Answer:
L =( 5.17 10⁻⁶ v) m
Explanation:
This is a diffraction exercise, which is described by the equation
a sin θ = n λ
where a is the width of the slit a = 200 nm = 200 10⁻⁹ m, n is the order of diffraction n = 1 and lan the wavelength in this case we use the DeBroglie relation to find the wavelength
λ = h´ / p = h´ / mv
with h´ = h / 2π
let's use trigonometry for distance,
sin θ = y / L
tan θ = sin θ / cos θ
since these experiments the angle is very small, we approximate
tan θ = sin θ = y / L
we substitute
a y / L = n λ
L = a y / n λ
L = a y 2π mv / n h
we calculate
L = 200 10⁻⁹ 0.300 10⁻² 2π 9.1 10⁻³¹ v / (1 6.63 10⁻³⁴)
L = 517.44 10⁻⁸ v m
L =( 5.17 10⁻⁶ v) m
for a specific value we must know the speed of the electrons
"You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 9.00 s after it was thrown. What is the speed of the rock just before it reaches the water 21.0 m below the point where the rock left your hand
Answer:
48.54 m/s
Explanation:
If the rock takes 9 seconds to reach your position after being thrown, it reaches its maximum height in 4.5 seconds.
The height the rock reaches above your position is ...
h = (1/2)gt^2 = (4.9 m/s^2)(4.5 s)^2 = 99.225 m
This height is an additional 21 m above the water, so the maximum height above the water is ...
99.225 m +21.0 m = 120.225 m
The velocity (v) achieved when falling from this distance is found from ...
v^2 = 2gh
v = √(2(9.8)(120.225)) = √2356.41 ≈ 48.543 . . . . m/s
The speed of the rock when it hits the water is about 48.54 m/s.
When we apply the energy conversation principle to a cylinder rolling down an incline without sliding, we exclude the work done by friction because: A. there is no friction present B. the angular velocity of the center of mass about the point of contact is zero C. the coefficient of kinetic friction is zero D. the linear velocity of the point of contact (relative to the inclined surface) is zero E. the coefficient of static and kinetic friction are equal
Answer:
D. the linear velocity of the point of contact (relative to the inclined surface) is zero
Explanation:
The force of friction emerges only when there is relative velocity between two objects . In case of perfect rolling , there is no sliding so relative velocity between the surface and the point of contact is zero . In other words the velocity of point of contact becomes zero , even though , the whole body is in linear motion . It happens due point of contact having two velocities which are equal and opposite . One of the velocity is in forward direction and the other velocity which is due to rotation is in backward direction . So net velocity of point of contact becomes zero . Due to absence of sliding , displacement due to friction becomes zero . Hence work done by friction becomes zero.
A person drives north 6 blocks, then turns west, and drives 6 blocks. The driver then turns south and drives 6 blocks. How could the driver have made the distance shorter while maintaining the same displacement?
Answer:
Considering that there is no obstructions, he could go west from the start.
Explanation:
Each propeller of the twin-screw ship develops a full-speed thrust of F = 285 kN. In maneuvering the ship, one propeller is turning full speed ahead and the other full speed in reverse. What thrust P must each tug exert on the ship to counteract the effect of the ship's propellers?
Answer:
tug_tug = 570 10³ l
Explanation:
In this problem, each propeller creates a force that makes the boat rotate, so the tugs have to create a die of equal magnitude rep from the opposite direction
∑ τ = 0
F1 la+ (-F1) (-l) = τ-tug
τ-tug = 2 f1 l
τ-tug = 2 28510³ l
tug_tug = 570 10³ l
where the is the distance from the propane axis to the point where the ship turns
This force may be less depending on where the tug is.
Which of the following actions would decrease the energy stored in a parallel plate capacitor when a constant potential difference is applied across the plates? (Choose from: Increasing the area of the plates; Decreasing the area of the plates; Increasing the separation between the plates; Decreasing the separation between the plates; Inserting a material with a higher dielectric constant between the plates
Answer:
increasing the separation between the plates
Explanation:
The increase in the vacuum/separation between the plates in a parallel plate capacitor connected to a constant potential difference decreases the energy stored in the capacitor. the increase in the separation of the plates of a parallel plate capacitor reduces the capacitance of the capacitor because
Q(charge) = CV V = VOLTAGE , c = capacitance
E = 1/2 eAV^2/ D ( ENERGY STORED )
where D = distance between plates, e = dielectric, A = area of capacitor , V = potential difference
A turntable has a moment of inertia of 3.00 x 10-2 kgm2 and spins freely on a frictionless bearing at 25.0 rev/min. A 0.300 kg ball of putty is dropped vertically on the turntable and sticks at a point 0.10m from the center. The total moment of inertia of the system increases, and the turntable slows down. But by what factor does the angular momentum of the system change after the putty is dropped onto the turntable
Answer:
There will be no change in the angular momentum of the system.
