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.
An airplane flies 1000 miles in 2 hours. What is its average speed in miles per hour?
Answer:
500km per hour
Explanation:
if in 2 hours the airplane flies 1000 km then 1000 divided by 2 is 500km per hour.
A substance whose shape can easily change is a
Which of the following could be an example of chemical weathering?
a. rocks tumbling against each other
b. water seeping into the ground, dissolving the limestone to form a cave
c. a waterfall boring out a whole in a rock under it
Answer: B
Explanation:
Answers A and C are examples of physical weathering while B is chemical weathering when water and lime mix it creates a reaction
I beg you plz help me asap!!!
Q. How does new planet change our understanding about the universe?
Answer:
a.
Explanation:
there would be a new planet is our solar system which could cause different gravitation pull on all the planets also there could be possible be new life form or other valuable metals that haven't been discovered on this planet. hope this helps somewhat
A horizontal 2.00\ m2.00 m long, 5.00\ kg5.00 kg uniform beam that lies along the east-west direction is acted on by two forces. At the east end of the beam, a 200\ N200 N forces pushes downward. At the west end of the beam, a 200\ N200 N force pushed upward. What is the angular acceleration of the beam
Answer: [tex]240\ rad/s^2[/tex]
Explanation:
Given
Length of beam [tex]l=2\ m[/tex]
mass of beam [tex]m=5\ kg[/tex]
Two forces of equal intensity acted in the opposite direction, therefore, they create a torque of magnitude
[tex]\tau =F\times l=200\times 2=400\ N.m[/tex]
Also, the beam starts rotating about its center
So, the moment of inertia of the beam is
[tex]I=\dfrac{ml^2}{12}=\dfrac{5\times 2^2}{12}\\\\I=\dfrac{5}{3}\ kg.m^2[/tex]
Torque is the product of moment of inertia and angular acceleration
[tex]\Rightarrow \tau=I\alpha\\\\\Rightarrow 400=\dfrac{5}{3}\times \alpha\\\\\Rightarrow \alpha =240\ rad/s^2[/tex]
Hand pushes on a table with a force of 35n forward.reaction force
Answer:
como ías
Explanation:
At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM
Answer:
[tex]4.47\ \text{km/h}[/tex]
Explanation:
[tex]\dfrac{da}{dt}[/tex] = Rate at which the distance between A and starting point of B is changing = -20 km/h
[tex]\dfrac{db}{dt}[/tex] = Rate at which the distance of B is changing = 15 km/h
[tex]\dfrac{dc}{dt}[/tex] = Rate at which the distance between A and B is changing
Time after which the rate at which the distance between A and B is changing is 4 hours
Distance covered by A in 4 hours = [tex]20\times 4=80\ \text{km}[/tex]
a = Distance remaining to the start point of B = [tex]110-80=30\ \text{km}[/tex]
b = Distance covered by B in 4 hours = [tex]15\times 4=60\ \text{km}[/tex]
Distance between A and B after 4 hours
[tex]c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}[/tex]
[tex]c^2=a^2+b^2[/tex]
Differentiating with respect to time we get
[tex]c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}[/tex]
The rate at which the distance between the ships is changing at 4 PM is [tex]4.47\ \text{km/h}[/tex].
To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average whisper of 20.0 dB . So you decide to move closer to give the conversation a sound level of 60.0 dB instead. How close should you come
Answer:
r₂ = 0.316 m
Explanation:
The sound level is expressed in decibels, therefore let's find the intensity for the new location
β = 10 log [tex]\frac{I}{I_o}[/tex]
let's write this expression for our case
β₁ = 10 log \frac{I_1}{I_o}
β₂ = 10 log \frac{I_2}{I_o}
β₂ -β₁ = 10 ( [tex]log \frac{I_2}{I_o} - log \frac{I_1}{I_o}[/tex])
β₂ - β₁ = 10 [tex]log \frac{I_2}{I_1}[/tex]
log \frac{I_2}{I_1} = [tex]\frac{60 - 20}{10}[/tex] = 3
[tex]\frac{I_2}{I_1}[/tex] = 10³
I₂ = 10³ I₁
having the relationship between the intensities, we can use the definition of intensity which is the power per unit area
I = P / A
P = I A
the area is of a sphere
A = 4π r²
the power of the sound does not change, so we can write it for the two points
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
we substitute the ratio of intensities
I₁ r₁² = (10³ I₁ ) r₂²
r₁² = 10³ r₂²
r₂ = r₁ / √10³
we calculate
r₂ = [tex]\frac{10.0}{\sqrt{10^3} }[/tex]
r₂ = 0.316 m
Please solve for 15 points. Please don’t input a link.
Answer:
a). Single replacement.
