This question is not complete, the complete question is;
A railroad handcar is moving along straight, frictionless tracks with negligible air resistance.
In the following cases, the car initially has a total mass (car and contents) of 170 kg and is traveling east with a velocity of magnitude 4.60 m/s.
Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks.
Part A
An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car's initial velocity.
Part B
An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.
Answer:
Part A) the final velocity of the car is 4.6 m/s
Part B) the final velocity of the car is 5.28 m/s
Explanation:
Given the data in the question;
Total mass (m₁+m₂) = 170 kg
velocity of magnitude Vx = 4.60 m/s
PART A)
An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car's initial velocity,
i.e
m₂ = 22.0 kg
so m₁ = 170 - 22 = 148 kg
so, we apply conservation of momentum
since the object thrown out of the car, it has nothing to do with the car's velocity.
(m₁+m₂)Vx = m₁Vx₁ + m₂Vx₂
we substitute
(170)4.60 = 148Vx₁ + 22(4.60)
782 = 148Vx₁ + 101.2
148Vx₁ = 782 - 101.2
148Vx₁ = 680.8
Vx₁ = 680.8 / 148
Vx₁ = 4.6 m/s
Therefore, the final velocity of the car is 4.6 m/s
Part B)
An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.
Vx = V(m₁+m₂) / ((m₁+m₂) - m₁)
we substitute
Vx = 4.60(170) / ((170) - 22)
Vx = 782 / 148
Vx = 5.28 m/s
Therefore, the final velocity of the car is 5.28 m/s
A planet with a mass one-half that of Earth has a radius that is 3 times that of Earth's radius. What is the gravitational field strength of the planet?
The gravitational acceleration of a planet is proportional to the planet's mass, and inversely proportional to square of the planet's radius.
So when you stand on the surface of this particular planet, you feel a force of gravity that is
(1/2) / (3²)
of the force that you feel on the surface of the Earth.
That's (1/18) as much as on Earth.
The acceleration of gravity there would be about 0.545 m/s².
This is about 12% less than the gravity on Pluto.
what colors of light are absorbed when white light falls on a green object?
If soldiers march across the bridge with a cadence equal to the bridge’s natural frequency and impart 1x104 J of energy each second, how long does it take for the bridge’s oscillations to go from 0.1 m to 0.5 m amplitude?
Answer: Hello, Mark me as Brainliest! :)
If soldiers march across the bridge with a cadence equal to the bridge's natural frequency and impart $$1.00 × 10^4 J$$ of energy each second, how long does it take for the bridge's oscillations to go from 0.100 m to 0.500 m amplitude. $ 5 \times 10^7 \text{J} $ . \\ b) $ 12 \times 10^4 \text{s}$ .
Your Welcome!
Explanation:
1
Select the correct answer.
Which type of energy is thermal energy a form of?
A
chemical energy
B.
kinetic energy
C. magnetic energy
D. potential energy
Reset
Next
Answer:
B. kinetic energy
Explanation:
Thermal energy (It’s a low form of energy ) is a form of kinetic energy as it is produced as a result of motion of particles either if they vibrate at their position or they move along longer paths.
prove that d1=R(d1-d2) in relative density
A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be considered a uniform disk of mass 4.5 kgkg and diameter 0.30 mm . The potter then throws a 2.8-kgkg chunk of clay, approximately shaped as a flat disk of radius 8.0 cmcm , onto the center of the rotating wheel. Part A What is the frequency of the wheel after the clay sticks to it
Answer:1.7 rev/s
Explanation:
Given
Frequency of wheel [tex]N_1=2\ rev/s[/tex]
angular speed [tex]\omega_1=2\pi N_1=4\pi\ rad/s[/tex]
mass of wheel [tex]m_1=4.5\ kg[/tex]
diameter of wheel [tex]d_1=0.30\ m=30\ cm[/tex]
radius of wheel [tex]r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm[/tex]
mass of clay [tex]m_2=2.8\ kg[/tex]
the radius of the chunk of clay [tex]r_2=8\ cm[/tex]
Moment of inertia of Wheel
[tex]I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2[/tex]
Combined moment of inertia of wheel and clay chunk
[tex]I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2[/tex]
Conserving angular momentum
[tex]\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi[/tex]
Common frequency of wheel and chunk of clay is
[tex]\Rightarrow N_2=\dfrac{4\pi \times 0.849}{2\pi}=1.698\approx 1.7\ rev/s[/tex]
A uniform-density 7 kg disk of radius 0.20 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 42 N through a distance of 0.9 m. Now what is the angular speed
Answer:
The angular speed is 23.24 rad/s.