Explanation:
Total angular momentum of the system will remain unchanged . We can apply law of conservation of momentum because no external torque is acting on the system . There is increase in the momentum of inertia due to dropping of ball of putty . In order to conserve angular momentum , the system decreases its angular velocity . Hence the final angular momentum remains unchanged .
A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 3.7 mm will experience only elastic deformation when a tensile load of 1890 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.
Answer:
L= 276.4 mm
Explanation:
Given that
E= 180 GPa
d= 3.7 mm
F= 1890 N
ΔL= 0.45 mm
We know that ,elongation due to load F in a cylindrical bar is given as follows
[tex]\Delta L =\dfrac{FL}{AE}[/tex]
[tex]L=\dfrac{\Delta L\times AE}{F}[/tex]
Now by putting the values in the above equation we get
[tex]L=\dfrac{0.45\times 10^{-3}\times \dfrac{\pi}{4}\times (3.7\times 10^{-3})^2\times 108\times 10^9}{1890}\ m[/tex]
L=0.2764 m
L= 276.4 mm
Therefore the length of the specimen will be 276.4 mm
2. Air at a temperature of 20 ºC passes through a pipe with a constant velocity of 40 m/s. The pipe goes through a heat exchanger in which hot gases outside the pipe heat up the air to 820 ºC. It then enters a turbine with a velocity of 40 m/s and expands till the temperature falls to 620 ºC. The air stream loses 4.3 kW heat in the turbine. If the air flow rate is 2.5 kg/s, calculate (a) How much heat is transferred to the air in the heat exchanger. (b) The power output of the turbine.
Answer:
a) Q = 1436 kW
b) P ≈ 776 kW
Explanation:
Let's begin by listing out the given parameters:
T1 = 20 °C, u = 40 m/s, T2 = 820 °C, P = 4.3 kW, m = 2.5 kg/s, T3 = 510 °C, V1 = 40 m/s,
V2 = 40 m/s, V3 = 55 m/s, ṁ = 2.5 kg/s
To solve the question, we make this assumption that the size of the pipe is constant
a) No change in velocity implies that heat added is isochoric
Q = m * C * ΔT
Cv of air at 300 K(≈20 °C) = 0.718
Q = 2.5 * 0.718 * (820 − 20)
Q = 1436 kW
b) P = ṁ * Cp * ΔT + ṁ * (V2² - V3²) ÷ 2000] - Ql
V2² - V3² = 55² - 40² = 1425
ΔT = T2 - T3 = 820 - 510 = 310 °C
Cp of air at 300 K(≠20 °C) = 1.005 kJ/kgK
Ql = 4.3 kW
P = 2.5 * (1.005 * 310) + 2.5 * (1425 ÷ 2000) - 4.3
P = 778.875 + 1.78125 - 4.3 = 776.35625
P ≈ 776 kW
A"boat"is"moving"to"the"right"at"5"m/s"with"respect"to"the"water."A"wave"moving"to"the"left,"opposite"the"motion"of"the"boat."The"waves"have"2.0"m"between"the"top"of"the"crests"and"the"bottom"of"the"troughs."The"period"of"the"wave"is"8.3"s"and"their"wavelength"is"110"m."At"one"instant"the"boat"sits"on"a"crest"of"the"wave,"20"seconds"later,"what"is"the"vertical"displacement"of"the"boat
Answer:
0.99m
Explanation:
Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:
[tex]v=\lambda f=\lambda\frac{1}{T}=(110m)\frac{1}{8.3s}=13.25\frac{m}{s}[/tex]
the relative velocity is:
[tex]v'=13.25m/s-5m/s=8.25\frac{m}{s}[/tex]
This velocity is used to know which is the distance traveled by the boat after 20 seconds:
[tex]x'=v't=(8.25m/s)(20s)=165m[/tex]
Next, you use the general for of a wave:
[tex]f(x,t)=Acos(kx-\omega t)=Acos(\frac{2\pi}{\lambda}x-\omega t)[/tex]
you take the amplitude as 2.0/2 = 1.0m.