Explanation:
Because one element replaces another element in a compound
Please help guy review question.
Answer:
66.67 km/h
Explanation:
20 + 30 = 50
50/.75 = 66.67
The liquid emerges into a vertical jet as it drains from the container, with the velocity profile in the jet remaining uniform. The outlet of the container is located 2.0 m above ground, and the radius of the emerging liquid jet changes with vertical distance from the bottom of the container as it accelerates under the action of gravity. Neglecting viscous losses and surface tension effects in the liquid jet, what is the velocity of the water jet as it strikes the ground when the container begins to drain
Answer:
6.26 m/s
Explanation:
Since we are neglecting viscous losses and surface tension effects in the liquid jet, by conservation of energy, the potential energy loss of the jet = kinetic energy gain of the jet
So, mgh = 1/2mv² where m = mass of water in jet, g = acceleration due to gravity = 9.8 m/s², h = height of outlet = 2.0 mand v = velocity of liquid jet
So, mgh = 1/2mv²
gh = 1/2v²
v² = 2gh
v = √(2gh)
v = √(2 × 9.8 m/s² × 2.0 m)
v = √(39.2 m²/s²)
v = 6.26 m/s
a disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular acceleration?
b. how long did it take the disk to reach this velocity?
Answer:
a) The angular acceleration is 1.875 radians per square second.
b) The time taken by the disk to reach the final angular speed is 8 seconds.
Explanation:
a) Let suppose that the disk accelerates uniformly, given that initial and final angular speed ([tex]\omega_{o}[/tex], [tex]\omega_{f}[/tex]), in radians per second, and change in angular position ([tex]\Delta \theta[/tex]), in radians, are known. The angular acceleration ([tex]\alpha[/tex]), in radians per square second, are found by using this expression:
[tex]\alpha = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \Delta \theta}[/tex] (1)
If we know that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega_{f} = 15\,\frac{rad}{s}[/tex] and [tex]\Delta \theta = 60\,rad[/tex], then the angular acceleration of the disk is:
[tex]\alpha = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \Delta \theta}[/tex]
[tex]\alpha = 1.875\,\frac{rad}{s^{2}}[/tex]
The angular acceleration is 1.875 radians per square second.
b) The time taken by the disk to reach the final angular velocity is determined by the following kinematic formula:
[tex]t = \frac{\omega_{f}-\omega_{o}}{\alpha}[/tex] (2)
Where [tex]t[/tex] is the time, in seconds.
If we know that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega_{f} = 15\,\frac{rad}{s}[/tex] and [tex]\alpha = 1.875\,\frac{rad}{s^{2}}[/tex], then the time taken by the disk is:
[tex]t = \frac{\omega_{f}-\omega_{o}}{\alpha}[/tex]
[tex]t = 8\,s[/tex]
The time taken by the disk to reach the final angular speed is 8 seconds.
A solenoid of 2100 turns, area 10 cm2, and length 30 cm carries a current of 4.0 A. (a) Calculate the magnetic energy stored in the solenoid from 1/2 LI 2. J [2 points] 0 attempt(s) made (maximum allowed for credit
Answer:
E = 0.1472 J
Explanation:
Given that,
The number of turns in the solenoid, N = 2100
Area of the solenoid, A = 10 cm² = 0.001 m²
The length of the solenoid, l = 30 cm = 0.3 m
Current in the solenoid, I = 4 A
We need to find the magnetic energy stored in the solenoid. The expression for the stored energy is :
[tex]E=\dfrac{1}{2}LI^2[/tex]
Where
L is self inductance of the solenoid,
[tex]L=\dfrac{\mu_oN^2A}{l}\\\\L=\dfrac{4\pi \times 10^{-7}\times 2100^2\times 0.001}{0.3}\\\\L=0.0184\ H[/tex]
So,
[tex]E=\dfrac{1}{2}\times 0.0184\times 4^2\\\\E=0.1472\ J[/tex]
So, 0.1472 J of energy is stored in the solenoid.
An enormous thunderstorm covers Dallas-Ft. Worth. Your best friend Clark is a storm chaser and heads to the center of the storm to take some readings while you stay dry at home. While Clark is at the center of the storm, he sees and hears lightning strike a tree that is 150 m from where he is standing. You are 127 km from the tree. How long does it take for the sound to reach Clark
Answer:
t = 0.437 s
Explanation:
The speed of sound is a constant that is worth v = 343 m / s
v = d / t
t = d / v
the time it takes for the sound to reach Clark at d = 150 m is
t = 150/343
t = 0.437 s
This same sound takes much longer to reach you
t₂ = 127 10³/343
t₂ = 370 s
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:
The angle between reflected ray and the normal line is
Answer:
Explanation:
angle of incidence.
