dharna is said to be concentration.. it is true
c) If the ice block (no penguins) is pressed down even with the surface and then released, it will bounce up and down, until friction causes it to settle back to the equilibrium position. Ignoring friction, what maximum height will it reach above the surface
Answer:
y = 20.99 V / A
there is no friction y = 20.99 h
Explanation:
Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy
when the block of ice is submerged it is subjected to two forces its weight hydrostatic thrust
F_net= ∑F = B-W
the expression stop pushing is
B = ρ_water g V_ice
where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice
We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³
W = ρ-ice g V
F_net = (ρ_water - ρ_ ice) g V
this is the net force directed upwards, we can find the potential energy with the expression
F = -dU / dy
ΔU = - ∫ F dy
ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy
ΔU = - (ρ_water - ρ_ ice) g A y² / 2
we evaluate between the limits y = 0, U = 0, that is, the potential energy is zero at the surface
U_ice = (ρ_water - ρ_ ice) g A y² / 2
now we can use the conservation of mechanical energy
starting point. Ice depth point
Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2
final point. Highest point of the block
[tex]Em_{f}[/tex] = U = m g y
as there is no friction, energy is conserved
Em₀ = Em_{f}
(ρ_water - ρ_ ice) g A y² / 2 = mg y
let's write the weight of the block as a function of its density
ρ_ice = m / V
m = ρ_ice V
we substitute
(ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y
y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A
let's substitute the values
y = 0.913 / (1 - 0.913) 2 V / A
y = 20.99 V / A
This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction
V / A = h
where h is the height of the block
y = 20.99 h
1. Rank the transformers on the basis of their rms secondary voltage. Rank from largest to smallest.
Vp = 240 V; Np = 1000 turns; Ns = 2000 turns
Vp = 480 V; Np = 4000 turns; Ns = 2000 turns
Vp = 480 V; Np = 2000 turns; Ns = 1000 turns
Vp = 120 V; Np = 500 turns; Ns = 2000 turns
Vp = 240 V; Np = 1000 turns; Ns = 500 turns
2. 100 A of rms current is incident on the primary side of each transformer. Rank the transformers on the basis of their rms secondary current. Rank from largest to smallest.
Vp = 240 V; Np = 1000 turns; Ns = 2000 turns
Vp = 480 V; Np = 2000 turns; Ns = 1000 turns
Vp = 240 V; Np = 1000 turns; Ns = 500 turns
Vp = 120 V; Np = 500 turns; Ns = 2000 turns
Vp = 480 V; Np = 4000 turns; Ns = 2000 turns
Answer:
1. Transformer 3> Transformer 1 and 2 > Transformer 4
2. Transformer 2,3,5 > Transformer 1 > Transformer 4
Explanation:
From;
Vs/Vp = Ns/Np
Where;
Vp= voltage in primary coil
Vs= voltage in secondary coil
Ns= number of turns in secondary coil
Np= number of turns in primary coil.
Vs= Ns×Vp/Np
Vs= 480 ×2000/4000
Vs= 240 V
Vs= 480 ×1000/2000
Vs=240 V
Vs= 120 × 2000/500
Vs= 480 V
Vs= 240 × 500/1000
Vs= 120 V
2. Ns/Np= Ip/Is
Is= Np×Ip/Ns
Is= 1000 × 100/2000
Is= 50 A
Is= 2000 × 100/1000
Is= 200 A
Is= 1000 × 100/500
Is= 200 A
Is= 500 × 100/2000
Is= 25 A
Is= 4000 × 100/2000
Is= 200 A
The same force is applied to two hoops. The hoops have the same mass, but the larger hoop has twice the radius. How are the angular accelerations of the hoops related
Answer:
The angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex].