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{8.3s}=0.75\frac{rad}{s}[/tex]
by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:
[tex]f(165,20)=1.0m\ cos(\frac{2\pi}{110m}(165)-(0.75\frac{rad}{s})(20s))\\\\f(165,20)=0.99m[/tex]
When a star explodes as a supernova, the expanding gas can pass through the interstellar medium triggering...
A.) an increase in variable star activity
B.) formation of new stars
C.) life to develop
D.) additional supernovas
B. Formation of new stars
A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed uniformly throughout the volume of the shell. What is the magnitude of the electric field at a distance r = 11.2 cm from the center of the shell? (ε0 = 8.85 × 10-12 C2/N ∙ m2) (Give your answer to the nearest 0.01 MN/C)
Answer:
E = 1580594.95 N/C
Explanation:
To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:
[tex]\int EdS=\frac{Q_{in}}{\epsilon_o}[/tex] (1)
dS: differential of the Gaussian surface
Qin: charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2
The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:
[tex]\int EdS=ES=E(4\pi r^2)[/tex] (2)
Qin is calculate by using the charge density:
[tex]Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho[/tex] (3)
Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.
The charge density is given by:
[tex]\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}[/tex]
Next, you use the results of (3), (2) and (1):
[tex]E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})[/tex]
Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:
[tex]E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}[/tex]
hence, the electric field is 1580594.95 N/C
The space probe Deep Space 1 was launched on October 24th, 1998 and it used a type of engine called an ion propulsion drive. An ion propulsion drive generates only a weak force (or thrust), but can do so for long periods of time using only a small amount of fuel. Suppose the probe, which has a mass of 474 kg is travelling at an initial speed of 275 m/s. No forces act on it except the 5.60 x 10⁻² N thrust from the engine. This external force is directed PARALLEL to the displacement. The displacement has a magnitude of 2.42 x 10⁹ m. {PART A} Calculate the INITIAL kinetic energy of the probe [2 marks] {PART B} Find the work done BY THE ENGINE on the space probe [2 marks] {PART C} Calculate the FINAL KINETIC ENERGY of the probe (Hint W=∆E) [2 marks] {PART D} Determine the final speed of the probe, assuming that its mass remains constant [3 marks]
Answer:
Explanation:
mass of probe m = 474 Kg
initial speed u = 275 m /s
force acting on it F = 5.6 x 10⁻² N
displacement s = 2.42 x 10⁹ m
A )
initial kinetic energy = 1/2 m u² , m is mass of probe.
= .5 x 474 x 275²
= 17923125 J
B )
work done by engine
= force x displacement
= 5.6 x 10⁻² x 2.42 x 10⁹
= 13.55 x 10⁷ J
C ) Final kinetic energy
= Initial K E + work done by force on it
= 17923125 +13.55 x 10⁷
= 1.79 x 10⁷ + 13.55 x 10⁷
= 15.34 x 10⁷ J
D ) If v be its velocity
1/2 m v² = 15.34 x 10⁷
1/2 x 474 x v² = 15.34 x 10⁷
v² = 64.72 x 10⁴
v = 8.04 x 10² m /s
= 804 m /s
Yellow light with wavelength 600 nm is travelling to the left (in the negative x direction) in vacuum. The light is polarized along the z direction. (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave. (c) If the wave were to represent blue light instead of yellow light, how would your pictures in parts a and b change? If there is no change, say so explicitly.
Answer: (a) and (b) => check attached file.
(c). Picture (a) and (b) will both remain the same.
Explanation:
IMPORTANT: The solution to the question (a) and (b) that is (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave is there in the ATTACHED FILE/PICTURE.
It is also worthy of note to know that in anything Electromagnetic wave, the magnetic field, the Electric Field and their direction of propagation are perpendicular to each other.
Therefore, knowing the fact above we can say that in yellow light, the magnetic field is in the y-direction and the Electric Field is in the z-direction.
Hence, the solution to option C is given below;
(C).If the wave were to represent blue light instead of yellow light, picture (a) will remain the same because both light are Electromagnetic wave, although the wavelength will have to change. Picture (b) will also remain the same because they are both Electromagnetic waves and possess similar properties.
Three packing crates of masses, M1 = 6 kg, M2 = 2 kg
and M3 = 8 kg are connected by a light string of
negligible mass that passes over the pulley as shown.
