Answer: Maintain the flight altitude obtained at that moment.
Explanation:
An autopilot is a system used without the pilot's direct assistance to guide an aircraft. Only a steady heading and altitude could be maintained by early autopilots, but modern autopilots are capable of monitoring any aspect of the flight envelope from just after take-off until landing.
Write down the formula of expansion
Answer:
well for me it's
Explanation:
L2-L1/L1(O2-O1)
O is the change in temperature
Problem 7:esvoe71017425b601f756a59bc7b76d2f39502730941492c25f1dd66c7ed01e4bb38ceovse 0N86-C1-52-40-A837-22820 If object A is twice the mass of object B what can be said about their kinetic energies just as they hit the ground
Answer:
the kinetic energy of A will be twice that of B
Explanation:
The formula for calculating kinetic energy is expressed using the formula
KE = 1/2mv²
m is the mass of the object
v is the velocity
For object A:
KEA = 1/2mAvA²
For object B:
KEB = 1/2mBvB² ... 1
If object A is twice the mass of object B, then mA = 2mB
From 1:
KEA = 1/2mAvA²
Substitute mA = 2mB
KEA = 1/2(2mB)vA² .... 2
Divide 1 by 2
KEB/KEA = 1/2mBvB²/mBvA²
KEB/KEA = 1/2vB²/vA²
Assuming they have the same velocities then vA ,= VB
The equation becomes:
KEB/KEA = 1/2vB²/vA²
KEB/KEA = 1/2vB²/vB²
KEB/KEA = 1/2
KEA = 2EB
Hence the kinetic energy of A will be twice that of B
You are planning a bicycle trip for which you want to average 24 km/h. You cover the first half of the trip at an average speed of 21 km/h. What must your average speed be in the second half of the trip to meet your goal?
Answer:
27km/h that's the answer
A scientist heated a tank containing 50 g of water. The specific heat of water is 4.18 J/gºC. The temperature of the water increased from 25ºC to 37ºC. How much heat energy did the water absorb?
1: 2,508 joules
2: -2,508 joules
3: 5,225 joules
4: 7,733 joules
Answer: a) 2,508
Explanation:
as speed increases, what happens to potential energy ?
Answer:
As the Speed increases , the potential energy decreases
Answer:
Since the object only has so much total internal energy, and the faster an object moves the larger the kinetic energy is, hence the potential energy will decrease as a result of speed increasing.
Explanation:
Hope this helped!
Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially elastic. Car B is stopped at a light when it is struck. Car A has mass m and speed v before the collision. After the collision:______
a. each car has half the momentum.
b. car A stops and car B has momentum m v.
c. car A stops and car B has momentum 3m v.
d. the momentum of car B is three times as great in magnitude as that of car A.
e. each car has half of the kinetic energy.
Answer:
D. The momentum of Car B is three times as great in magnitude as that of car A.
Explanation:
I majored in Physics
What is Intracellular communication
Answer:
Intercellular communication refers to the communication between cells. Membrane vesicle trafficking has an important role in intercellular communications in humans and animals, e.g., in synaptic transmission, hormone secretion via vesicular exocytosis.
If you had to choose one device to use with the battery, what would it be? Do you think such a battery could be invented
someday? Explain
Answer:
I would choose refrigerator.
Explanation:
In my opinion, such a battery could be invented by changing the device as a low power mode. Plus, it is a must in order to decrease the CFC gas spreading through the atmosphere. That's all, thank you.
Points P and Q are located at (0, 2, 4) and (-3, 1,5). Calculate
a) The position vector of P
b) The distance vector from P to Q
c) The distance between P and Q
d) A vector parallel to PQ with magnitude of 10
Answer:
a) The position vector of P is [tex]\vec P =(0, 2,4)[/tex].
b) The distance vector from P to Q is [tex]\overrightarrow{PQ} = (-3,-1,1)[/tex].
c) The distance between P and Q is [tex]\|\overrightarrow{PQ}\|=\sqrt{11}[/tex].
d) A vector parallel to PQ with magnitude of 10 is [tex]\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)[/tex].
