Answer:
[tex]f=4.70\times 10^{14}\ Hz[/tex]
Explanation:
Given that,
The wavelength of light, [tex]\lambda=6.38\times 10^{-7}\ m[/tex]
We need to find the frequency of the light. We know that,
[tex]c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.38\times 10^{-7}}\\\\f=4.70\times 10^{14}\ Hz[/tex]
So, the required frequency of light is equal to [tex]4.70\times 10^{14}\ Hz[/tex].
Calculate the terminal velocity of
the following nain drops faning
through air (a) one with a diameter
of 0.3cm 6 one with a a diameter
of o. Olm. Take the density of
water to be looo Kym3 and the
eis cosity of air to be ixlos pas.
The buoyancy effect of the air
may be ignored)
A wave has a wavelength of 1.5 meters and period of 0.083s. What is the waves speed?
HELP URGENT PLEASE!!!!!!!
Answer:
I think c I dont know sorry if I'm wrong
Convert (a) 50 oF, (b) 80 oF, (c) 95 oF to Celsius
Why is it harder to breathe on a
mountain?
A. The air pressure is so high the lungs can't expand.
B. The air is denser and oxygen can't flow easily into the
lungs.
C. The denser oxygen molecules sink below the
surrounding air.
D. The air is less dense so there are fewer oxygen
molecules.
4) Which statement about teamwork is not true?
A) Team members should not have to make personal sacrifices for the success of the team.
B) To be successful, all team members need to agree about how to achieve the goal.
C) To achieve agreement, teams must be able to communicate and negotiate.
D) Team members need to be ready to resolve conflicts in an open and honest way
Answer: A) Team should not have to make personal sacrifices for the success of the team.
Explanation:
A 2.0 kg breadbox on a fric-
tionless incline of angle u 40 is
connected, by a cord that runs over a
pulley, to a light spring of spring con-
stant k 120 N/m, as shown in
Fig. 8-43. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10 cm down the in- cline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box’s acceleration at the instant the box momentarily stops?
What causes the Coriolis effect?
A
The sun's position relative to Earth
B.
Earth's orbit around the sun
с
Moon phases
D
Earth's rotation
A 1.65-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 16.0 N. (a) What are the frequencies of the first three allowed modes of vibration
Answer:
Explanation:
mass per unit length ρ = .100 / 1.65 = .0606 . kg /m
length of wire L = 1.65 m
For fundamental frequency , the expression is as follows
n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]
L = 1.65 , T = 16 n and m = .0606
n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]
= 4.9 /s .
This is fundamental frequency .
other mode of vibration ( first three ) will be as follows
4.9 x 2 = 9.8 /s ,
4.9 x 3 = 14.7 /s .
What is the order of the events for the water cycle on a typical warm day?
А
rain, snow, sleet
B
precipitation, evaporation, rain
с
evaporation, condensation, precipitation
D
condensation, evaporation, precipitation
Which of the following actions will increase the current induced in a wire by a
magnetic field?
Answer:
The induced current can be increased in the coil in the following ways: By increasing the strength of the magnet. By increasing the speed of the magnet through the coil.
Explanation:
When a 20 kg explosive detonates and sends a 5 kilogram piece traveling to the right at 105 m/s
what is the speed and direction of the other 15 kilogram piece of the explosive!
Answer:
speed: 35m/s
direction: left
Explanation:
Assuming the right side is the positive direction:
before explosion:
P = mv = 0
after explosion:
P' = 15P + 5P
(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')
P' = 0.75mv1' + 0.25mv2'
P' = (15kg)v' + (5kg)(105m/s)
P' = 525kg/m/s + (15kg)v1'
P = P'
525kg/m/s + (15kg)v1' = 0
(15kg)v1' = -525kg/m/s
v1' = -35m/s
speed = |-35| = 35m/s
direction is to the left since the right side is the positive direction.
Your boss asks you to design a drone that begins its flight near the surface and rises to 9600 m. At the surface it will fly through air having a density of 1.23 kg per cubic meter and at its highest altitude the air density will become 0.62 kg per cubic meter. If the flight velocity near sea level is 45 mph, then how fast will in need to go at its highest altitude to maintain the same lift. Assume the coefficient of lift remains constant.
Answer:
[tex]63.38\ \text{mph}[/tex]
Explanation:
L = Lift force
[tex]\rho[/tex] = Density of air
A = Surface area
v = Velocity
[tex]v_1[/tex] = 45 mph
[tex]\rho_1=1.23\ \text{kg/m}^3[/tex]
[tex]\rho_2=0.62\ \text{kg/m}^3[/tex]
Coefficient of lift is given by
[tex]CL=\dfrac{2L}{\rho v^2A}\\\Rightarrow \rho=\dfrac{2L}{CL v^2A}[/tex]
So
[tex]\rho\propto \dfrac{1}{v^2}[/tex]
[tex]\dfrac{\rho_1}{\rho_2}=\dfrac{v_2^2}{v_1^2}\\\Rightarrow v_2=\sqrt{\dfrac{\rho_1}{\rho_2}}\times v_1\\\Rightarrow v_2=\sqrt{\dfrac{1.23}{0.62}}\times 45\\\Rightarrow v_2=63.38\ \text{mph}[/tex]
The velocity at the required altitude should be [tex]63.38\ \text{mph}[/tex] to maintain the same lift.
