Answer: See explanation
Explanation:
From the information given, we are informed that an instrument rated pilot is planning a flight under IFR on July 10, this year.
It should be noted that before conducting the flight, the pilot must have performed and logged the prescribed tasks and repetitions that are required for instrument currency no earlier than January, 10 for the year.
If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 30 mi/h, what is
its final speed when it hits the ground? (Disregard wind resistance. Do not reflect negative direction in your answer.)
ft/s
110 fus
100 ft/s
097 ft/s
91 ft/s
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Answer:
We can use 2 g H = v2^2 - v1^2 or
v2^2 = 2 g H + v1^2
Since 88 ft/sec = 60mph we have 30 mph = 44 ft/sec
The object will return with the same speed that it had initially so the object
starts out with a downward speed of 44 ft/sec
Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2
v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2
v2 = 110 ft/sec
ive an example of a pair of variables that have negative correlation. A. The quantity of fertilizers used and crop yield. B. The color of a person's shirt and the number of meals sold at Chinese restaurants. C. The number of physicists and the speed of sound. D. The number of winter coats sold and the temperature outside.
Answer:
D. The number of winter coats sold and the temperature outside.
Explanation:
A negative correlation can be described as a relationship existing between two variables such that as one of these variables goes up, the other variable would go down and vice versa.
With this explanation in mind, as the temperature outside increases, less winter coats would be sold because an increase in temperature would mean more heat and hotness. People would be uncomfortable wearing thick clothing.
What happen to the frequency of transverse vibration of a stretched string if its tension is halved and the area of cross section of the string is doubled?
Answer:
The fundamental frequency of the stretched string is:
[tex]f=[/tex] [tex]\frac{1}{2} \sqrt{\frac{T}{L} }[/tex] [ T = Tension and μ = mass per unit length]
Here,
μ = [tex]\frac{m}{L} = \frac{Vp}{L} = Ap[/tex]
[tex]f= \frac{1}{2} \sqrt{\frac{T}{Ap} }[/tex]
If T is halved and A is doubled,
[tex]f= \frac{1}{2} \sqrt{\frac{T'}{A'p} } = \sqrt{\frac{1}{2* 2* A* p} } = \frac{1}{2} (\frac{1}{2} \sqrt{\frac{T}{Ap} } = \frac{1}{2} f[/tex]
Thus, the frequency is reduced to half if its tension is halved and the area of cross-section of the string is doubled.
When the tension is halved and the area of cross section is doubled, the frequency increases by a factor of 1.414.
The frequency of transverse vibration of a stretched string is calculated as follows;
[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]
where;
T is the tension on the stringμ is the mass per unit length[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }\\\\f = \frac{1}{2l} \sqrt{\frac{Tl}{m} } \\\\f = \frac{l^2}{2l} \sqrt{\frac{T}{m} } \\\\f = \frac{A}{2l} \sqrt{\frac{T}{m}}[/tex]
when the tension is halved and the area of cross section is doubled, the frequency is calculated as;
[tex]\frac{f_1}{A_1 \sqrt{T_1} } = \frac{f_2}{A_2\sqrt{T_2} } \\\\f_2 = \frac{f_1 A_2\sqrt{T_2}}{A_1 \sqrt{T_1} }\\\\f_2 = \frac{f_1 \times 2A_1\sqrt{0.5T_1} }{\sqrt{T_1} }\\\\f_2 = \frac{f_1 \times 2A_1 \times 0.7071\sqrt{T_1} }{\sqrt{T_1} }\\\\f_2 = 1.414 f_1[/tex]
Thus, when the tension is halved and the area of cross section is doubled, the frequency increases by a factor of 1.414.
Learn more about tension in a string here: https://brainly.com/question/25743940
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Answer:
01.
Explanation:
Half the acceleration. Its heavier and moves slower. If it moved the same acceleration, the forces would also have to be doubled since the mass was.
How much current flows in the circuit if two resistors of 6 connected in parallel are
supplied with potential difference of 12 volts?