Explanation:
a) The position vector of a point is the vector displacement from the origin to the location of the point. That is:
[tex]\vec P = (0,2,4)-(0,0,0)[/tex]
[tex]\vec P = (0-0, 2-0, 4-0)[/tex]
[tex]\vec P =(0, 2,4)[/tex]
The position vector of P is [tex]\vec P =(0, 2,4)[/tex].
b) First, we calculate the position vector of point Q:
[tex]\vec Q = (-3,1,5)-(0,0,0)[/tex]
[tex]\vec Q = (-3-0,1-0,5-0)[/tex]
[tex]\vec Q =(-3,1,5)[/tex]
The distance vector from P to Q is define by the following vectorial expression:
[tex]\overrightarrow{PQ} = \vec Q - \vec P[/tex] (1)
[tex]\overrightarrow{PQ} = (-3,1,5)-(0,2,4)[/tex]
[tex]\overrightarrow{PQ} =(-3-0,1-2,5-4)[/tex]
[tex]\overrightarrow{PQ} = (-3,-1,1)[/tex]
The distance vector from P to Q is [tex]\overrightarrow{PQ} = (-3,-1,1)[/tex].
c) There are two approaches to calculate the distance between P and Q:
First Method - Pythagorean Theorem:
[tex]\|\overrightarrow{PQ}\| = \sqrt{(-3)^{2}+(-1)^{2}+1^{2}}[/tex]
[tex]\|\overrightarrow{PQ}\|=\sqrt{11}[/tex]
Second Method - Dot Product:
[tex]\|\overrightarrow{PQ}\| = \sqrt{\overrightarrow{PQ}\,\bullet\,\overrightarrow{PQ}}[/tex] (2)
[tex]\|\overrightarrow{PQ}\| = \sqrt{(-3,-1,1)\,\bullet (-3,-1,1)}[/tex]
[tex]\|\overrightarrow{PQ}\|=\sqrt{11}[/tex]
The distance between P and Q is [tex]\|\overrightarrow{PQ}\|=\sqrt{11}[/tex].
d) To determine a vector parallel to PQ with a given magnitude is determined by the following expression:
[tex]\vec v = \frac{k}{\|\overrightarrow{PQ}\|} \cdot \overrightarrow{PQ}[/tex] (3)
Where [tex]k[/tex] is the scale factor.
If we know that [tex]\overrightarrow{PQ} = (-3,-1,1)[/tex], [tex]\|\overrightarrow{PQ}\|=\sqrt{11}[/tex] and [tex]k = 10[/tex], then the vector is:
[tex]\vec v = \frac{10}{\sqrt{11}}\cdot (-3,-1,1)[/tex]
[tex]\vec v = \left(-\frac{30}{\sqrt{11}},-\frac{10}{\sqrt{11}},\frac{10}{\sqrt{11}}\right)[/tex]
[tex]\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)[/tex]
A vector parallel to PQ with magnitude of 10 is [tex]\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)[/tex].
Why does the transition of light waves from water to air make it seem as if fish and other things in a pond are shallower than they actually are?
Answer:
When light gets in contact with water, it emerges into the air, and then it speeds up, which makes it look shallower than it actually is.
Explanation:
Answer:
Refraction
Explanation:
Light rays reflected from the fish are refracted at the surface of the water, but the eyes and brain trace the light rays back into the water as thought they had not refracted, but traveled away from the fish in a straight line. This effect creates a "virtual" image of the fish that appears at a shallower depth.
Please help which letter goes where ?
Answer:
See list below
Explanation:
5. positively charged particles : D
6. dense center of the atom: C
7. negatively charged particles: A
8. articles with NO charge: B
If a force does a negative amount of work on an object, does the object's speed increase, decrease, or remain the same? Justify your answer.
Answer:
if a force does a negative amount of work on an object ,so the speed of the object got decrease ,because the acceleration becomes negative so if the acceleration becomes negative so the speed is decreased.
Explanation:
Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B is negligible.
Answer:
The force P required is 1759.22 N
Explanation:
The missing diagram is seen in the first image below.
From the second image, we can see the schematic diagram of the engine hanging over the pulley.