A bullet of mass 4.00 g is fired horizontally into a wooden block of mass 1.30 kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.170. The bullet remains embedded in the block, which is observed to slide a distance 0.240 m along the surface before stopping. Part A What was the initial speed of the bullet
Answer:
[tex]291.67\ \text{m/s}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of bullet = 4 g
[tex]m_2[/tex] = Mass of block = 1.3 kg
[tex]\mu[/tex] = Coefficient of friction = 0.17
[tex]s[/tex] = Displacement of block = 0.24 m
[tex]v_1[/tex] = Velocity of bullet
[tex]v[/tex] = Velocity of combined mass
[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
The energy balance of the system is given by
[tex]\dfrac{1}{2}(m_1+m_2)v^2=\mu(m_1+m_2)gs\\\Rightarrow v=\sqrt{2\mu gs}[/tex]
As the momentum is conserved in the system we have
[tex]m_1v_1=(m_1+m_2)v\\\Rightarrow m_1v_1=(m_1+m_2)\sqrt{2\mu gs}\\\Rightarrow v_1=\dfrac{(m_1+m_2)\sqrt{2\mu gs}}{m_1}\\\Rightarrow v_1=\dfrac{(4\times 10^{-3}+1.3)\times \sqrt{2\times 0.17\times 9.81\times 0.24}}{4\times 10^{-3}}\\\Rightarrow v_1=291.67\ \text{m/s}[/tex]
The initial speed of the bullet is [tex]291.67\ \text{m/s}[/tex].
Basic Science!! Helppp
How would increasing the pressure of this reaction affect the equilibrium
Explanation:
c because there is element
Answer:
C. H2 and N2 would react to produce more NH3
Explanation:
A.P.E.X
Which device converts electric energy into mechanical energy?
O A. An electromagnet
O B. A motor
O C. A transformer
O D. A generator
Answer:
B motor
Explanation:
A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surface, making it look rainbow-like (an effect known as iridescence). She aims a spectrometer at a particular spot and measures the wavelength to be 750 nm (in air). The index of refraction of water is 1.33.
Part A: The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot? t= 313nm
Part B: Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now? t=125nm
Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t[tex]_{min[/tex] = λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t[tex]_{min[/tex] = 750 / 2(1.20)
t[tex]_{min[/tex] = 750 / 2.4
t[tex]_{min[/tex] = 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t[tex]_{min[/tex] = λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t[tex]_{min[/tex] = 750 / 4(1.50)
t[tex]_{min[/tex] = 750 / 6
t[tex]_{min[/tex] = 125 nm
Therefore, the minimum thickness be now will be 125 nm
Which of the following happens to
density as air pressure decreases?
С C
A. Density increases.
B. Density stays the same.
C. Density decreases.
D. There is no correlation between air pressure and
density.
Explanation:
As pressure increases, with temperature constant, density increases. Conversely when temperature increases, with pressure constant, density decreases. Air density will decrease by about 1% for a decrease of 10 hPa in pressure or 3 °C increase in temperature.
The low pressure area near Earth's equator is filled by cool air moving in from
А
Europe and South America
B
the North and South Pole
с
the Prime Meridian
D
the Atlantic and Pacific Ocean
2. Plastic is a great conductor of charge so it moves quicker.
True
False
Answer:
the answer is false
Explanation:
plastic doesnt conduct anything
A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at a rate of 2ft/sec. Find the velocity of the top of the ladder at time t=1.
Answer: 0.516 ft/s
Explanation:
Given
Length of ladder L=20 ft
The speed at which the ladder moving away is v=2 ft/s
after 1 sec, the ladder is 5 ft away from the wall
So, the other end of the ladder is at
[tex]\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft[/tex]
Also, at any instant t
[tex]\Rightarrow l^2=x^2+y^2[/tex]
differentiate w.r.t.
[tex]\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s[/tex]
How many gallons of water does it take to produce the following:
a. Cheeseburger
b. Pound of butter
c. A pair of jeans
Answer:
a. 660 gallons
b.665 gallons
c. 1,800
Fairly easy question I’ll give extra points help.
1. third law
2. first law
3. third law
4. second law
The augue
1) What will be number of image if the angle
between two mirroro is
a) 45
b:36
The moment of inertia of the club head is a design consideration for a driver in golf. A larger moment of inertia about the vertical axis parallel to the club face provides more resistance to twisting of the club face for off-center hits. The mass of one club head is 200 g and its moment of inertia is 5000 g cm2 . What is the radius of gyration of this club head
Answer:
Explanation:
Moment of inertia I = M k² , where M is mass and k is radius of gyration .
Putting the given values in the equation
5000 = 200 x k²
k² = 25
k = 5 cm .
Radius of gyration is 5 cm .