Answer:
4 A
Explanation:
Firstly we need to calculate the equivalent resistance. Since, two resistors of 6 connected in parallel, so equivalent resistance will be given by,
[tex]\longrightarrow[/tex] 1/R = 1/R1 + 1/R2
[tex]\longrightarrow[/tex] 1/R = (1/6 + 1/6) Ω
[tex]\longrightarrow[/tex] 1/R = (1 + 1)/6 Ω
[tex]\longrightarrow[/tex] 1/R = 2/6 Ω
[tex]\longrightarrow[/tex] R = 6/2 Ω
[tex]\longrightarrow[/tex] R = 3 Ω
We know that,
[tex]\longrightarrow[/tex] V = IR
V denotes p.d (12V)I denotes currentR denotes resistance (3 Ω)[tex]\longrightarrow[/tex] 12 = I × 3
[tex]\longrightarrow[/tex] 12 ÷ 3 = I
[tex]\longrightarrow[/tex] 4 Ampere = I
5- Clasifica los siguientes cambios de la materia, anotando delante de cada uno cambio físico (F) o cambio químico (Q): • Disolver azúcar en agua • Freir una chuleta • Arrugar un papel • El proceso de la digestión • Secar la ropa al sol • Congelar una paleta de agua • Hacer un avión de papel • Oxidación del cobre • Romper un lápiz • Prender fuegos artificiales • Excavar un hoyo • Quemar basura
Answer:
1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.
2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.
3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.
4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.
5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.
6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.
7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.
8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.
9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.
10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.
11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.
12) Quemar basura - Cambio químico - Reacción de combustión.
Explanation:
A continuación, veremos que representa cada caso:
1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.
2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.
3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.
4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.
5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.
6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.
7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.
8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.
9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.
10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.
11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.
12) Quemar basura - Cambio químico - Reacción de combustión.
the bodies in this universe attract one another name the scientist who propounded this statement
Answer:
It was proposed by Isaac Newton
Explanation:
The law of universal attraction of expression
F = [tex]G \ \frac{m_1m_2}{ r^2}[/tex]G m1m2 / r ^ 2
where G is a constant, m₁ and m₂ are the masses of the bodies and r the distance between them.
It was proposed by Isaac Newton
With this law Newton explained that the force that pulls the moon towards the earth is the same as that which attracts an apple towards the earth
Find the sum. Express the answer in scientific notation. (1.54 x 10^6)+(6.15 x 10^6)
Answer:
[tex] { \tt{(1.54 \times {10}^{6}) + (6.15 \times {10}^{6}) }} \\ = { \tt{(1.54 + 6.15) \times {10}^{6} }} \\ = { \tt{7.69 \times {10}^{6} }}[/tex]
at which point is the Kinetic energy the lowest?
Answer:
the lowest kinetic energy point is option D
A stone is dripped into the well 44 meter dip the spalash in heard in 3.12 s find speed of sound in air.
Answer:
The speed of sound is 366.67 m/s.
Explanation:
height, h = 44 m
total time, T = 3.12 s
Let the time taken by the stone to hit the water is t.
use second equation of motion
[tex]h = u t + 0.5 gt^2\\\\44 = 0.5\times 9.8 t^2\\\\t = 3 s[/tex]
Time taken by sound to g up
t'=T - t = 3.12- 3 = 0.12 s
The speed of sound is
[tex]v = \frac{h}{t'}\\\\v = \frac{44}{0.12}\\\\v = 366.7 m/s[/tex]
Determine the amount of potential energy of a 4.2 kg book that is placed on a shelf with a height of 0.9 meters. Round your answer to one decimal place
Answer:
that is a correct answer maybe .
Una grúa eleva un tubo de concreto de
400 kg, con Movimiento rectilíneo uniforme,
con el
cable ABC. Determine la tensión que pueden
soportar los cables AB, BC y BD, sabiendo
que los cables AB=BC y la tensión que
soporta el cable AB es de 150 N.
Explanation:
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what is chemical change ??
Answer:
the combination,decomposition or replacement that occurs in the molecules of matter during chemical change called chemical reaction
Answer:
A chemical change happens when one chemical substance is transformed into one or more different substances, such as when iron becomes rust. ... A chemical change is different from a physical change, which doesn't rearrange atoms or molecules and produce a completely new substance.
An unstrained horizontal spring has a length of 0.40 m and a spring constant of 340 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.033 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.
Answer:
(a) Both the charges are positive or negative.
(b) Teh value of each charge is 1.53 x 10^-5 C.
Explanation:
Spring constant, K = 340 N/m
Natural length, L = 0.4 m
stretch, y = 0.033 m
(a) Let the charge on each sphere is q and they repel each other so the nature of charge of either sphere may be both positive or both negative.
(b) The electrostatic force is balanced by the spring force.
[tex]\frac{kq^2}{(L + y)^2}=Ky\\\\\\\frac{9\times 10^9 q^2}{(0.4 +0.033)^2} = 340\times0.033\\\\q= 1.53\times 10^{-5} C[/tex]
[tex]what \: is \: reflection \: of \: light \: \: \: \: {?}[/tex]
You can see stars brighter than magnitude +6 with your naked eye under dark sky conditions, and you can also see the full moon at magnitude –13. (You can glance at the Sun, which is at magnitude –27, but you can't safely look at the Sun.) Given these limits, what is the dynamic range of the naked eye?
Quickly pls!!! A wave with a wavelength of 0.5 m moves with a speed of 1.5 m/s. What is the frequency of the wave?
A. 2.0 Hz
B. 1.0 Hz
C. 0.33 Hz
D. 3.0 Hz
9. In a __________ collision, 100% of both vehicles' speed is directed towards the point of impact. A. head-on B. rear-end C. side-impact
Answer: A
Explanation:
:)
This question is divided into two parts. This is part (a) of the question. A proton accelerates from rest in a uniform electric field of 580 N/C. At some later time, its speed is 1.00 x 106 m/s. (a) Find the magnitude of the acceleration of the proton. (Mass of the proton is 1.67 x 10-27 kg and charge is 1.60 x 10-19 C) (in the following options 10^10 m/s^2 is 1010 m/s2)
Answer:
The acceleration of proton is 5.56 x 10^10 m/s^2 .
Explanation:
initial velocity, u = 0
Electric field, E = 580 N/C
final speed, v = 10^6 m/s
(a) Let the acceleration is a.
According to the Newton's second law
F = m a = q E
where, q is the charge of proton and m is the mass.
[tex]a= \frac{q E}{m}\\\\a = \frac{1.6\times10^{-19}\times 580}{1.67\times 10^{-27}}\\\\a= 5.56\times 10^{10} m/s^2[/tex]
E
How does the construction of dams positively affect natural resources?
by providing water for irrigation and restoring trees to areas where forests once existed
by creating reservoirs, preventing flooding, and renewing destroyed ecosystems
by preventing flooding, creating reservoirs, and providing water for irrigation
by renewing destroyed ecosystems and restoring trees to areas where forest once ersted
Answer:
by preventing flooding, creating reservoirs, and providing water for irrigation
Dams and waterways store and provide water for irrigation so farmers can use the water for growing crops. This idea goes way back into history. Irrigation is an important part of using water. In areas where water and rain are not abundant (like the desert), irrigation canals from rivers and dams are used to carry water.
Dams help in preventing floods. They catch extra water so that it doesn’t run wild downstream. Dam operators can let water out through the dam when needed.
A boy travels 12km east wards to a point B and then 5km southwards to another point C. Calculate the difference between the magnitude of the displacement of the boy and the distance travelled by him
The difference b/w the displacement and total distance traveled is 4km.
Explanation.
▪ total distance - displacement
= 17 km - 13 km
= 4 km...answer
The electric motor in the car is powered by a battery.
To charge the battery, the car is plugged into the mains supply at 230 V
The power used to charge the battery is 6.9 kW
Calculate the current used to charge the battery.
Answer:
I = 30 A.
Explanation:
Given that,
The voltage of the battery, V = 230 V
Power used to charge the battery, P = 6.9 kW
We need to find the current used to charge the battery. The formula for the power is given by :
P = VI
Where
I is current
So,
So, the required current is 30 A.
scientists are seen very busy in designing the solar power equipments, why?
Answer:
it is because solar energy is the perpetual source of energy and by using it non renewable sources of energy can be conserved for the future
6. An object is fired from the gound at 275 m/s at an angle of 55° N of E.
a. How far away did the object first hit the ground?
b. what is the maximum height that the object reaches?
there u go fella hope u understood
¿ Sobre que superficie se desplazara mas rápidamente un tejo, sobre cemento o sobre cerámica? ¿por que?
Answer:
cemento potque es plano
The reservoir stores 6 500 000 m3 of water. The density of the water is 998 kg/m3. Calculate the mass of water in the reservoir. Give your answer in standard form.
