = F / A
F / (¼ π d²) = F / (¼ π d²)
F / d² = F / d²
5 N / (0.05 m)² = F / (1 m)²
F = 2000 N
None of the options are correct.
In Young's 2-slit interference experiment, the wavelength of laser light can be determined. The two slits are separated by 0.16 mm. The screen is 1.4 m from the slits. It is observed that the second bright band is located 11 mm from the center of the pattern. Given this information, what is the wavelength of the laser light?
a. 1258 nm
b. 419 nm
c. 500 nm
d. 629 nm
Answer:
d. 629 nm
Explanation:
slit separation d = .16 x 10⁻³ m
distance of screen D = 1.4 m
distance of second bright band = 11 x 10⁻³
distance of second bright band = 2 x band width
= 2 x λ D /d
Putting the values given ,
11 x 10⁻³ = 2 x λ x 1.4 / .16 x 10⁻³
λ = 1.76 x 10⁻⁶ / 2.8
= .6285 x 10⁻⁶
= 628.5 x 10⁻⁹
= 629 nm approx .
Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.75 keV from the collision. What is the wavelength of the scattered photon?
0.6764*10^-10m
Explanation:
Using E= hc/wavelength
(4.14x10^-15)x(3.0x10^8)/(65x10^-12)=0.1911x10^5 eV=19.1 keV
So subtract the calculated energy from the given energy of scattered photons
9.11-0.75=18.36 keV
To find wavelength
Wavelength= hc/ E
[(4.14x 10^-15)x (3.0x10^8)]/(18.36*10^3) =0.6764^-10 m
What is the maximum speed with which a 1200-kg car can round a turn of radius 94.0 m on a flat road if the coefficient of static friction between tires and road is 0.50?
Answer:
v= 21.47m/sExplanation:
For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road
we will use the expression relating centripetal force and static friction below
let U represent the coefficient of static friction
Given that
U= 0.50
mass m= 1200-kg
radius r= 94.0 m
Assuming g= 9.81 m/s^2
[tex]U*m*g=\frac{mv^2}{r}[/tex]
[tex]U*g=\frac{v^2}{r}[/tex]
substituting our given data in to expression we can solve for the speed V
[tex]0.5*9.81=\frac{v^2}{94}[/tex]
making v the subject of formula we have
[tex]0.5*9.81=\frac{v^2}{94}\\\v= \sqrt{0.5*9.81*94} \\\\v= \sqrt{461.07} \\\\v= 21.47[/tex]
v= 21.47m/s
hence the maximum velocity of the car is 21.47m/s
Someone help me with these questions please!
Answer:
a 25 and b 25
2. 26
60n
Scientists today learn about the world by _____. 1. using untested hypotheses to revise theories 2. observing, measuring, testing, and explaining their ideas 3. formulating conclusions without testing them 4. changing scientific laws
Answer:
Option 2 (observing, measuring, testing, and explaining their ideas) is the correct choice.
Explanation:
A traditional perception of such a scientist is those of an individual who performs experiments in some kind of a white coat. The reality of the situation is, a researcher can indeed be described as an individual interested in the comprehensive as well as a recorded review of the occurrences occurring in nature but perhaps not severely constrained to physics, chemistry as well as biology alone.The other three choices have no relation to a particular task. So the option given here is just the right one.