You are at a train yard observing trains (because why not). You see a train car (let's call it car 1) moving to the right ( x direction) towards a stationary train car (let's call this one car 2). Car 1 has an initial velocity of 15.0 m/s. A helpful train employee tells you that Car 1 also has a mass of 1,825 kg and Car 2 has a mass of 2,645 kg. Car 1 gently collides with Car 2, allowing them to connect. After the collision the two train cars stay connected. You can assume that there is no friction in the system. If you have never see train cars connect, you can watch the first 25ish seconds of this video to see two train cars couple. However, these cars have friction, so they stop - unlike our problem. What is the Final Velocity of the system consisting of Car 1 and Car 2
Answer:
6.12 m/s
Explanation:
Using the law of conservation of momentum
momentum before collision = momentum after collision
m₁v₁ + m₂v₂ = (m₁ + m₂)V (since the train cars become attached to each other) where m₁ = mass of car 1 = 1,825 kg, m₂ = mass of car 2 = 2,645 kg, v₁ = initial velocity of car 1 = + 15.0 m/s (positive since it is moving in the positive x direction), v₂ = initial velocity of car 2 = 0 m/s (since it is initially stationary) and V = velocity of both cars after collision,
So, m₁v₁ + m₂v₂ = (m₁ + m₂)V
m₁v₁ + m₂(0 m/s) = (m₁ + m₂)V
m₁v₁ + 0 = (m₁ + m₂)V
V = m₁v₁/(m₁ + m₂)
substituting the values of the other variables into the equation, we have
V = 1,825 kg × 15.0 m/s/(1,825 kg + 2,645 kg)
V = 27375 kgm/s/ 4470kg
V = 6.124 m/s
V ≅ 6.12 m/s
Predicted height and total energy
Answer:
The predicted height is 2.809 meters, writing this in centimeters we get (1m = 100cm):
h = 2.809 m = (2.809)*(100cm) = 280.9 cm
And the total energy is:
E = 6.696 J
Explanation:
First let's see the problem.
We have an object of mass m = 274g which is thrown upwards with an initial velocity v0 = 6.991 m/s, in a place with a gravitational acceleration of g = 8.7 m/s^2
When the object is on the air, the only force acting on it will be the gravitational force, then the acceleration of the object will be equal to the gravitational acceleration, then we can write:
a(t) = -8.7 m/s^2
Where the negative sign is because this acceleration points down.
Now to get the velocity of the object we can integrate over time to get:
v(t) = (-8.7 m/s^2)*t + v0
Where v0 is a constant of integration, which is the initial velocity, then we can write this as:
v(t) = (-8.7 m/s^2)*t + 6.991 m/s
Now we can integrate again over the time to get the position equation.
p(t) = (1/2)*(-8.7 m/s^2)*t^2 + (6.991 m/s)*t + p0
Where p0 is the initial position, because the ball is being thrown from the ground, the initial position is 0.
Then the position equation is:
p(t) = (1/2)*(-8.7 m/s^2)*t^2 + (6.991 m/s)*t
Ok, now we know all the movement equations for the object.
The first thing we want to know is the maximum height of the object.
We know that the object reaches its maximum height when the velocity is zero (this is, the velocity stops being positive, meaning that the object stops going up, then in that time we have the maximum height)
We need to solve:
v(t) = 0m/s = (-8.7 m/s^2)*t + 6.991 m/s
(8.7 m/s^2)*t = 6.991 m/s
t = 6.991 m/s/( (8.7 m/s^2) = 0.804 seconds
The maximum height of the object is given by:
p(0.804s) = (1/2)*(-8.7 m/s^2)*(0.804)^2 + (6.991 m/s)*(0.804) = 2.809 m
The maximum height of the object is 2.809 meters.
Now let's find the maximum energy.
Remember that the energy of an object can be written as the sum of the potential energy U and the kinetic energy K.
E = K + U
Such that for an object of mass m and velocity v, the kinetic energy is:
K = (1/2)*m*v^2
And for an object of mass m, at a height h from the ground and with gravitational acceleration g, the potential energy is:
U = m*g*h
Now, when the object is at its maximum height, the velocity is zero.
Then K = 0
And for conservation of energy, the total energy of the object becomes potential energy.
E = 0 + U
E = U
So if we find the potential energy at the maximum height of the object's path, we can find the total energy of the object.
We know that:
mass = m = 274g = 0.274 kg (here i used that 1kg = 1000g)
height = h = 2.809 meters.
gravitational acceleration = g = 8.7 m/s^2
Then the potential energy at this point is:
U = 0.274 kg*(2.809 meters)*(8.7 m/s^2) = 6.696 J
This means that the total energy of the object is:
E = 6.696 J
would it be m/s or kg?
Answer:
m.s
Explanation:
Two students are sitting 1.50 m apart. One student has a mass of 70.0 kg and
the other has a mass of 52.0 kg. What is the gravitational force between them?
A. 8.01 x 10-9
B. 1.08 x 10-2
C. 2.28 x 10-8
Answer:
B
Explanation: