Answer:
y = 1.19 m and λ = 8.6036 10⁻⁷ m
Explanation:
This is a slit interference problem, the expression for destructive interference is
d sin θ = m λ
indicate that for the angle of θ = 35º it is in the third order m = 3 and the separation of the slits is d = 4.50 10⁻⁶ m
λ = d sin θ / m
let's calculate
λ = 4.50 10⁻⁶ sin 35 /3
λ = 8.6036 10⁻⁷ m
for the separation distance from the central stripe, we use trigonometry
tan θ= y / L
y = L tan θ
the distance L is measured from the slits, it indicates that the light source is at x = 0.30 m from the slits
L = 2 -0.30
L = 1.70 m
let's calculate
y = 1.70 tan 35
y = 1.19 m
Explain why it takes much more effort to stop a freight train compared with a car?
Answer:
Train wheels and rails are both made of steel, and the steel-steel friction coefficient is around 0.25. As a result, the stopping time and distance will be three to four times that of a car.
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F?
Answer:
The correct answer is - low pitch
Explanation:
Now for the case it is mentioned that the tube closed on one end frequency is:
f = v/2l
Where,
l = length of the tube
v = velocity of longitudinal wave of gas filled in the tube
if frequency increases then pitch will be increase as well as pitch depends on frequency.
Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases.
As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.
question 1+1677-789909
Answer:
your answer is -788231
Explanation:
1+1677=1678
1678-789909=-788231
A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediately after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction relative to the x-axis does the third piece move
Answer:
Explanation:
There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.
momentum of first piece = .320 kg x 2 m/s
= 0.64 kg m/s along x -axis.
momentum of second piece = .355 kg x 1.5 m/s
= 0.5325 kg m/s along y- axis .
Let the velocity of third piece be v and it is making angle of θ with x -axis .
Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ
vertical component of its velocity = .100 kg x v sinθ = .1 v sinθ
For making total momentum in the plane zero
.1 v cosθ = 0.64 kg m/s
.1 v sinθ = 0.5325 kg m/s
Dividing
Tanθ = .5325 / .64 = .83
θ = 40⁰.
The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.
Answer:
8.3 m/s, 2196 degree from + X axis
Explanation:
m = 320 g , u = 2 m/s along X axis
m' = 355 g, u' = 1.5 m/s along Y axis
m'' = 100 g, u'' = v
Let the speed of the third piece is v makes an angle A from the X axis.
use conservation of momentum along X axis
0 = 320 x 2 + 100 x v cos A
v cos A = - 6.4 ..... (1)
Use conservation of momentum along Y axis
0 = 355 x 1.5 + 100 x v sin A
v sinA = - 5.3 ... (2)
Squaring and adding
[tex]v^2 = (-6.4)^2 +(-5.3)^2\\\\v= 8.3 m/s[/tex]
The angle is given by
[tex]tan A = \frac{-5.3}{-6.4}\\\\A = 219.6 degree[/tex] from + X axis
What is the current in milliamperes produced by the solar cells of a pocket calculator through which 9.00 C of charge passes in 8.50 h
Answer:
Current = 0.000294 A
Explanation:
Below is the given values:
Given the charge = 9.00 C
Time = 8.50 h
Use the below formula to find the current:
Current = Q / t
Now plug the values:
Current = 9 / (8.5 x 3600)
Current = 0.000294 A
Work-Energy Theorem & Power
A 0.5 kg mass sitting on smooth ice is accelerated from rest by a force until is
acquires a speed of 8 m/s. The force acts while the mass moves through a
displacement of 2 m.
A. Calculate the kinetic energy of the mass after the force acts.
B. Calculate the work done by the force.
C. Calculate the magnitude of the force that accelerated the mass.
Answer:
A. 16 J
B. 16 J
C. 8 N
Explanation:
A. Determination of the kinetic energy.
Mass (m) = 0.5 Kg
Velocity (v) =. 8 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.5 × 8²
KE = ½ × 0.5 × 64
KE = 0.5 × 32
KE = 16 J
B. Determination of the Workdone by the force.
Kinetic energy (KE) = 16 J
Workdone =.?
Workdone and kinetic energy has the same unit of measurement. Thus,
Workdone = kinetic energy
Workdone = 16 J
C. Determination of the force.
Workdone (Wd) = 16 J
Displacement (s) = 2 m
Force (F) =?