Explanation:
Given;
mass of the disk, m = 7 kg
radius of the disk, r = 0.2 m
applied force, F = 42 N
distance moved by disk, d = 0.9 m
The torque experienced by the disk is calculated as follows;
τ = F x d = I x α
where;
I is the moment of inertia of the disk = ¹/₂mr²
α is the angular acceleration
F x r = ¹/₂mr² x α
The angular acceleration is calculated as;
[tex]\alpha = \frac{2Fr}{mr^2} \\\\\ \alpha = \frac{2F}{mr}\\\\\alpha = \frac{2 \times 42 }{7 \times 0.2} \\\\\alpha = 60 \ rad/s^2[/tex]
The angular speed is determined by applying the following kinematic equation;
[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta[/tex]
initial angular speed, ωi = 0
angular distance, θ = d/r = 0.9/0.2 = 4.5 rad
[tex]\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta} \\\\\omega _f = \sqrt{2 \times 60 \times 4.5} \\\\\omega _f = 23.24 \ rad/s[/tex]
Therefore, the angular speed is 23.24 rad/s.
The amount of light that enters the pupil is controlled by the:
retina.
lens.
inis.
Answer: The amount of light that enters the pupil is controlled by the Iris
Explanation:
Solve the below problems being sure to provide the correct significant figures.
1) 1000 ÷ 4.886 = __________
2) 240 ÷ 12.3 = __________
3) 80 x 4.6 = __________
4) 4.527 x 30 = __________
5) 86 x 63.855 x 8000 = __________
6) 700 x 91.186 = __________
7) 7.1 x 348 = __________
8) 50 ÷ 29.1 = __________
9) 98.773 x 24.891 x 409 = __________
10) 0.065 x 3 x 3007 = __________
Answer:
1) 204.6663938
2) 19.51219512
3) 368
4) 135.81
5) 43932240
6) 63830.2
7) 2470.8
8) 1.718213058
9) 1005550.526
10) 586.365
Most of the questions you asked were in repeating decimal form.
Explanation:
A motorist is driving at 15 m/s when she sees that a traffic light 315m ahead has just turned red. She knows that this light stay red for 25 s, and she wants to be 20 m from the ligt when it turns green again. Taht way, she will still be able to stop if the light stays red longer than expected. She applies the brake gradually such that her acceleration is ax(t)= c + bt, where c and b are constant. Assume she starts with a constant speed at the origin.
Find the values of c a b and any other unknown constants in order to answer the following questions.
1. Given the motorist's acceleration as a function of time, what are her position and velocity fucntions? - Do not use numbers for any constant here. Only derive the position and velocity functions.
2. What is her speed as she reaches the light?
Answer:
1) x = x₀ + vot - ½ c t² - 1/6 bt³, v = v₀ - ct - ½ b t²
2) v₁ = 5.25 m/s, v₂ = -8 m/s
Explanation:
1) For this exercise, the relationship of the body is not constant, so you must use the definition of speed and position to find them.
acceleration is
a = c + bt
a) the relationship between velocity and acceleration
a = [tex]\frac{dv}{dt}[/tex]
dv = -a dt
The negative sign is because the acceleration is contrary to the speed to stop the vehicle.
