Answer:an inspirational statement of an idealistic emotional future
Explanation:
I don't know if it's right tho
Answer:
The document that state tge currentand future objectives of an organization.
what are the importance of informal education?
Answer:
Informal education is important because it can help individual to learn how to react and control situations.
It help individual to improve on its existing knowledge, new skills or ideas. This kind of education can happen any where and it can add values to the learner.
Explanation:
Informal education is a type of education that is learned from different life experiences, happenings outside a structured curriculum.
Informal education is important because it can help individual to learn how to react and control situations.
It help individual to improve on its existing knowledge, new skills or ideas. This kind of education can happen any where and it can add values to the learner.
From the anatomical position, the digits are ____ to the brachium.
A. Anterior
B. Inferior
C. Superior
D. Medical
which model best represents a pattern
Answer:
A
Explanation:
Use the image below to answer the question.
What does the arrow 'B' represent?
Question 7 options:
wavelength
refraction
frequency
compression
What kind of electricity does turning wheel generates? Please help!
Answer: Kinetic Energy to Electrical.
Explanation: The magnet is rotated as a result of the spinning wheels, and this results in a powerful stream of electrons, therefore converting kinetic to electrical.
A ray moving in water at 55.5 deg
enters plastic, where it bends to
48.7 deg. What is the index of
refraction for the plastic?
( water n = 1.33, Air n = 1.00 )
Answer:
Refractive index of the plastic = 1.46
Explanation:
By Snell's law,
[tex]\frac{\text{sin}\theta _{2} }{\text{sin}\theta _{1}}=\frac{n_1}{n_2}[/tex]
Here, [tex]\theta _1[/tex] = Angle of incidence in medium 1 (Plastic)
[tex]\theta_2[/tex] = Angle of refraction in medium 2 (Water)
[tex]n_1[/tex] = Refractive index of medium 1 (Plastic)
[tex]n_2[/tex] = Refractive index of medium 2 (Water)
By substituting values in the formula,
[tex]\frac{\text{sin}(48.7)}{\text{sin}(55.5)}=\frac{1.33}{n_2}[/tex]
[tex]n_2=\frac{1.33\times \text{sin}(55.5)}{\text{sin}(48.7)}[/tex]
= 1.46
Therefore, refractive index of the plastic = 1.46
g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket, and the drive
Answer:
The minimum coefficient of static friction required, µ = 0.10
Note. The question is incomplete. The complete question is given below:
While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.
The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.
Explanation:
First, velocity in mph is converted to m/s
1 mph = 0.447 m/s
55 mph ≈ 24.6 m/s
The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²
Force that can be generated by the truck, F = ma
F = 8850kg * 1.07 m/s² = 9469.5 N
However, with the added mass of the log on it, the acceleration of the truck will become;
a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²
Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N
Normal reaction on the truck due to the weight of the log, R = mg
R = 929 kg * 9.8m/s² = 9104.2 N
Coefficient of static friction, µ = F/R
µ = 901.13/9104.2
µ = 0.098 ≈ 0.10
Therefore, the minimum static friction required is µ = 0.10
Which of the following examples best represents an object with balanced forces acting upon it?
A - A boat accelerating through the water.
B - A book sitting at rest on a high shelf.
C- A wagon rolling down a steep hill.
D- A baseball thrown into the air.
Answer:the answer is “A book sitting on a high shelf” :)
Explanation:
Define a rotation of the earth answer fast
Answer:
here u go
Explanation:
Earth's rotation is the rotation of planet Earth around its own axis. Earth rotates eastward, in prograde motion. As viewed from the north pole star Polaris, Earth turns counterclockwise.
What effect does atmospheric pressure have on density?
Answer:sometimes it commits to how it will happen
Explanation:
Answer:
Atmospheric pressure increases density
Explanation:
Atmospheric pressure decreases the volume of space, since volume is inversely proportional to density, density will increase.