To start with determining the value of the angle ∝;
[tex]tan \ \alpha = \dfrac{CD}{BD}[/tex]
where;
BD = AB-AD
Then;
[tex]tan \ \alpha = \dfrac{CD}{AB-AD}[/tex]
[tex]\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )[/tex]
replacing their respective values, where;
CD = 2 sin 30° m, AB = 2m and AD = 2 cos 30° m
[tex]\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )[/tex]
[tex]\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )[/tex]
[tex]\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )[/tex]
[tex]\alpha = tan^{-1} \bigg(3.73\bigg )[/tex]
[tex]\alpha \simeq 75^0[/tex]
From the third diagram attached below:
The tension occurring in the thread BC is equal to force P
[tex]T_{BC} = P[/tex]
Using the force equilibrium expression along the horizontal direction.
[tex]\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0[/tex]
replacing the value of [tex]\alpha \simeq 75^0[/tex]
[tex]-T_{AC} \ cos 30^0 + P cos 75^0 = 0[/tex]
[tex]P \ cos \ 75^0 = T_{AC} \ cos \ 30^0[/tex]
[tex]P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - - (1)[/tex]
Along the vertical direction, the force equilibrium equation can be expressed as:
[tex]\sum F_y =0[/tex]
[tex]-W + P \ sin \alpha + T_{AC} \ sin \ 30^0 = 0[/tex]
[tex]W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0[/tex]
replacing [tex]\alpha \simeq 75^0[/tex] and [tex]P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}[/tex]
[tex]W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0[/tex]
Also, replacing W for (200 × 9.81) N
[tex]200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0[/tex]
[tex]200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0[/tex]
[tex]1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)[/tex]
[tex]1962= T_{AC} \ (0.8660\times 3.732 + 0.5)[/tex]
[tex]1962= T_{AC} \ (3.231912 + 0.5)[/tex]
[tex]1962= T_{AC} \ (3.731912)[/tex]
[tex]T_{AC} = \dfrac{1962}{ \ (3.731912)}[/tex]
[tex]T_{AC} = 525.736 \ N[/tex]
From [tex]P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}[/tex]
[tex]P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}[/tex]
[tex]P =\dfrac{ 525.736 \times0.866}{0.2588}[/tex]
P = 1759.22 N
Thus, the force P required is 1759.22 N
A spring with a spring constant value of 2500 StartFraction N over m EndFraction is compressed 32 cm. A 1.5-kg rock is placed on top of it, then the spring is released. Approximately how high will the rock rise? 9 m 17 m 27 m 85 m
Answer:
it's 9m
Explanation:
Answer:
9m
Explanation:
edge2o2o
Why the tyres are in circular shape?
A rock is attached to the left end of a uniform meter stick that has the same mass as the rock. How far from the left end of the stick should the triangular object be placed so that the combination of meter stick and rock is in balance
Answer:
M₂ = M then L₂ = L
M₂> M then L₂ = \frac{M}{M_{2}} L
Explanation:
This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive
∑ τ = 0
The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂
The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.
M L + M₁ 0 - m₂ L₂ = 0
M L - m₂ L₂ = 0
L₂ = [tex]\frac{M}{M_{2}}[/tex] L
From this answer we have several possibilities
* if the two masses are equal then L₂ = L
* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L
The triangular object should be placed at the distance of [tex]\dfrac{M}{M_{2}} \times L[/tex] from the left end of the stick.
The given problem can be resolved using the concept of static equilibrium. We must consider a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive. So, static equilibrium says,
[tex]\sum \tau = 0[/tex]
Here, [tex]\tau[/tex] is the torque.
Let the mass of the rock is M and it is placed at a distance L. Such that the mass of the rod is [tex]M_{1}[/tex] which is considered to be placed in its center of mass. By uniformity in its geometric center (x = 0) and the triangular mass [tex]M_{2}[/tex] lies at the distance [tex]L_{2}[/tex].
Also, triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.