Wd = F × s
16 = F × 2
Divide both side by 2
F = 16 / 2
F = 8 N
The upward normal force exerted by the floor is 710 N on an elevator passenger who weighs 720 N . You may want to review (Pages 107 - 110) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Weighing yourself in an elevator. Part A What is the reaction force to the upward normal force exerted by the floor
Answer:
If the person is to remain the floor the reaction force will be equal to the normal force exerted by the floor.
F(normal) - F(reaction) = 0
That means the person is not moving with respect to the elevator.
Expanding the applied forces we have:
Fw - Fn = 720 - 710 = 10 N where the positive direction is chosen as down
Fw is the weight of the person and Fn the force exerted on the person by the elevator,
The acceleration of the person the becomes F = m a = m * 10 N and will be downward agreeing with our choice of coordinate axes.
A boy with a mass of 140 kg and a girl with a mass of 120 kg are on a merry go round. Th merry go round has a radius of 5 meters and its moment of inertia is 986 kg m 2. Beginning from rest the merry go round accelerates with an angular acceleration of 0.040 rad/s2 for 30 seconds then has a constant angular speed.
1. How many revolutions do the kids make before the constant operational speed is reached ?
2. What's the angular speed and magnitude of the tangential of the kids if they are standing at a distance of 1.5m and 2.4 m from the center of the ride.
3. During the ride the kids switch places what is the angular speed and magnitude of the tangential velocities ?
Answer:
we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer
a fixed mass of gas occupies a volume of 1000 CM3 at 0 degree celsius if it is heated at constant pressure of 100 degree celsius calculate the new volume
Answer:
P V = N R T ideal gas equation
V1 = k * T1 if P is constant and also N and R will be constant
V2 = k * T2 where k is some constant
Or V2 = (T2 / T1) * V1 also known as "Charles Law" for expansion at
constant pressure
V2 = (373 / 273) * 1000 cm^3 = 1366 cm^3 where T is absolute temperature
What process provides the sun with its energy
Answer:
nuclear fusion
The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom. During the fusion process, radiant energy is released.Answer:
nuclear fusion
Explanation:
The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom.
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. What is the distance covered before the car comes to a stop
Answer:
The correct solution is "122.2211".
Explanation:
Given:
deceleration,
a = 22 ft/sec²
Initial velocity,
[tex]V_i=50 \ m/h[/tex]
Now,
[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]
[tex]=73.333 \ ft/sec[/tex]
Now,
Final velocity,
[tex]V_f=0[/tex]
Initial velocity,
[tex]V_{initial} = 73.333 \ ft/sec[/tex]
hence,
⇒ [tex]V_f^2=V_i^2+2aD[/tex]
By putting the values, we get
[tex]0=(73.333)^2+2\times( -22) D[/tex]
[tex]44D=(73.333)^2[/tex]
[tex]D=\frac{(73.333)^2}{44}[/tex]
[tex]=122.2211[/tex]
Physics question on picture
Answer:
B. according to Newton's Third Law of Motion, the force of the Moon on the Earth and the force of the Earth on the Moon are equal in magnitude and opposite in direction
A glass block in air has critical angle of 49. What will happen to a ray of light coming through the glass when it is incident at and angle of 50 at the glass air boundary? Illustrate with a diagram
Answer:
b
Explanation:
what's impulse of
force
Answer:
The impulse experienced by the object equals the change in momentum of the object. In equation form, F.t = m. Δv. In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum.
The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns
A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.
Answer:
C
Explanation:
Similarity
gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be
Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:
[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]
where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,
[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]
d = 68.5 x 10⁻⁶ m = 68.5 μm
A vehicle is used to transport material down a straight aisle. The max acceleration of the vehicle is 1 m/s/s and the max speed of the vehicle is 5m/s. The vehicle starts at the beginning of the aisle. How long does it take to move down the aisle and come to a stop at the other end if:
a) the aisle is 100 meters long?
b) the aisle is 9 meters long?
Answer:
(a) 14.14 s
(b) 4.24 s
Explanation:
maximum acceleration, a = 1 m/s^2
maximum speed, v = 5 m/s
initial speed, u = 0 m/s
(a) distance, s = 100 m
Let the time is t.
Use second equation of motion
[tex]s = u t 0.5 at^2\\\\100 = 0 + 0.5 \times 1 \times t^2\\\\t = 14.14 s[/tex]
(b) distance, s = 9 m
Let the time is t'.