we integrate
∫ dv = - ∫ a dt
∫ dv = -∫ (c + bt) dt
v = -c t - ½ b t²
This must be valued from the lower limit, the velocity is vo, up to the upper limit, the velocity is v for time t
v - v₀ = -c (t-0) - ½ b (t²-0)
v = v₀ - ct - ½ b t²
b) the velocity of the body is
v = [tex]\frac{dx}{dt}[/tex]
dx = v dt
we replace and integrate
∫ dx = ∫ (v₀ - c t - ½ bt²) dt
x-x₀ = v₀ t - ½ c t² - ½ b ⅓ t³
Evaluations from the lower limit the body is at x₀ for t = 0 and the upper limit the body is x = x for t = t
x - x₀ = v₀ (t-0) - ½ c (t²-0) + [tex]\frac{1}{6}[/tex] (t³ -0)
x = x₀ + vot - ½ c t² - 1/6 bt³
2) The speed when you reach the traffic light
Let's write the data that indicates, the initial velocity is vo = 15 m / s, the initial position is xo = 315m, let's use the initial values to find the constants.
t = 25 s x = 20
we substitute
20 = 315 + 15 25 - ½ c 25² - 1/6 b 25³
0 = 295 + 375 - 312.5 c - 2604.16 b
670 = 312.5 c + 2604.16 b
we simplify
2.144 = c + 8.33 b
Now let's use the equation for velocity,
v = v₀ - ct - ½ b t²
v = 15 - c 25 - ½ b 25²
v = 15 - 25 c - 312.5 b
let's write our two equations
2.144 = c + 8.33 b
v = 15 - 25 c - 312.5 b
Let's examine our equations, we have two equations and three unknowns (b, c, v) for which the system cannot be solved without another equation, in the statement it is not clear, but the most common condition is that if the semaphore does not change, it follows with this acceleration (constant) to a stop
a = c + b 25
from the first equation
c = 8.33 / 2.144 b
C = 3.885 b
we substitute in the other two
v = 15 - 25 (3.885 b) - 312.5 b
v = 15 - 409.6 b
final acelearation
a = 28.885 b
let's use the cinematic equation
[tex]v_{f}^2[/tex]= v² - 2 a x
0 = v² - 2a 20
0 = v² - (28.885b) 40
v² = 1155.4 b
we write the system of equations
v = 15 - 409.6 b
v² = 1155.4 b
resolve
v²= 1155.4 ( [tex]\frac{15 -v }{409.6}[/tex] )
v² = 2.8 ( 15 -v)
v² + 2.8 v - 42.3 = 0
v= [ -2.8 ±[tex]\sqrt {2.8^2 + 4 \ 42.3) }[/tex] ]/2 = [-2.8 ± 13.3]/2
v₁ = 5.25 m/s
v₂ = -8 m/s
(will give brainliest to whoever answers first and explains reasoning) A 10kg object is spun around in a circle with a centripetal acceleration of 3.5m/s^2. What is the centripetal force acting on the object?
Answer:
35 N
Explanation:
F = ma
centripetal force = 10(3.5) = 35 N
A 38.0 kg box initially at rest is pushed 4.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between box and floor is 0.300, find the following. (a) the work done by the applied force J (b) the increase in internal energy in the box-floor system due to friction J (c) the work done by the normal force J (d) the work done by the gravitational force J (e) the change in kinetic energy of the box J (f) the final speed of the box m/s
Answer:
a) Wapp = 520 N
b) ΔUf = 447 N
c) Wn = 0
d) Wg = 0
e) ΔK = 73 J
f) v = 1.96 m/s
Explanation:
a)
Applying the definition of work, as the dot product between the applied force and the displacement, since both are parallel each other, the work done on the box by the applied force can be written as follows:[tex]W_{app} = F_{app} * \Delta X = 130 N * 4.0 m = 520 N (1)[/tex]
b)
The change in the internal energy due to the friction, is numerically equal to the work done by the force of friction.This work is just the product of the kinetic force of friction, times the displacement, times the cosine of the angle between them.As the friction force opposes to the displacement, we can find the work done by this force as follows:[tex]W_{ffr} = F_{fr} * \Delta X * cos (180) (2)[/tex]
The kinetic force of friction, can be expressed as the product of the kinetic coefficient of friction times the normal force.If the surface is level, this normal force is equal to the weight of the object, so we can write (2), as follows:[tex]W_{ffr} = F_{fr} * \Delta X * cos (180) = -\mu_{k} * m* g* \Delta X = \\ -0.300*38.0kg9.8 m/s2*4.0m = -447 J (3)[/tex]