Explanation:
Net force on the hoop is given by;
[tex]F_{net} = ma[/tex]
where;
a is linear acceleration
m is the mass
Net torque on the hoop is given by;
[tex]\tau_{net} =I\alpha[/tex]
where;
I is moment of inertia
α is the angular acceleration
But, τ = Fr
[tex]Fr = I \alpha\\\\\alpha = \frac{Fr}{I} \\\\\alpha = \frac{Fr}{mr^2} \\\\\alpha = \frac{F}{mr} \\\\\alpha = \frac{1}{r} (\frac{F}{m} )\\\\(since\ the \ force\ and \ mass \ are \ the \ same, \frac{F}{m} = constant=k)\\\\ \alpha = \frac{k}{r}\\\\k = \alpha r[/tex]
[tex]\alpha _1 r_1= \alpha_2 r_2[/tex]
let the angular acceleration of the smaller hoop = α₁
let the radius of the smaller hoop = r₁
then, the radius of the larger loop, r₂ = 2r₁
let the angular acceleration of the larger hoop = α₂
[tex]\alpha _1 r_1= \alpha_2 r_2\\\\\alpha_2= \frac{ \alpha _1 r_1}{r_2} \\\\\alpha_2=\frac{\alpha _1 r_1}{2r_1} \\\\\alpha_2= \frac{\alpha _1}{2} \\\\\alpha _1 = 2\alpha_2[/tex]
Therefore, the angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex]
Suppose you want a telescope that would allow you to see distinguishing features as small as 3.5 km on the Moon some 384,000 km away. Assume an average wavelength of 550 nm for the light received.Required:What is the minimum diameter mirror on a telescope?
Explanation:
[tex]\theta=1.22 \frac{\lambda}{D}[/tex]
And, from equation ( 2 ), we get
[tex]\theta=\frac{x}{d}[/tex]
Thus,
[tex]\frac{x}{d} &=1.22 \frac{\lambda}{D}[/tex]
[tex]D &=1.22 \frac{\lambda d}{x}[/tex]
[tex]=1.22 \frac{550 \times 10^{-9} 3.84 \times 10^{8}}{5 \times 10^{3}}[/tex]
[tex]=0.0515 \mathrm{m}[/tex]
Thus, the diameter of the telescope's mirror that would allow us to see details as small as is
A person is being pulled by gravity with a force of 500 N. What is the force with which the person pulls Earth?
1,000 N
O100 N
500 N
0 250 N
Answer:
The correct answer is 500 N
Explanation:
This is an exercise in Newton's third law or law of action and reaction
The Earth exerts a force on the person, which we call a weight of 500 N directed downwards, we can call this action and the person exerts a force on the Earth of equal magnitude 500N and in the opposite direction, that is directed upwards.
Which force we call action does not matter, the analysis and conclusions are the same
The correct answer is 500N
A 13.6 kg block is tied at the top of an incline to a tree. If the incline is 35.5 degrees and the coefficient of friction between the sled and the incline is .45, What is the tension force between the block and the tree
Answer:
Explanation:
ASSUMING that block = sled AND that the rope is parallel to the slope.
The force acting parallel due to the weight is
13.6(9.81)sin35.5 = 77.475 N
The maximum friction force is
(0.45)13.6(9.81)cos35.5 = 48.877 N
If rope tension is T
77.475 - 48.877 < T < 77.475 + 48.877
28.6 N < T < 126 N
28.6 N will occur if the block is on the verge of sliding downhill
126 N will occur if the block is on the verge of sliding uphill
Could be any value between them.
The magnitude of the magnetic field at point P for a certain electromagnetic wave is 2.12 μT. What is the magnitude of the electric field for that wave at P? (c = 3.0 × 108 m/s)
Answer:
The electric field is [tex]E = 636 \ V/m[/tex]
Explanation:
From the question we are told that
The magnitude of magnetic field is [tex]B = 2.12 \mu T = 2.12*10^{-6} \ T[/tex]
The value for speed of light is [tex]c = 3.0 *10^8 \ m/s[/tex]
Generally the magnitude of the electric field at point P is
[tex]E = B * c[/tex]
substituting values
[tex]E = 2.12 *10^{-6} * 3.0 *10^{8}[/tex]
[tex]E = 636 \ V/m[/tex]
The magnitude of electric field for the wave at point P is 636 V/m.
Given data:
The strength of magnetic field at point P is, [tex]B = 2.12 \;\rm \mu T=2.12 \times 10^{-6} \;\rm T[/tex].
The speed of light is, [tex]c = 3.0 \times 10^{8} \;\rm m/s[/tex].
The given problem is based on the concept of electric field and magnetic field. The electromagnetic wave works on the principle of oscillating magnetic field and electric field at the same region. We can find any of the two using the expression,
[tex]E = B \times c[/tex]
here,
E is the strength of electric field.
Solving as,
[tex]E = (2.12 \times 10^{-6}) \times (3 \times 10^{8})\\\\E = 636 \;\rm V/m[/tex]
Thus, we can conclude that the magnitude of electric field for the wave at point P is 636 V/m.