Which is true of low resistance wires
Answer:
the cooler the wire, the less the resistance
A small, 300 g cart is moving at 1.10 m/s on an air track when it collides with a larger, 4.00 kg cart at rest. After the collision, the small cart recoils at 0.890 m/s. What is the speed of the large cart after the collision (answer in m/s please)?
Answer:
0.0158m/s
Explanation:
Using the law of conservation of energy which states that the sum of momentum before collision is equal to the sum after collision. It is expressed mathematically as;
m1u1 + m2u2 = m1v1 + m2v1
m1 and m2 are the masses of the object
u1 and u2 are the initial velocities
v1 and v2 are the final velocities
Given
m1 = 300g = 0.3kg
u1 = 1.10m/s
m2 = 4.00kg
u2 = 0m/s (at rest)
v1 = 0.890
v2 = ?
Substitute the given values into the formula;
0.3(1.10) + 0 = 0.3(0.89) + 4v2
0.33 = 0.267 + 4v2
0.33-0.267 = 4v2
0.063 = 4v2
v2 = 0.063/4
v2 = 0.0158m/s
Hence the speed of the large cart after the collision is 0.0158m/s
According to the question:
Mass,
[tex]m_1 = 300 \ g = 0.3 \ kg[/tex][tex]m_2 = 4.00 \ kg[/tex]Final velocity,
[tex]u_1 = 1.10 \ m/s[/tex][tex]u_2 = 0 \ m/s[/tex]Initial velocity,
[tex]v_1 = 0.890 \ m/s[/tex][tex]v_2 = \ ?[/tex]By using the law of conservation, we get
→ [tex]m_1 u_1 +m_2 u_2 =m_1 v_1 +m_2 v_1[/tex]
By substituting the values, we get
→ [tex]0.3(1.10)+0 = 0.3(0.89)+4v_2[/tex]
→ [tex]0.33=0.267+4 v_2[/tex]
[tex]0.33-0.267 = 4 v_2[/tex]
[tex]0.063 =4v_2[/tex]
[tex]v_2 = \frac{0.063}{4}[/tex]
[tex]= 0.0158 \ m/s[/tex]
Thus the response above is appropriate.
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a 2 kg block is attached to a horizontal ideal spring with a spring constant 200N/m. when the
A circular wire loop lies inside a region of space containing a magnetic field. The direction of the magnetic field is out of the screen and parallel to the central axis of the loop. The magnitude of the magnetic field increases as a function of time. A circular loop oriented parallel to the plane of the screen lies inside a region containing magnetic field B. The field is directed out of the screen and is increasing. What is the direction of the induced current in the loop
Answer:
clockwise
Explanation:
According to the law given by Lenz, known as the Lenz law, it is said that a current induced in the circuit which is due to the change in the magnetic field and is so directed so as to oppose the change in the flux and to apply a force in the opposite direction if the force.
Here, as the magnetic field is directed out of the screen, the current flows in the direction which is clockwise in the loop and it opposes the increasing magnetic field.
The clockwise induced current will produce magnetic field in to the screen.
A piece of copper wire with thin insulation, 200 m long and 1.00 mm in diameter, is wound onto a plastic tube to form a long solenoid. This coil has a circular cross section and consists of tightly wound turns in one layer. If the current in the solenoid drops linearly from 1.60 A to zero in 0.120 seconds, an emf of 80.0 mV is induced in the coil. What is the length of the solenoid, measured along its axis
Answer:
N= 3
Explanation:
For this exercise we must use Faraday's law
E = - dФ / dt
Ф = B . A = B Acos θ
tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives
E = - A cos θ (B - B₀) / t
The angle enters the magnetic field and the normal to the area is zero
cos 0 = 1
A = π r²
In the length of the wire there are N turns each with a length L₀ = 2π r
L = N (2π r)
r = L / 2π N
we substitute
A = L² / (4π N²)
The magnetic field produced by a solenoid is
B = μ₀ N/L I
for which
B₀ = μ₀ N/L I
The final field is zero, because the current is zero
B = 0
We substitute
E = - (L² / 4π N²) (0 - μ₀ N/L I) / t
E = μ₀ L I / (4π N t)
N = μ₀ L I / (4π t E)
The electromotive force is E = 0.80 mV = 0.8 10⁻³ V
let's calculate
N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]
N = 320 10⁻⁷ / 9.6 10⁻⁶
N = 33.3 10⁻¹
N= 3
what is the difference between alcoholic and Mercury thermometer based on their function?