So, the increase in the internal energy in the box-floor system due to the friction, is -Wffr = 447 Jc)
Since the normal force (by definition) is normal to the surface, and the displacement is parallel to the surface, no work is done by the normal force.d)
Since the surface is level, the displacement is parallel to it, and the gravitational force is always downward, we conclude that no work is done by the gravitational force either.e)
The work-energy theorem says that the net work done on the object, must be equal to the change in kinetic energy.We have two forces causing net work, the applied force, and the friction force.So the change in kinetic energy must be equal to the sum of the work done by both forces, that we found in a) and b).So, we can write the following expression:[tex]\Delta K = W_{app} + W_{ffr} = 520 J - 447 J = 73 J (4)[/tex]
f)
Since the object starts at rest, the change in kinetic energy that we got in e) is just the value of the final kinetic energy.So, replacing in (4), we finally get:[tex]\Delta K = 73 J = \frac{1}{2}*m*v^{2} (5)[/tex]
Solving for v:[tex]v_{f} = \sqrt{\frac{2*\Delta K}{m} } = \sqrt{\frac{2*73J}{38.0kg}} = 1.96 m/s (6)[/tex]a) The work done by the applied force [tex]W_{AP}=520\ J[/tex]
b) The change in the internal energy [tex]\Delta U=447\ J[/tex]
c) Work done by normal force [tex]W_n=0[/tex]
d) Work done by gravitation [tex]W_g=0[/tex]
e) The change in KE will be [tex]\Delta KE=73\ J[/tex]
f) The final speed v = 1.96 m/s
What will be the work done?The work done on any object can be defined as the force applied on the object and its displacement due the effect of the force.
If the object achieve movement due to the work then the energy in the object will be kinetic energy.
If the object attains some height against the gravity then the energy in the object will be potential energy.
Now it is given in the question that
The horizontal force [tex]F_h=130\N[/tex]
mass of the object m= 38 kg
Coefficient of friction [tex]\mu=0.3[/tex]
Displacement of the object [tex]\delta x=4\ m[/tex]
(a) The work done will be
[tex]W=F_h\tines \Delta x[/tex]
[tex]W=130\times 4=520\ J[/tex]
(b) The increase in the internal energy
The increase in the internal energy of the box is due to the energy generated by the force of friction so
[tex]W_f=F_f\times \Delta x\times Cos(180)[/tex]
here [tex]F_f[/tex] is the frictional force and is given as
[tex]\mu=\dfrac{F_f}{R}[/tex]
Here R is the normal reaction and its value will be weight of the box in opposite direction.
[tex]\mu=\dfrac{F_f}{-mg}[/tex]
[tex]F_f=-mg\times \mu[/tex]
[tex]W_f=F_f\times \Delta x\times Cos180=-mg\times\mu \times cos180[/tex]
[tex]W_f=-38\times 9.81\times 0.3\times4=-447\J\ J[/tex]
(c) The work done by the normal force will be zero because the displacement is horizontal against the normal work so the work done will be zero.
(d) The work done by the gravitational force will also be zero. Because the displacement is horizontal and the gravitational force acts downward.
(e) The change in the KE of the box.
The change in the KE of the box will be the net energy of the box so from the work energy theorem the net energy will be
[tex]\Delta KE =W_{AP}-W_f=520-447=73\ J[/tex]
(f) The speed of the box
The KE of the box will be
[tex]KE=\dfrac{1}{2} mv^2[/tex]
[tex]v=\sqrt{ \dfrac{2\times KE}{m}[/tex]
[tex]v=\sqrt{\dfrac{2\times73}{38} }=1.96\ \dfrac{m}{s}[/tex]
Thus
a) The work done by the applied force [tex]W_{AP}=520\ J[/tex]
b) The change in the internal energy [tex]\Delta U=447\ J[/tex]
c) Work done by normal force [tex]W_n=0[/tex]
d) Work done by gravitation [tex]W_g=0[/tex]
e) The change in KE will be [tex]\Delta KE=73\ J[/tex]
f) The final speed v = 1.96 m/s
To know more about Work and energy follow
https://brainly.com/question/25959744
NEED HELP ASAP, ILL GIVE YOU BRAINLIEST IF CORRECT (30POINTS)
Drag each label to the correct location on the image. Each label can be used more than once.