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If a nucleus decays by successive b, a, a emissions, its mass number will Group of answer choices decrease by seven. decrease by two. decrease by four. decrease by eight. increase by four.
Answer:
The mass number will decrease by eight (8).
Explanation:
Given;
successive beta (b), alpha (a), alpha (a) emissions.
Generally, when a radioactive element emits a beta-particle (b), its mass number doesn't increase but its atomic number increases by 1 . [tex](^{0}_{-1}\beta )[/tex]
Also when a radioactive element emits an alpha-particle (a), its mass number decreases by 4, while its atomic number decrease by 2. [tex](^4_2\alpha)[/tex]
For the given question, a successive beta (b), alpha (a), and alpha (a) emissions = (0) + (-4) + (-4) = -8
Thus, when a radioactive element emit a successive beta (b), alpha (a), alpha (a) particles, the mass number will decrease by eight (8).
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the molecule of magnet are independent _____________
Answer:
The first is the electric field, which describes the force acting on a stationary charge and gives the component of the force that is independent of motion. The magnetic field, in contrast, describes the component of the force that is proportional to both the speed and direction of charged particles.
What is the direction of the net gravitational force on the mass at the origin due to the other two masses?
Answer:
genus yds it's the
Explanation:
xmgxfjxfjxgdfjusufzjyhmfndVFHggssjtjhryfjftjsrhrythhrsrhrhsfhsgdagdah vhj
What is the mass of a rectangular block of
density 2.5 ×10³ k gm³that measures 10cm by 5 cm by 4 cm?
A. 0.002 kg
B. 0.080 kg
C. 0.200 kg
D. 0.500 kg
E. 1.000 kg
Answer:
Option (D) : 0.5 kg
Explanation:
[tex]mass = density \times volume[/tex]
[tex]mass = {2500} \times 0.1 \times 0.05 \times 0.04[/tex]
Mass of block = 0.5 kg
the mass of a rectangular block of density 2.5 ×10³ k gm³ that measures 10cm by 5 cm by 4 cm is 0.5 kg.
What is density ?Density is the ratio of mass to volume. it tells how much mass a body is having for its unit volume. for example egg yolk has 1027kg/m³ of density, means if we collect numbers of egg yolk and keep it in a container having volume 1 m³ then total amount of mass it is having will be 1027kg. Density is a scalar quantity. when we add egg yolk into the water, egg yolk has greater density than water( 997 kg/m³), because of higher density of egg yolk it contains higher mass in same volume as water. hence due to higher mass higher gravitational force is acting on the egg yolk therefore it goes down on the inside the water. water will float upon the egg yolk. same situation we have seen when we spread oil in the water. ( in that case water has higher density than oil. thats why oil floats on the water)
The Volume of the block is,
V = LBD, where L = length, B = breadth , D = depth of the block.
V = 10 × 5 × 4 = 200 cm³
Density of Block = 2.5 ×10³ kg/m³
Density = Mass / Volume
2.5 ×10³ kg/m³ = Mass / 200 cm³
2.5 ×10³ kg/m³ × 200 cm³ = Mass
2.5 ×10³ kg/m³ × 0.2 × 10⁻³ m³ = Mass
Mass = 0.5 kg
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Calculate the electromotive force produced by each of the battery combinations shown in the figure, if the emf of each is 1.5 V.
Answer:
A) 1.5 V
B) 4.5 V
Explanation:
A) Batteries in parallel have the same voltage as an individual battery.
V = 1.5 V
B) Batteries in series have a voltage equal to the sum of the individual batteries.
V = 1.5 V + 1.5 V + 1.5 V
V = 4.5 V
A concrete slab shown in Figure 5 is being lifted by using three cables connected to the slab at points A, B and C. The slab is in the xy plane. The vertical force required to lift this slab is 60 kN (F 60 kN). Find the tensions in cables DA, DB and DC (show all your workings that you do to find these)
Answer:
Fad = 28.8 kN
Fbd = 16.4 kN
Fcd = 28.1 kN
Explanation:
First, find the length of each cable.
AD = √((2 m)² + (0.5 m)² + (2.5 m)²)
AD = √10.5 m
AD ≈ 3.24 m
BD = √((1.5 m)² + (1 m)² + (2.5 m)²)
BD = √9.5 m
BD ≈ 3.08 m
CD = √((1 m)² + (1 m)² + (2.5 m)²)
CD = √8.25 m
CD ≈ 2.87 m
Next, use similar triangles to find the x, y, and z components of each tension force.