calculate the force necessary to keep a mass of 2 kg moving on a circular path of radius 0.2 m with a period of 0.5 second. what is the direction of force
Answer:
jjjjj
Explanation:
dkcdzjc
How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 7.1 % increase in the emitted frequency?
Answer:
v0 = 24.42 m/s (Approx)
Explanation:
Given:
Increase in frequency = 7.1% =
Computation:
Assume n = 100%
n1 = [(v+v0)/(v+v1)]n
[100 + 7.1] = [(344+v0)/(344+0)]100
107.1 = [(344+v0)/(344)]100
v0 = 24.42 m/s (Approx)
How fast should a rocket ship move for its mass to be dilated to 167% of its rest
mass?
Answer:
The two forces acting on rockets at the moment of launch are the thrust upwards and the weight downwards. Weight is the force due to gravity and is calculated (at the Earth’s surface) by multiplying the mass (kilograms) by 9.8.The resultant force on each rocket is calculated using the equation resultant force = thrust – weight.
Hopefully, this answer helps you! :)
The space shuttle releases a satellite into a circular orbit 630 km above the Earth.
How fast must the shuttle be moving (relative to Earth's center) when the release occurs?
Answer:
7,539 m/s
Explanation:
Let's use this equation to find the gravitational acceleration of this space shuttle:
[tex]\displaystyle g=\frac{GM}{r^2}[/tex]We know that G is the gravitational constant: 6.67 * 10^(-11) Nm²/kg².
M is the mass of the planet, which is Earth in this case: 5.972 * 10^24 kg.
r is the distance from the center of Earth to the space shuttle: radius of Earth (6.3781 * 10^6 m) + distance above the Earth (630 km → 630,000 m).
Plug these values into the equation:
[tex]\displaystyle g=\frac{(6.67\cdot 10^-^1^1 \ Nm^2kg^-^2)(5.972\cdot 10^2^4 \ kg)}{[(6.3781\cdot 10^6 \ m)+(630000 \ m)]^2}[/tex]Remove units to make the equation easier to read.
[tex]\displaystyle g=\frac{(6.67\cdot 10^-^1^1 )(5.972\cdot 10^2^4 )}{[(6.3781\cdot 10^6)+(630000 )]^2}[/tex]Multiply the numerator out.
[tex]\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(6.3781\cdot 10^6)+(630000 )]^2}[/tex]Add the terms in the denominator.
[tex]\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(7008100)]^2}[/tex]Simplify this equation.
[tex]\displaystyle g=8.11045189 \ \frac{m}{s^2}[/tex]The acceleration due to gravity g = 8.11045189 m/s². Now we use the equation for acceleration for an object in circular motion which contains v and r.
[tex]\displaystyle a = \frac{v^2}{r}[/tex]a = g, v is the velocity that the space shuttle should be moving (what we are trying to solve for), and r is the radius we had in the previous equation when solving for g.
Plug these values into the equation and solve for v.
[tex]\displaystyle 8.11045189 \ \frac{m}{s^2} = \frac{v^2}{7008100 \ m}[/tex]Remove units to make the equation easier to read.
[tex]\displaystyle 8.11045189 = \frac{v^2}{7008100}[/tex]Multiply both sides by 7,008,100.
[tex]56838857.89=v^2[/tex]Take the square root of both sides.