Masses M1 and M3 lies on a 30o
incline plane which
slides down the plane. The coefficient of kinetic friction
on the incline plane is 0.28.
Determine the acceleration of the system.
Answer:
a = 2.5 m / s²
Explanation:
This is an exercise of Newton's second law, in this case we fix a coordinate system with the x axis parallel to the plane with positive direction
Let's write the second law for bodies in the inclined plane
W₁ₓ + W₃ₓ - fr = (m₁ + m₃) a
N₁ - [tex]W_{1y}[/tex] + N₃- W_{3y} = 0
N₁ + N₃ = W_{1y} + W_{3y}
let's use trigonometry to find the weight components
sin 30 = Wₓ/ W
Wₓ = W sin 30
cos 30 = W_{y} / W
W_{y} = W cos 30
we substitute
N₁+ N₃ = W₁ cos 30 + W₃ cos 30
W₁ₓ + W₃ₓ - μ (m₁ + m₃) g cos30 = (m₁ + m₃) a
a = (m₁g sin 30 + m₃g sin 30 - μ (m₁ + m₃) g cos 30) / (m₁ + m₃)
a = g sin 30 - μ g cos30
let's calculate
a = 9.8 sin 30 - 0.28 9.8 cos 30
a = 4.9 - 2,376
a = 2.5 m / s²
You could use an analytical or triple beam balance to determine a ___ called ____
A)
physical property; mass.
B)
chemical property, mass.
C)
physical property; weight.
D)
physical property; density.
Answer:
a and b are the correct answers
Explanation:
Answer:
A) physical property; mass.
Explanation:
took the test
Now that you've done your research on the law of supply, you understand that it basically asserts that how much coffee you'd be willing to supply depends on how much money you can make for each cup.
A 20-kilogram box is sitting on an inclined plane with a 30-degree slope. If the box is at rest, what is the force
on the box due to friction?
Answer:
Explanation:
The box is at rest because , the component of weight along the surface downwards is equal to frictional force .
frictional force = mgsinθ
= 20 x 9.8 sin 30
= 98 N .
A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2300 kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.75, and the cars slide to a halt at a point 5.54 m west and 6.19 m south of the impact point. How fast was sedan traveling just before the collision? How fast was SUV traveling just before the collision?
Answer:
Explanation:
momentum of sedan of 1600 kg = 1600x v , where v is its velocity
momentum of suv of 2300 kg = 2300 x u where u is its velocity .
force of friction = ( 1600 + 2300 ) x 9.8 x .75 ( fiction = μ mg )
= 28665 N
distance by which friction acted = √ (5.54² + 6.19²)
= 8.3 m
work done by friction
= 28665 x 8.3
= 237919.5 J
Total kinetic energy of cars = work done by friction
1/2 x 1600 x v² + 1/2 x 2300 u² = 237919.5
16 v² + 23 u² = 4758.4
1600 x v / 2300 u = 6.19 / 5.54
v / u = 1.6
v = 1.6 u
putting this equation in fist equation
40.96 u² + 23 u² = 4758.4
= 63.96 u² = 4758.4
u² = 74.4
u = 8.62 m /s
v = 13.8 m /s
Exercise should challenge your body and be at a greater intensity than your usual bif daily activity. Discuss
Answer:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
Explanation:
The main component of all computer memory is
Answer: R.A.M
Explanation:
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 210 degrees Fahrenheit when freshly poured, and 2.5 minutes later has cooled to 191 degrees in a room at 64 degrees, determine when the coffee reaches a temperature of 156 degrees.
Answer:
Explanation:
The problem is based on Newton's law of cooling .
According to Newton's law
dQ / dt = k ( T - T₀ ) ,
dT / dt = k' ( T - T₀ ) ; dT / dt is rate of fall of temperature.
T is average temperature of hot body , T₀ is temperature of surrounding .
In the first case rate of fall of temperature = (210 - 191) / 2.5
= 7.6 degree / s
average temperature T = (210 + 191) /2
= 200.5
Putting in the equation
7.6 = k' ( 200.5 - 64 )
k' = 7.6 / 136.5
= .055677
In the second case :---
In the second case, rate of fall of temperature = (191 - 156) / t
= 35 / t , t is time required.
average temperature T = (156 + 191) /2
= 173.5
Putting in the equation
35 / t = .05567 ( 173.5 - 64 )
t = 5.74 minute .
the medium of a mechanical wave is the matter through wich it travels? tru or false
Answer:
it can only travel through matter
Explanation:
the mechanical wave can be any type of matter
A landscaper is shopping for landscaping materials. She wants to use materials through which water flows easily.