Fadx = 2/3.24 Fad = 0.617 Fad
Fady = 0.5/3.24 Fad = 0.154 Fad
Fadz = 2.5/3.24 Fad = 0.772 Fad
Fbdx = 1.5/3.08 Fbd = 0.487 Fbd
Fbdy = 1/3.08 Fbd = 0.324 Fbd
Fbdz = 2.5 / 3.08 Fbd = 0.811 Fbd
Fcdx = 1/2.87 Fcd = 0.348 Fcd
Fcdy = 1/2.87 Fcd = 0.348 Fcd
Fcdz = 2.5/2.87 Fcd = 0.870 Fcd
Now sum the forces in the x, y, and z directions:
∑Fx = ma
-0.617 Fad + 0.487 Fbd + 0.348 Fcd = 0
∑Fy = ma
-0.154 Fad − 0.324 Fbd + 0.348 Fcd = 0
∑Fz = ma
60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0
To solve this system of equations algebraically, start by subtracting the first two equations, eliminating Fcd.
-0.463 Fad + 0.811 Fbd = 0
0.811 Fbd = 0.463 Fad
Fbd = 0.571 Fad
Substitute into either of the first two equations:
-0.617 Fad + 0.487 (0.571 Fad) + 0.348 Fcd = 0
-0.617 Fad + 0.278 Fad + 0.348 Fcd = 0
-0.339 Fad + 0.348 Fcd = 0
0.348 Fcd = 0.339 Fad
Fcd = 0.975 Fad
Now substituting into the third equation:
60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0
60 kN − 0.772 Fad − 0.811 (0.571 Fad) − 0.870 (0.975 Fad) = 0
60 kN − 0.772 Fad − 0.463 Fad − 0.849 Fad = 0
60 kN − 2.083 Fad = 0
Fad = 28.8 kN
Solving for the other two tension forces:
Fbd = 0.571 Fad = 16.4 kN
Fcd = 0.975 Fad = 28.1 kN
Answer:
Tensions of:
DA = 28.81 KN
DB = 16.45 KN
DC = 28.07 KN
Explanation:
see attached
What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×10-34 Js.)
Answer:
9.82 × [tex]10^{-35}[/tex] Hz
Explanation:
De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:
λ = [tex]\frac{h}{mv}[/tex]
where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.
Given that: h = 6.63 ×[tex]10^{-34}[/tex] Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;
λ = [tex]\frac{h}{mv}[/tex]
= [tex]\frac{6.63*10^{-34} }{2.5*2.7}[/tex]
= [tex]\frac{6.63 * 10^{-34} }{6.75}[/tex]
= 9.8222 × [tex]10^{-35}[/tex]
The wavelength of the object is 9.82 × [tex]10^{-35}[/tex] Hz.
You need to repair a broken fence in your yard. The hole in your fence is
around 3 meters in length and for whatever reason, the store you go to
has oddly specific width 20cm wood. Each plank of wood costs $16.20,
how much will it cost to repair your fence? (Hint: 1 meter = 100 cm) *
Answer:
cost = $ 243.00
Explanation:
This exercise must assume that it uses a complete table for each piece, we can use a direct ratio of proportions, if 1 table is 0.20 m wide, how many tables will be 3.00 m
#_tables = 3 m (1 / 0.20 m)
#_tables = 15 tables
Let's use another direct ratio, or rule of three, for cost. If a board costs $ 16.20, how much do 15 boards cost?
Cost = 15 (16.20 / 1)
cost = $ 243.00
Megan accelerates her skateboard from 0 m/s to 8 m/s in 2 seconds. What is the magnitude of the acceleration of the skateboard?
O 8 m/s^2
O 16 m/s^2
O 2 m/s^2
O 4 m/s^2
Answer:
chk picture for eqn
Explanation:
In a velocity selector having electric field E and magnetic field B, the velocity selected for positively charged particles is v= E/B. The formula is the same for a negatively charged particles.
a. True
b. False
Answer:
True or False
Explanation:
Because.....
easy 50% chance you are right
10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ????? How far will this pulse travel in the same time if the tension is doubled?
Answer: Tension = 47.8N, Δx = 11.5×[tex]10^{-6}[/tex] m.