Identify the parts of the barred spiral galaxy.
SPIRAL ARM, NUCLEUS, BAR
NOTE I JUST FILLED IN THE SPOTS FOR YOU TO SEE, THEY ARE NOT CORRECT
Answer:
the bar is the top and bottem. the nucleas in the middle and the Spiral arm is the last space
Explanation:
Answer:
look pkch
Explanation:
I don’t understand this
Answer:
true
Explanation:
force or powerbecause he pushes a disk
a boy throws a ball straight up into the air it reaches the highest point of its flight after 4 seconds how fast was the ball going when it left the boy's hand
Answer:
Gravity pulls down on the ball at g=-9.81 m/s^2. Up is positive, down is negative. The ball started at a certain initial velocity of Vi m/s. Time it took is t=4s. Final velocity is Vf=0 m/s, because at the highest point the ball stops moving.
When6-2 He He-6 undergoes beta decay, the daughter is?
Answer: The daughter is named Susie.
Explanation: LIL SUSIE!!!
HUH? DIDN'T UNDERSTAND THE QUESTION!
HAVE A GREAT DAY!!!!!
Answer:6/3 Li
Explanation:
I’m not sure what the person under me is talking about but yeah
If it takes 560 joules of work to move an object 10 meters what force was needed
Answer:
F = 56 N
Explanation:
Assuming that the applied force is parallel to the displacement, according to the definition of work, we can write the following expression for the force required:[tex]F = \frac{W}{\Delta x} =\frac{560J}{10m} = 56 N (1)[/tex]
What is the scientific study of how animals are classified?
Answer:
biology
Explanation:
Animals and plants
A bump should primarily be used in which situation?
A. when the ball is going out of bounds
B. when returning a serve
C. to start game play
(5 Points)
a) At ground level, the pressure of the helium in a balloon is 1x105
Pa. The volume occupied by the helium is 9.6m The balloon is
released and it rises quickly through the atmosphere. Calculate
the pressure of the helium when it occupies a volume of 12m3.
(3 Marks)
b) A box is 15m below the surface of the sea. The density of sea-
water is 1020 kg/m.
Calculate the pressure on the box due to the sea-water.
(2 Marks)
Answer:
1. [tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa
2. P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]
Explanation:
1. From Boyles' law;
[tex]P_{1}[/tex][tex]V_{1}[/tex] = [tex]P_{2}[/tex][tex]V_{2}[/tex]
[tex]P_{1}[/tex] = 1 x [tex]10^{5}[/tex] Pa
[tex]V_{1}[/tex] = 9.6 [tex]m^{3}[/tex]
[tex]V_{2}[/tex] = 12 [tex]m^{3}[/tex]
Thus,
1 x [tex]10^{5}[/tex] x 9.6 = [tex]P_{2}[/tex] x 12
[tex]P_{2}[/tex] = [tex]\frac{100000 x 9.6}{12}[/tex]
= 80000
[tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa
2. Pressure, P = ρhg
where: ρ is the density of the fluid, h is the height/ depth and g is the acceleration due to gravity (9.8 m/[tex]s^{2}[/tex]).
Thus,
P = 1020 x 15 x 9.8
= 149940
P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]
Mechanical energy is the most concentrated form of energy.
a. true
b. false
a cohesive force between the liquids molecules is responsible for the fluids is called
Answer:
static force
Explanation:
mark me brainliest
If we always drop the balls from 1-m
height in each trial, what type of variable
is this in this experiment?
Constant Variable
Dependent Variable
Independent Variable
Answer:
height
weight of ball
time of ball falling
As a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 4.12 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W.
Answer:
102.5N
Explanation:
Given that a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 4.12 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W
The power = 4.12 × 746 = 3073.52 W
Using the formula
Power = force × velocity
3073.52 = force × 30
Force = 3073.52 / 30
Force = 102.5 N
Since most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road, therefore,
the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s is 102.5 N
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength ?
Complete question:
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?