Use second equation of motion
[tex]s = u t + 0.5 at^2\\\\9= 0 + 0.5 \times 1 \times t'^2\\\\t' = 4.24 s[/tex]
A police car travels towards a stationary observer at a speed of 15m/s. the siren on the car emits a sound of frequency 250Hz. Calculate the observer frequency. the speed of sound is 340m/s
Observer Frequency = sound frequency x ( speed of sound / speed of sound - speed of car)
= 250 x (340/( 340-15))
= 261.54 Hz
describe the movement of the man when the resultant horizontal force is 0 N
can anyone help in both questions please
Answer:
Force A newton Law first law
F = M.A which Force in 0 N as you Questions Above
Force B
Newton Law 3
Action = -Reaction
Hope you can explain this formula as you want to scribe to explaining
An inductive circuit contains resistance of 20 ohm and an inductance of 20 H. If an ac voltage of 120 V and frequency 60 Hz is applied to this circuit, the current would be
A 0.0159
A 0.017
A 0.02
A 0.16
Answer:
answer : option (b) 0.016 amp
explanation : resistance of resistor , R = 10 Ω
inductance of inductor , X_LX
L
= 20H
voltage of AC circuit , V = 120volts
frequency, ff =60Hz
so, angular frequency, \omega=2\pi fω=2πf = 2 × π × 60 = 120π rad/s
now, current , i=\frac{V}{\sqrt{R^2+\omega^2L^2}}i=
R
2
+ω
2
L
2
V
= 120/√{10² + (120π)² × 20²}
= 120/√{100 + 14400π² × 400}
after solving this we get, i = 0.016 amp
your neighbour is throttling his recent bought motorbike to show off. the sound intensity measured at your window 16m away is 0.25W/m^2. what is the sound intensity level in dB at your friend's house, a distance of 28m away from the noisy bike?
Answer:
hi your pinterest I'd or Twitter I'd pls.
What is the electric field strength between two parallel conducting plates separated by 10 cm and having a potential difference between them of 2000 V?
a.
2000 V/m
b.
200 V/m
c.
20 kV/m
d.
200000 V/m
Answer:
• Potential Difference (V) = 2000 V
• Distance b/w the two parallel plates (d) = 10 cm = 10/100 = 1/10 = 0.1 m
• Electric field (E) = ?
[tex]\implies V = E.d[/tex]
[tex]\implies E = \dfrac{V}{d} [/tex]
[tex]\implies E = \dfrac{2000}{0.1} [/tex]
[tex]\implies E = \dfrac{2000}{ {10}^{ - 1} } [/tex]
[tex]\implies E = 2000 \times {10}^{1} [/tex]
[tex]\implies\bf E = 20000 \:V/m[/tex]
[tex]\implies\bf E = 20\:kV/m[/tex]
Hence, option C) the correct answer.
The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy stored in C3 and C4.
Answer:
A) C_{eq} = 15 10⁻⁶ F, B) U₃ = 3 J, U₄ = 0.5 J
Explanation:
In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.
In this case let's start by finding the equivalent capacitance.
A) Let's solve the part where C1 and C3 are. These two capacitors are in serious
[tex]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}[/tex] (you has an mistake in the formula)
[tex]\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}[/tex]
[tex]\frac{1}{C_{eq1}}[/tex] = 0.1 10⁶
[tex]C_{eq1}[/tex] = 10 10⁻⁶ F
capacitors C₂, C₄ and C₅ are in series
[tex]\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}[/tex]
[tex]\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6[/tex]
[tex]\frac{1}{C_{eq2} }[/tex] = 0.2 10⁶
[tex]C_{eq2}[/tex] = 5 10⁻⁶ F
the two equivalent capacitors are in parallel therefore
C_{eq} = C_{eq1} + C_{eq2}
C_{eq} = (10 + 5) 10⁻⁶
C_{eq} = 15 10⁻⁶ F
B) the energy stored in C₃
The charge on the parallel voltage is constant
is the sum of the charge on each branch
Q = C_{eq} V
Q = 15 10⁻⁶ 6
Q = 90 10⁻⁶ C
the charge on each branch is
Q₁ = Ceq1 V
Q₁ = 10 10⁻⁶ 6
Q₁ = 60 10⁻⁶ C
Q₂ = C_{eq2} V
Q₂ = 5 10⁻⁶ 6
Q₂ = 30 10⁻⁶ C
now let's analyze the load on each branch
Branch C₁ and C₃
In series combination the charge is constant Q = Q₁ = Q₃
U₃ = [tex]\frac{Q^2}{2 C_3}[/tex]
U₃ =[tex]\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}[/tex]
U₃ = 3 J
In Branch C₂, C₄, C₅
since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅
U₄ = [tex]\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}[/tex]
U₄ = 0.5 J
Two people, who have the same mass, throw two different objects at the same velocity. If the first object is heavier than the second, compare the velocities gained by the two people as a result of recoil.
a. The first person will gain more velocity as a result of recoll.
b. The second person will gain more velocity as a result of recoll.
c. Both people will gain the same velocity as a result of recoll.
d. The velocity of both people will be zero as a result of recoil
Answer:
The first person will gain more velocity as a result of recoil.