Which materials should she choose? Select three options.
clay
gravel
granite
rocks with cracks
loosely packed soil
Next
Mark this and retum
Save and Exit
Submit
Answer:
He needs clay gravel and rocks with cracks
Which of the following is correct? When a current carrying wire is in your left hand, thumb in the direction of the magnetic field lines, your fingers point in the direction of the current. When a current carrying wire is in your left hand, thumb in the direction of the current, your fingers point in the direction of the magnetic field lines. When a current carrying wire is in your right hand, thumb in the direction of the magnetic field lines, your fingers point in the direction of the current. When a current carrying wire is in your right hand, thumb in the direction of the current, your fingers point in the direction of the magnetic field lines.
Answer:
When a current carrying wire is in your left hand, thumb in the direction of the magnetic field lines, your fingers point in the direction of the current
Explanation:
This is in line with the left hand rule
B.) When the current flows up the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
1. Deri had a large tank of oil (s-0.8) and was requested to determine the viscosity of that fluid. To assist with the process, she was given a 0.25-inch-diameter steel ball (sphere, s=8.0) to conduct the test. From the tests, she found that the terminal velocity of the sphere was 2.5 fpm. What is the viscosity of the oil? Remember, the volume of a sphere is (pi D3 /6). ANS. viscosity is 0.258 lb-s/ft2
Answer:
0.25916 lb-s/ft^2
Explanation:
Given:-
- The specific gravity of oil, SGo = 0.8
- The specific gravity of sphere, SGo = 8
- Terminal velocity of sphere, v = 2.5 fpm
- The diameter of sphere, D = 0.25 in
Find:-
What is the viscosity of the oil?
Solution:-
- Consider a sphere completely submerged into oil and travelling with terminal velocity ( v ).
- Develop a free body diagram for the sphere. There are forces acting on the sphere.
- The downward acting force is due to the weight of the sphere ( W ):
[tex]W = m_s*g[/tex]
Where,
The mass ( m_s ) of the sphere is given as:
[tex]m_s = S.G_s*p_w*V_s[/tex]
Where,
ρ_w : Density of water = 1.940 slugs/ft3
V_s: The volume of object ( sphere )
- The volume of sphere is expressed as a function of radius:
[tex]V_s = \frac{\pi *D^3}{6}[/tex]
Hence,
[tex]W = S.G_s*p_w*\frac{\pi*D^3 }{6}* g\\\\W = 8*1.940*\frac{\pi*(0.25/12)^3 }{6}*32\\\\W = 0.00235 lb[/tex]
- One of the upward acting force is the buoyant force ( Fb ) that is proportional to the volume of fluid displaced by the immersed object.
- The buoyant force ( Fb ) is given by:
[tex]F_b = S.G_o*p_w*V_s*g[/tex]
- Therefore the buoyant force ( Fb ) becomes:
[tex]F_b = 0.8*1.94*\frac{\pi*(0.25/12)^3 }{6} *32\\\\F_b = (4.73451*10^-^6)*(49.664)\\\\F_b = 0.00023 lb[/tex]
- The other upward acting force is the frictional drag ( F_d ) i.e the resistive frictional force acting on the contact points of the sphere and the fluid oil.
- From stokes formulations the drag force acting on a spherical object which is completely immersed in a fluid is given as:
[tex]F_d = 3*\pi*D*u*v[/tex]
Where,
μ: The viscosity of fluid
v : The velocity of object
Therefore,
[tex]F_d = 3*\pi*\frac{0.25}{12} *u*0.041666\\\\F_d = 0.00818*u\\[/tex]
- Apply Newton's second law of motion for the sphere travelling in the fluid:
[tex]F_n_e_t = m_s*a[/tex]
Where,
a: Acceleration of object = 0 ( Terminal velocity condition )
[tex]F_n_e_t = 0[/tex]
- Plug in the three forces acting on the metal sphere:
[tex]F_d + F_b - W = 0\\\\F_d = W - F_b\\\\0.00818*u = 0.00235 - 0.00023\\\\u = \frac{0.00212}{0.00818} = 0.25916 \frac{lb-s}{ft^2}[/tex]
An experiment based at New Mexico’s Apache Point observatory uses a laser beam to measure the distance to the Moon with millimeter precision. The laser power is 120 GW, although it’s pulsed on for only 90 ps. The beam emerges from the laser with a diameter of 7.0 mm. It’s then beamed into a telescope aimed at the Moon. When the beam leaves the telescope, it has the telescope’s full 3.5-mm diameter. By the time it reaches the Moon, the beam has expanded to a diameter of 6.5 km.
a. Find the intensity of the beam as it leaves the laser. Express your answer with the appropriate units.
b. Find the intensity of the beam as it leaves the telescope. Express your answer with the appropriate units.