[tex]v=7539.154985[/tex]The shuttle should be moving at a velocity of about 7,539 m/s when it is released into the circular orbit above Earth.
A 300 g bird flying along at 6.0 m/s sees a 10 g insect heading straight toward it with a speed of 30 m/s. the bird opens its mouth wide and enjoys a nice lunch. What is the bird's speed immediately after swallowing?
Answer:
6.77m/s
Explanation:
Using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses of the object
u1 and u2 are the velocities before collision
v is the final collision
Given
m1 = 300g = 0.3kg
u1 = 6.0m/s
m2 = 10g = 0.01kg
u2 = 30m/s
Required
The bird's speed immediately after swallowing v
Substitute the given values into the formula
m1u1 + m2u2 = (m1+m2)v
0.3(6) + 0.01(30) = (0.3+0.01)v
1.8+0.3 = 0.31v
2.1 = 0.31v
v = 2.1/0.31
v = 6.77m/s
Hence the bird's speed immediately after swallowing is 6.77m/s
A mass m is located at the origin; a second mass m is at x = d. A third mass m is above the first two so the three masses form an equilateral triangle. What is the net gravitational force on the third mass?
Answer:
√3 * Gm²/d²
Explanation:
m1 = m, m2= m, distance = d. hence:
F = Gm²/d²
Let the origin be A, the point x = d be B and the point above the first two is C.
The net force acting on the third mass (point C) [tex]F_{net}[/tex] = [tex]F_A+F_B[/tex]
Let j represent the vertical component and i the horizontal component. Hence:
[tex]F_B=-F_j\\\\F_A=-F(icos\frac{\pi}{6}+jsin\frac{\pi}{6} )=-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )\\\\F_{net} =F_A+F_B\\\\F_{net} =-F_j+{-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )}\\\\F_{net} =-\frac{F}{2} \sqrt{3}(i+j\sqrt{3} )\\\\The\ magnitude\ of\ the\ net\ force\ is:\\\\|F_{net}|=\frac{F}{2}\sqrt{3}(\sqrt{1^2+\sqrt{3}^2 })=\frac{F}{2} \sqrt{3}(\sqrt{4})\\\\|F_{net}|=\frac{F}{2} \sqrt{3}*2=F*\sqrt{3}\\\\|F_{net}|=\sqrt{3}*\frac{Gm^2}{d^2}[/tex]
Answer:
The net gravitational force on the third mass = [tex]3^{0.5} * \frac{Gm^2}{d^2}[/tex]Explanation:
For equilateral triangle,
[tex]\theta = 30^o[/tex]
Force between masses,
[tex]F_1 = \frac{G*m*m}{d^2}\\\\F_! = \frac{Gm^2}{d^2}[/tex]
Therefore,
[tex]F_net = 2F_1cos\theta\\\\F_net = 2 * \frac{Gm^2}{d^2} * cos30\\\\F_net = 3^{0.5} * \frac{Gm^2}{d^2}[/tex]
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Obesity and stress are confirmed risk factors for which of the following?
lost of nutrition in human
Answer:
True
Explanation:
Obesity and stress are confirmed risk factors for CHD
A car is moving at 25.5 m/s when it accelerates at 1.94 m/s^2 for 2.3 s. What is the car's final speed? (Keep in mind direction and round to 2 decimals)
Answer:
29.96m/s
Explanation:
Given parameters:
Initial speed = 25.5m/s
Acceleration = 1.94m/s²
Time = 2.3s
Unknown:
Final speed of the car = ?
Solution:
To solve this problem, we are going to apply the right motion equation:
v = u + at
v is the final speed
u is the initial speed
a is the acceleration
t is the time taken
Now insert the parameters and solve;
v = 25.5 + (1.94 x 2.3) = 29.96m/s
What is the magnitude of the Box's Acceleration?