Answer:
Mass = 64,870,000,000 kilograms
Explanation:
Given the following data;
Density = 998 kg/m³
Volume = 6,500,000 m³
To find the mass of water in the reservoir;
Density can be defined as mass all over the volume of an object.
Simply stated, density is mass per unit volume of an object.
Mathematically, density is given by the equation;
Density = mass/volume
Making mass the subject of formula, we have;
Mass = density * volume
Mass = 998 * 6,500,000
Mass = 64,870,000,000 kilograms
1. Use a micrometer to find the thickness of the foil you used. Do your data confirm the measured thickness? Determine the percentage difference between the measured thickness and the thickness you found in your experiment.
Answer:
Yes.
Explanation:
Yes, my data confirm the measured thickness because the micrometer measure precise thickness of any object. If the same instrument and right method is used for the measurement of the thickness of the foil then there is zero difference occur between the measured thickness and the thickness in your experiment so using right method can reduce or eliminate the difference between the measured thickness and the thickness in experiment.
Imagine you see Mars rising in the east at 6:30 pm. Six hours later what direction would you face (look) to see Mars when it is highest in the sky
Answer:
The Mars appears in the direction of South.
Explanation:
Mars is rising in the east at 6: 30 PM. The period of rotation of earth is 24 hours.
So, 6 hours is the one fourth of the period of rotation of earth. Earth rotates counter clockwise on its axis, so after 6 hours, we see the Mars in the direction of South.
A hot-air balloon is rising upward with a constant speed of 2.85 m/s. When the balloon is 2.50 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground
Answer:
Explanation:
You have to declare which way is plus -- up or down. I will use down.
vi = - 2.85 The balloon is going up. That is the minus direction.
a = 9.81
d = 2.50 meters distance in this case is from the object to the ground.
d = vi*t + 1/2 a t^2
-2.50 = -2.85*t + 1/2 * 9.81 * t^2
-2.50 = -2.85*t + 4.905 * t^2 transfer the left to the right.
-4.905 t^2 + 2.85*t + 2.50 = 0
Use the quadratic formula to solve for t.
It turns out that t = 1.06
A 5.0 kg block of ice is at rest at the top of a smooth inclined plane. The block is released and slides 2.0 m down the plane. Assuming there is no friction between the block and the surface, calculate
a) the gravitational potential energy at the top of the plane
b) the component of the weight parallel to the plane
c) the acceleration of the block
d) the velocity of the block at the bottom of the plane
e) the kinetic energy at the bottom of the plane.
Answer:
a) 98.1 Joules
b) 49.05 N × sin(θ)
c) 9.81 × sin(θ)
d) The velocity of the block at the bottom of the plane, v is approximately 6.264 m/s
e) 98.1 Joules
Explanation:
The given parameters of the block are;
The mass of the block, m = 5.0 kg
The distance down the plane the block slides, h = 2.0 m
The friction between the block and the surface = 0
Let θ represent the angle of inclination oof the plane
a) The gravitational potential energy, P.E. = m·g·h
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
∴ P.E. ≈ 5.0 kg × 9.81 m/s² × 2.0 m = 98.1 Joules
The gravitational potential energy, P.E. ≈ 98.1 Joules
b) The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex], is given as follows;
[tex]w_{\parallel}[/tex] = w × sin(θ) = m·g·sin(θ)
∴ [tex]w_{\parallel}[/tex] ≈ 5.0 kg × 9.81 m/s² × sin(θ) = 49.05 × sin(θ) N
The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex] ≈ 49.05 N × sin(θ)
c) The component of the weight along the inclined plane = The force with which the block moves along the inclined plane, therefore;
[tex]w_{\parallel}[/tex] = m·g·sin(θ) = m·a
Where a represents the acceleration of the block along the plane
Therefore, by comparison, we have;
g·sin(θ) = a
∴ a ≈ 9.81 × sin(θ)
d) Given that the motion of the block is 2.0 m downwards, we have;
The velocity of the block at the bottom of the plane, v² = 2·g·h
Therefore, v² ≈ 2 × 9.81 m/s²× 2.0 m = 39.24 m²/s²
v = √(39.24 m²/s²) ≈ 6.264 m/s
e) The kinetic energy at the bottom of the plane, K.E. = (1/2)·m·v²
∴ K.E. = (1/2) × 5.0 kg × 39.24 m²/s² = 98.1 J