Explanation:
Let us recall that from Newton's third law of motion, action and reaction are equation and opposite. A consequence of this law is the proposition that ''momentum can neither be created nor destroyed.''
Hence, when two people who have the same mass, throw two different objects at the same velocity but the first object is heavier than the second, the first object possesses greater momentum than the second object hence the first person will gain more velocity as a result of recoil.
The reason why a teacher is more important then a farmer
Answer:
A teacher is more important than a famer.
Explanation:
A teacher is more important than a famer because the knowledge of farming is gotten through the teacher. Thus, without a teacher; whether formal or informal, there cannot be farming, let alone farmers.
To get maximum current in a circuit, the resistance should be in _____
1)series
2)parallel
Answer:
no parallel is the correct answer
An initially motionless test car is accelerated uniformly to 105 km/h in 8.43 s before striking a simulated deer. The car is in contact with the faux fawn for 0.635 s, after which the car is measured to be traveling at 60.0 km/h. What is the magnitude of the acceleration of the car before the collision?
acceleration before collision:
3.45
m/s2
What is the magnitude of the average acceleration of the car during the collision?
average acceleration during collision:
19.68
m/s2
What is the magnitude of the average acceleration of the car during the entire test, from when the car first begins moving until the collision is over?
105 km/h ≈ 29.2 m/s
60.0 km/h ≈ 16.7 m/s
Before the collision the test car has an acceleration a of
a = (29.2 m/s - 0) / (8.43 s) ≈ 3.46 m/s²
During the collision, the car is slowed to about 16.7 m/s, so that its (average) acceleration is
a = (16.7 m/s - 29.2 m/s) / (0.635 s) ≈ -19.7 m/s²
i.e. with magnitude about 19.7 m/s².
Overall, the car has an average acceleration of
a = (16.7 m/s - 0) / (8.43 s + 0.635 s) ≈ 1.84 m/s²
1. A child slide down an inclined plane of length 10 m at an angle of 45°. If the coefficient friction between the child and the plane is 0.1, evaluate The velocity just before touching the bottom of the plane.
Answer:
The speed at the bottom is 11.2 m/s.
Explanation:
length, s = 10 m
Angle, A = 45 degree
coefficient of friction = 0.1
let the velocity is v.
The acceleration is given by
[tex]a = g sin A - \mu g cos A \\\\a = 9.8 (sin 45 - 0.1 cos 45)\\\\a = 6.24 m/s^2[/tex]
Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 6.24 \times 10 \\\\v = 11.2 m/s[/tex]
A measurement was made of the magnetic field due to a tornado, and the result was 13.00 nT to the north. The measurement was made at a position 8.90 km west of the tornado. What was the magnitude (in A) and direction of the current in the funnel of the tornado? Assume the vortex was a long, straight wire carrying a current.
Answer:
4
Explanation:
The blades of a fan running at low speed turn at 26.2 rad/s. When the fan is switched to high speed, the rotation rate increases uniformly to 36.5 rad/s in 5.75 seconds. What is the magnitude of the fan's angular acceleration
Answer: [tex]1.79\ rad/s^2[/tex]
Explanation:
Given
Initial angular speed is [tex]\omega_1=26.2\ rad/s[/tex]
Final angular speed is [tex]\omega_2=36.5\ rad/s[/tex]
Time period [tex]t=5.75\ s[/tex]
Magnitude of the fan's acceleration is given by
[tex]\Rightarrow \alpha=\dfrac{\omega_2-\omega_1}{t}[/tex]
Insert the values
[tex]\Rightarrow \alpha=\dfrac{36.5-26.2}{5.75}\\\\\Rightarrow \alpha=\dfrac{10.3}{5.75}\\\\\Rightarrow \alpha=1.79\ rad/s^2[/tex]
Thus, fan angular acceleration is [tex]1.79\ rad/s^2[/tex]
Answer:
The angular acceleration is given by 1.8 rad/s^2.
Explanation:
initial angular speed, wo = 26.2 rad/s
final angular velocity, w = 36.5 rad/s
time, t = 5.75 seconds
The first equation of motion is
[tex]w = wo + \alpha t\\\\36.5 = 26.2 + 5.75\alpha\\\\\alpha = 1.8 rad/s^2[/tex]