Tension = 95.6N, Δx = 15.4×[tex]10^{-5}[/tex] m
Explanation: A speed of wave on a string under a tension force can be calculated as:
[tex]|v| = \sqrt{\frac{F_{T}}{\mu} }[/tex]
[tex]F_{T}[/tex] is tension force (N)
μ is linear density (kg/m)
Determining velocity:
[tex]|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{0.00874 }[/tex]
[tex]|v| =[/tex] 0.0935 m/s
The displacement a pulse traveled in 1.23ms:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 9.35.10^{-2}*1.23.10^{-3}[/tex]
Δx = 11.5×[tex]10^{-6}[/tex]
With tension of 47.8N, a pulse will travel Δx = 11.5×[tex]10^{-6}[/tex] m.
Doubling Tension:
[tex]|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{2.0.00874 }[/tex]
[tex]|v| = \sqrt{0.01568}[/tex]
|v| = 0.1252 m/s
Displacement for same time:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 12.52.10^{-2}*1.23.10^{-3}[/tex]
[tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex]
With doubled tension, it travels [tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex] m
A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the resistor is
Answer:
At the moment contact is made with the battery, the voltage across the resistor is equal to the batteries terminal voltage
Explanation;
Because at series connection the battery and resistor have equal voltage
In which example is kinetic friction most involved? a sled stuck on a snowy hill a bottle of water wedged in a vending machine an explorer unsuccessfully pushing on a massive stone that is blocking the entrance to a cave a volleyball player sliding across the court while diving for the ball
Answer:
I believe the answer is A volleyball player sliding across the court while diving for the ball.
Explanation:
Kinetic friction is a body moving on the surface experiences a force in the opposite direction of its movement.
Hope this helps! （づ￣3￣）づ╭❤～
Describe how, using a positively-charged rod and two neutral metal spheres, we canmake one sphere positive without touching it to the rod. You might want to draw adiagram to help you.
Answer:
se the principle of induction.
place the two metallic spheres together, now we bring the positively charged bar closer to the first sphere.
The charge that was induced in the sphere is distributed as infirm as possible,
At this time I separate the spheres and move the bar away, by separating the spheres the excess positive
Explanation:
For this exercise we will use that the electric charge is not created, it is not destroyed and charges of the same sign repel.
Let's use the principle of induction. We place the two metallic spheres together, one in front of the other, now we bring the positively charged bar closer to the first sphere.
Here the positive charge of the bar repels the positive charge of the sphere, but as this is mocil it moves as far away as possible, until the negative charge that remains neutralizes the positive charge of the bar.
The charge that was induced in the sphere is distributed as infirm as possible, most of it in the furthest sphere, since the Coulomb force decreases.
At this time I separate the spheres and move the bar away, by separating the spheres the excess positive charge in the last sphere cannot be neutralized, therefore this sphere remains with a positive charge.
hi guys!!! i have no more points, can someone nice guess all of these for me? :)
1.What happens to the ocean water before the precipitation part of the water cycle
2.During which stage of the water cycle does water from the ocean form clouds?
3.what is a runoff??
4.Which statement about oceans is incorrect? A.Evaporation occurs when water is warmed by the sun. B.Most evaporation and precipitation occur over the ocean. C.97 percent of Earth's water is fresh water from the ocean. D.Water leaves the ocean by the process of evaporation
5.How does most ocean water return to the ocean in the water cycle
tysm to u who answers :)
1. The ocean water collects back in the ocean.
2. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.
3. an excessive amount of water flowing from downslope along earths surface
4. A.Evaporation occurs when water is warmed by the sun.
5. The water returns into the ocean by the water cycle . It evaporates , then it condensates , then it participates ( Rains ) and then goes back into the ocean.
Hope this answer correct ✌️
Light with an intensity of 1 kW/m2 falls normally on a surface with an area of 1 cm2 and is completely absorbed. The force of the radiation on the surface is
Answer:
The force of the radiation on the surface is 3.33 x 10⁻¹⁰ N
Explanation:
Given;
intensity of light, I = 1kw/m² = 1000 W/m²
area of the surface, A = 1 cm² = 1 x 10⁻⁴ m²
Since the light is completely absorbed, the force of the radiation is given by;
F = P/c
where;
c is the speed of light = 3 x 10⁸ m/s
But P = IA
F = IA /c
F = (1000 X 1 X 10⁻⁴) / 3 x 10⁸
F = 3.33 x 10⁻¹⁰ N
Therefore, the force of the radiation on the surface is 3.33 x 10⁻¹⁰ N
The force of radiation will be "3.33 × 10⁻¹⁰ N"
Intensity and ForceAccording to the question,
Intensity of force, I = 1 kW/m² or,
= 1000 W/m²
Area of surface, A = 1 cm² or,
= 1 × 10⁻⁴ m²
Speed of light, c = 3 × 10³ m/s
As we know the relation,
→ F = [tex]\frac{P}{c}[/tex]
or,
P = IA
or,
F = [tex]\frac{IA}{c}[/tex]
By substituting the values, we get
= [tex]\frac{1000\times 1\times 10^{-4}}{3\times 10^3}[/tex]
= 3.33 × 10⁻¹⁰ N
Thus the response above is correct.