Answer:
The electric field strength at the mid-point between the two rings is zero.
Explanation:
Given;
diameter of each ring, d = 10 cm = 0.1 m
distance between the rings, r = 21.0 cm = 0.21 m
charge of each ring, q = 40 nC = 40 x 10⁻⁹ C
let the midpoint between the two rings = x
The electric field strength at the midpoint between the two rings is given as;
[tex]E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0[/tex]
Therefore, the electric field strength at the mid-point between the two rings is zero.
Calculate the first and second order angles for light of wavelength 400. nm and 700. nm of the grating contains 1.00 x 104 lines/cm.
Answer:
[tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex]
[tex]44.43^{\circ}[/tex], second order does not exist
Explanation:
n = Number of lines grating = [tex]1\times10^4\ \text{Lines/cm}[/tex]
[tex]\lambda[/tex] = Wavelength
m = Order
Distance between slits is given by
[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{1\times 10^4}\\\Rightarrow d=10^{-6}\ \text{m}[/tex]
[tex]\lambda=400\ \text{nm}[/tex]
m = 1
We have the relation
[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}\\\Rightarrow \theta=\sin^{-1}\dfrac{1\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=23.58^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=53.13^{\circ}[/tex]
The first and second order angles for light of wavelength 400 nm are [tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex].
[tex]\lambda=700\ \text{nm}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 700\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=44.43^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 700\times 10^{-9}}{10^{-6}}[/tex]
Here [tex]\dfrac{2\times 700\times 10^{-9}}{10^{-6}}=1.4>1[/tex] so there is no second order angle for this case.
The first order angle for light of wavelength 700 nm are [tex]44.43^{\circ}[/tex].
Second order angle does not exist.
an iron Tyre of diameter 50cm at 288k is to be shrank on to a wheel of diameter 50.35cm.To what temperature must the tyre be heated so that it will slip over the wheel with a radial gap of 0.5mm.Linear expansivity of iron is 0.000012k-1
Answer:
The answer should be D
Explanation:
Two resistors have resistances R1 and R2. When the resistors are connected in series to a 12.6-V battery, the current from the battery is 2.07 A. When the resistors are connected in parallel to the battery, the total current from the battery is 8.98 A. Determine R1 and R2. (Enter your answers from smallest to largest.)
Answer:
When R1 = 2.193, R2 = 3.894
When R1 = 3.894, R2 = 2.193
Explanation:
We are told that when R1 and R2 are connected in series, the voltage is 12.6 V and the current is 2.07 A.
Formula for resistance is;
R = V/I
R = 12.6/2.07
R = 6.087 ohms
Since R1 and R2 are connected in series.
Thus; R1 + R2 = 6.087 ohms
R1 = 6.087 - R2
We are also told that when they are connected in parallel, the current is 8.98 A.
Thus, R = 12/8.98
R = 1.403 ohms
Thus;
(1/R1) + (1/R2) = 1/1.403
Let's put 6.087 - R2 for R1;
(1/(6.087 - R2)) + (1/R2) = 1/1.403
Multiply through by 1.403R2(6.087 - R2) to get;
1.403R2 + 1.403(6.087 - R2) = R2(6.087 - R2)
Expanding gives;
1.403R2 + 8.54 - 1.403R2 = 6.087R2 - (R2)²
(R2)² - 6.087R2 + 8.54 = 0
Using quadratic formula, we have;
R2 = 2.193 ohms or 3.894 ohms
Thus,
R1 = 6.087 - 2.193 or R1 = 6.087 - 3.894
R1 = 3.894 or 2.193
When R1 = 2.193, R2 = 3.894
When R1 = 3.894, R2 = 2.193
Force exerted on a body changes it's
Which of the following have frequencies greater than orange light Your answer:
radio waves
purple light
ultraviolet rays
red light
green light
gamma rays
microwaves
infrared rays
Answer:
Gamma Rays have the highest frequencies
Explanation:
This is because Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies compared to the other light frequency. Radio waves, on the other hand, have the lowest energies, longest wavelengths, and lowest frequencies of any type of EM radiation which means the answer has to be gamma rays. Brainly Please!!!! Here are screenshots that may help