Answer:
Sorry but i dont know
Explanation:
Which is a characteristic of diatoms?
A)whip-like tail
B)heterotrophic
C)multicellular
D)glass-ive cell wall
Answer:
b
Explanation:
Answer:
The correct answer is D, diatoms have glass-like cell walls.
Gerry is looking at salt under a powerful microscope and notices a crystalline structure. What can be known about the salt sample that Gerry is looking at?
1. The atoms have spread out from each other.
2.The atoms are sliding past each other.
3.The atoms have no particular pattern
4.The atoms are vibrating in place
Answer:
4.The atoms are vibrating in place
Explanation:
The answer is; The atoms are vibrating in place
We know that, molecules in the crystal have a definite position in the crystal and are bonded to each other by electrostatic forces. However, since the molecules have some energy, they vibrate in their positions. Their energy, however, is not high enough to cause them to overcome the strong bonding (unless the crystal is heated or the atoms are irradiated).
A convex mirror of focal length 33 cm forms an image of a soda bottle at a distance of 19 cm behind the mirror.If the height of the image is 7.0 cm,where is the object located,and how tall is it? What is the magnification of the image? Is the image virtual or real? Is the image inverted or upright? Draw a ray diagram to confirm your results.
Answer:
Image distance = 44.8cm, Image height = 16.5cm, Magnification = 0.42
The image is a virtual and upright image.
Explanation:
The nature of image formed by an object placed in front of a convex mirror is always diminished, virtual and erect.
The focal length f and the image distance are always NEGATIVE beacause the image is formed behind the mirror.
Given f = -33.0cm, v = -19.0cm
using thr mirror formula to get the object distance u, we have;
[tex]\frac{1}{f}=\frac{1}{u} + \frac{1}{v}\\ \frac{1}{u}=\frac{1}{f} - \frac{1}{v}\\\frac{1}{u}=\frac{1}{-33} - \frac{1}{-19}\\\frac{1}{u}=\frac{-19+33}{627} \\\frac{1}{u}=\frac{14}{627} \\u=\frac{627}{14} \\u = 44.8cm[/tex]
To calculate the image height, we will use the magnification formula
M = [tex]\frac{image\ height}{object\ height}=\frac{image\ distance}{object\ distance} \\[/tex]
M = [tex]\frac{Hi}{HI}=\frac{v}{u}[/tex]
Given Hi = 7.0cm
v = 19.0cm
u = 44.8cm
HI = 7*44.8/19
HI = 16.5cm
The object height is 16.5cm
Magnification = v/u = 19.0/44.8 = 0.42
SInce the image is formed behind the mirror, the image is a VIRTUAL and UPRIGHT image
a 1200 kg trailer is hitched to a 1400 kg car. the car and trailer are traveling at 72 km.h when the driver applies the brakes on both the car and the trailer. knowing that the braking forces exerted on the car and the trailer are 5000 N and 4000 N respectively, determine (a) the distance traveled by the car and trailer before they come to a stop and (b) the horizontal component of the force exerted by the trailer hitch
Answer:
a) 8.67m
b) 1000N
Explanation:
(a) To find the distance you use the second Newton Law for both car and trailer, in order to calculate the dis-acceleration of the system:
[tex]F=ma\\\\a_=\frac{F}{m}=\frac{5000N+4000N}{1400kg+1200kg}=3.46\frac{m}{s^2}[/tex]
once you have this value, you use the the following kinematic equation to calculate the distance traveled by both car and trailer:
[tex]v^2=v_o^2-2ax\\\\x=\frac{-v^2+v_o^2}{2a}[/tex]
v: final velocity=0
vo: initial velocity = 72km/h = 60 m/s
by replacing the values of these parameters you obtain for x:
[tex]x=\frac{-0m/s+60m/s}{2(3.46m/s^2)}\\\\x=8.67m[/tex]
(b) The horizontal component of the force exerted by the trailer hitch is given by:
[tex]F_T=5000N-4000N=1000N[/tex]