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A plano-convex glass lens of radius of curvature 1.4 m rests on an optically flat glass plate. The arrangement is illuminated from above with monochromatic light of 520-nm wavelength. The indexes of refraction of the lens and plate are 1.6. Determine the radii of the first and second bright fringes in the reflected light.
Given that,
Radius of curvature = 1.4 m
Wavelength = 520 nm
Refraction indexes = 1.6
We know tha,
The condition for constructive interference as,
[tex]t=(m+\dfrac{1}{2})\dfrac{\lambda}{2}[/tex]
Where, [tex]\lambda=wavelength[/tex]
We need to calculate the radius of first bright fringes
Using formula of radius
[tex]r_{1}=\sqrt{2tR}[/tex]
Put the value of t
[tex]r_{1}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]
Put the value into the formula
[tex]r_{1}=\sqrt{2\times(0+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]
[tex]r_{1}=0.603\ mm[/tex]
We need to calculate the radius of second bright fringes
Using formula of radius
[tex]r_{2}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]
Put the value into the formula
[tex]r_{1}=\sqrt{2\times(1+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]
[tex]r_{1}=1.04\ mm[/tex]
Hence, The radius of first bright fringe is 0.603 mm
The radius of second bright fringe is 1.04 mm.
You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 38.5∘ from the ground with an initial speed of 27.5 m/s. At what height above its initial position does the softball hit the wall? Ignore any effects of air resistance.
The ball's horizontal position in the air is
[tex]x=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t[/tex]
It hits the wall when [tex]x=17.5\,\mathrm m[/tex], which happens at
[tex]17.5\,\mathrm m=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t\implies t\approx0.813\,\mathrm s[/tex]
Meanwhile, the ball's vertical position is
[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ t-\dfrac g2t^2[/tex]
where [tex]g[/tex] is the acceleration due to gravity, 9.80 m/s^2.
At the time the ball hits the wall, its vertical position (relative to its initial position) is
[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ(0.813\,\mathrm s)-\dfrac g2(0.813\,\mathrm s)^2\approx\boxed{10.7\,\mathrm m}[/tex]
When the atmosphere is not quite clear, one may sometimes see colored circles concentric with the Sun or the Moon. These are generally not more than a few diameters of the Sun or Moon and invariably the innermost ring is blue. The explanation for these phenomena involves:_________
A) reflection
B) refraction
C) interference
D) diffraction
E) Doppler effect
Answer:
D) diffraction
Explanation:
Corona is an optical phenomenon produced by the diffraction of sunlight or moonlight, as light moves through water droplets in the atmosphere.
This phenomenon produces one or more diffuse concentric rings of light around the Sun or Moon, usually seen as colored circles.
Therefore, the explanation for these phenomena of colored concentric circles, sometimes seen with the Sun or the Moon involves diffraction.
What does E=mc2 stand for?
It stands for energy=mass times the speed of light squared.
3 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!
Explanation:
[tex]qV = \frac{1}{2}mv^2[/tex]
Multiply both sides by 2 and then divide by m to get
[tex]\dfrac{2qV}{m} = v^2[/tex]
Take the square root of both sides to get
[tex]v = \sqrt{\dfrac{2qV}{m}}[/tex]
An electron has an initial velocity to the south but is observed to curve upward as the result of a magnetic field. This magnetic field must have a component:___________
a) north
b) upwards
c) downwards
d) east
e) west
Answer:
e) west
Explanation:
According to Lorentz left hand rule, the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.
In this case, if we point the thumb to the south (towards your body), with the palm facing up, then the fingers will point west.
The average human walks at a speed of 5km per hour if your PE teacher asks you to walk for 30 minutes in gym class how far would you walk(km)?
Answer:
2.5 km
Explanation:
Answer:
2.5 km
Explanation:
Distance = speed x time
So =5 x 0.5