The Box's Acceleration : g sin θ
Further explanationNewton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object
∑F = m. a
F = force, N
m = mass = kg
a = acceleration due to gravity, m / s²
We plot the forces acting on the block (picture attached) according to the y-axis and the x-axis.
Because the motion of the block is in the same direction as the x-axis, ignoring the friction force with the inclined plane, then
[tex]\tt \sum F_x=m.a\\\\W.sin\theta=m.a\\\\mgsin\theta=m.a\\\\a=gsin\thet\theta[/tex]
The agonist in a movement is the muscle that provides the major force to complete the movement
(A)True
(B)False
Which has a greater force: a semi-truck at rest or a moving bicycle?
Although the semi truck certainly has a larger mass, it is not in motion and therefore does not have any momentum. The bicycle however has both mass and velocity and therefore has the larger momentum of the pair.
The ancestors of horses possess feet with five digits . Over time with the development of grassland the digits fused and former hooves . What evolutionary advantage did this transformation serve
Answer:
It enabled them to run faster from predators.
Explanation:
The evolution of hooves from a five-digit feet enabled horses to run faster from predators as well as support their larger weights and longer legs. Hope this helped and have an awesome day! :)
It enabled them to run faster from predators. ]
Explanation: The evolution of hooves from a five-digit feet enabled horses to run faster from predators as well as support their larger weights and longer legs.
giving brainiest to correct answer.
Imagine a 0.2 kg ball moving at 5 m/s colliding with a 0.3 kg ball moving at 2 m/s in the same direction. After the collision the 0.2 kg green ball stops moving and the 0.3 kg purple ball continues moving in the same direction.
giving brainiest to correct answer.
Answer:
[tex]5.33\ m/s[/tex]
Explanation:
[tex]We\ know\ that,\\Momentum=Mass*Velocity\\p=mv\\Hence,\\Lets\ first\ consider\ the\ case\ of\ the\ two\ balls\ 'Before\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Initial\ Velocity\ of\ the\ green\ ball=5\ m/s\\Initial\ Momentum\ of\ the\ green\ ball=5*0.2=1\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Initial\ Velocity\ of\ the\ pink\ ball=2\ m/s\\Initial\ Momentum\ of\ the\ pink\ ball=0.3*2=0.6\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'Before\ Collision'=1+0.6=1.6\ kg\ m/s[/tex]
[tex]Hence,\\Lets\ now\ consider\ the\ case\ of\ the\ two\ balls\ 'After\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Final\ Velocity\ of\ the\ green\ ball=0\ m/s\\Final\ Momentum\ of\ the\ green\ ball=0\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Final\ Velocity\ of\ the\ pink\ ball=v\ m/s\\Final\ Momentum\ of\ the\ pink\ ball=0.3*v=0.3v\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'After\ Collision'=0+0.3v=0.3v\ kg\ m/s[/tex]
[tex]As\ we\ know\ that,\\Through\ the\ law\ of\ conservation\ of\ momentum,\\In\ an\ isolated\ system:\\Total\ Momentum\ Before\ Collision=Total\ Momentum\ After\ Collision\\Hence,\\1.6=0.3v\\v=\frac{1.6}{0.3}=5.33\ m/s[/tex]
a passenger of mass m on a Ferris wheel moves in a vertical circle of radius r with constant speed v assuming that the seat remains upright during the motion which is true about the force the seat exerts on the passenger at the top of the circle
the upward force is equal in magnitude to the passenger's weight
the upward force is greater in magnitude than the passenger's weight
the upward force is smaller in magnitude than the passenger's weight
the upward force is zero
Answer:
the upward force is equal to the passenger weight
The answer that is the truth about the force that the seat exerts on the passenger is the upward force is equal in magnitude to the passenger's weight.
Why the upward force is equal in magnitude to weight
The acting forces on the body has been said to be vertical. This implies that the upward force and the downward force of gravity are the same.
Using the second law of Isaac Newton, the downward direction of the forces would be ascertained.
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