Example of the center of the gravity
Answer:
The example of the center of the gravity is the middle of a seesaw
Explanation:
I hope this will help you and plz mark me brainlist
Which of the following best describes wind?
A А
Sinking warm air moving a few feet above the ground
B
An air current formed by changes in ocean tides
с
Cool air rushing in to fill an area of low pressure
D
Rising warm air pushing cool air down toward Earth
The correct statement about the wind is:
Cool air rushing in to fill an area of low pressureWhat is the wind?Wind is the movement of air currents in relation to the Earth's surface, which is caused by pressure differences and air movement.
Characteristics of the windIt is a meteorological phenomenon originated in the movements of rotation and translation of the Earth.When the rising air cools and loses the moisture it was carrying, due to condensation and rain, the result is dry and cool air.Therefore, we can conclude that the wind is the current of air that occurs in the atmosphere due to natural causes, from high pressure areas to low pressure areas.
Learn more about Characteristics of the wind here: https://brainly.com/question/11463167
A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction
acting on the box is a constant 20 N. How much work is done by the gravitational force?
A. 03
OB. 10 J
C. 100
D. 50 J
Answer:
B i think
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.
The energy goes from _____ to the _____ above it.
Answer:
The energy goes from the ground state to the excited states above it.
2. Which of the following objects has the greatest momentum?
A) a 145 grain baseball hit at 45 m/s
B) a 1200 kg car driving 25 m/s
C) a 2.0 *10-2 kg snail moving al 3.0 * 104 m/s
D) a 500 kg horse galloping at 12.0 m/s
Answer:
B) a 1200 kg car driving 25 m/s
Explanation:
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = mass * velocity [/tex]
This ultimately implies that, the mass of an object or body is directly proportional to its momentum. Thus, the higher the mass of an object or body, the greater would be its momentum and vice-versa.
By mere inspection of the data given, we can see that the object with the greatest amount of mass and velocity is the car weighing 1200 kilograms and moving at 25 meters per seconds.
Substituting into the formula, we have;
[tex] Momentum = 1200 * 25 [/tex]
Momentum = 30,000 Kgm/s
A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby weighs 20
kg. The carriage has
energy. Calculate it
Answer:
Energy in carriage (Potential energy) = 4,116 J
Explanation:
Given:
Mass of baby = 20 kg
Height = 21 m
Find:
Energy in carriage (Potential energy)
Computation:
The energy accumulated in an object as a result of its location relative to a neutral level is known as potential energy.
In carriage accumulated energy is potential energy.
Energy in carriage (Potential energy) = mgh
Energy in carriage (Potential energy) = (20)(9.8)(21)
Energy in carriage (Potential energy) = 4,116 J
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
A wheel rotates about a fixed axis with an initial angular velocity of 24 rad/s. During a 4 s interval the angular velocity decreases to 14 rad/s. Assume that the angular acceleration is constant during the 4 s interval. How many radians does the wheel turn through during the 4 s interval
Answer:
[tex]\theta=76\ rad[/tex]
Explanation:
Hoven that,
Initial angular velocity of the wheel = 24 rad/s
Final angular velocity = 14 m/s
Time, t = 4 s
We need to find how many radians does the wheel turn through during the 4 s interval. Let the displacement is [tex]\theta[/tex]. Using second equation of rotational kinematics to find it such that,
[tex]\theta=\omega_i t+\dfrac{1}{2}\alpha t^2[/tex]
Where
[tex]\alpha[/tex] is angular acceleration
[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{14-24}{4}\\\\\alpha =-2.5\ rad/s^2[/tex]
So,
[tex]\theta=24\times 4+\dfrac{1}{2}\times (-2.5)\times 4^2\\\\\theta=76\ rad[/tex]
So, it will turn 76 radian during the 4 s interval.
a glass rod is given a positive charge by rubbing it with silk. how did the rod become positive ?
Answer:
When the glass rod is rubbed with silk, the silk strips electrons from the rod, leaving it a positive charge. When the hard rubber rod is rubbed with wool, it gains electrons from the wool, gaining a negative charge.
Explanation:
one negative change you may encounter as a student or an employee
Answer:
you may get bullied or teased for being a differrent race, ethnic.
Answer:
Workload
Explanation:
You may feel stressed about a heavier workload or more too do. You will also be paid as an employee unlike, a student.
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:
A mom pushes her 19.3 kg daughter on the swing. If she gives her an initial velocity of 7.5 m/s at the bottom of the swing and the swing sits 0.6 m above the ground at it's lowest point, what height does she reach above the ground?
Answer:
3.17333333333? I hope I get it right
Explanation:
..................hello
Help me with this review question please.
Answer:
K E=( mv²)/2
=(60×3.5²)/2
=367.5J