Then,
[tex]ML+M_{1} \times 0 -M_{2} \times L_{2}=0\\\\ML=M_{2}L_{2}\\\\L_{2}=\dfrac{M}{M_{2}} \times L[/tex]
Thus, we can conclude that triangular object should be placed at the distance of [tex]\dfrac{M}{M_{2}} \times L[/tex] from the left end of the stick.
Learn more about the equilibrium of forces here:
https://brainly.com/question/24018969
Waves that hit a fixed boundary return to the starting point on the ____
side that it began
opposite
same
Answer:
Opposite
Explanation:
Waves that hit a fixed boundary return to the starting point on the opposite side that it began.
After striking, the wave gets reflected and it moves in the direction opposite to the initial direction.
Hence, it would mean that waves that hit a fixed boundary return to the starting point on the opposite direction.
Define the Element of Art, value. If you were using a pencil, how would you create a darker value?
Elements of art are stylistic features that are included within an art piece to help the artist communicate. The seven most common elements include line, shape, texture, form, space, colour and value, with the additions of mark making, and materiality.
You would create a darker value if you shaded it, shading it makes it darker.
Energy Conservation With Conservative Forces: If a spring-operated gun can shoot a pellet to a maximum height of 100 m on Earth, how high could the pellet rise if fired on the Moon, where g
Answer:
h' = 603.08 m
Explanation:
First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)
h = height of pellet = 100 m
Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)
Vi = Initial Velocity of Pellet = ?
Therefore,
(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²
Vi = √(1960 m²/s²)
Vi = 44.27 m/s
Now, we use this equation at the surface of moon with same initial velocity:
2g'h' = Vf² - Vi²
where,
g' = acceleration due to gravity on the surface of moon = 1.625 m/s²
h' = maximum height gained by pellet on moon = ?
Therefore,
2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²
h' = (1960 m²/s²)/(3.25 m/s²)
h' = 603.08 m
what type of kinetic energy does flowing water have?
-rotational
-vibrational
-translational
-chemical
Answer:
none it's Hydroelectric energy
describe the relationship between the direction of the velocity vector and the direction of the acceleration for a body moving in a circle at coinstant speed
Answer:
a = v²/r
Explanation:
The acceleration of a body moving in a circular path is known as the centripetal acceleration. This is the acceleration of a body that keeps the body within the circular path. It is written in terms of the linear velocity v and the radius of the circle of rotation as shown;
a = v²/r where
v is the linear velocity
r is the radius
a is the centripetal acceleration
what is the secret of a delightful pastry
Answer:
✨ love ✨
words words words words words words words words words words words
Answer:
butter
Explanation:
a car drives under a bridge. as it passes underneath a man jumps down onto the top of the car. the man and the car continue moving together. what type of collision is this?
Answer:
Inelastic collision.
Explanation:
Because both the momentum of the man and that of the car, is conserved because they are both moving with the same velocity.
Also kinetic energy is not conserved because the car moved with different kinetic energy, as well as the man.
When they came into contact, the final kinetic energy is different from the initial kinetic energy.
Answer:
perfectly inelastic
Explanation:
WHO WANts TO HAVE SOME GLIZZY ACTION
What is Newton 2nd law
Answer:
Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass.
A heat engine receives 6000 J of heat from its combustion process and loses 4000 J through the exhaust and friction. What is its efficiency
Answer:
The efficiency of the heat engine is 33.33%.
Explanation:
Given;
input energy of the heat engine, = 6000 J
energy lost by the heat engine, = 4000 J
output energy of the heat engine = 6000 J - 4000 J = 2000 J
The efficiency of the heat engine is given as;
Efficiency = output energy / input energy
Efficiency = 2000 / 6000
Efficiency = 0.3333
Efficiency (%) = 33.33%
Therefore, the efficiency of the heat engine is 33.33%.
Focusing on irelevant information can negatively impact the ability to reason and problem solve effectively.
Answer:
true
Explanation:
as water warms on the stove, what is happening to the molecules in the liquid?
Answer:
The water molecules start moving faster.
Explanation:
As the water is heated, the water molecules absorb heat and moves faster. The longer the water is kept on the stove, the more heat and kinetic energy the molecules gain. Soon, the hydrogen bonds break down completely because of the high kinetic energy and the water molecules escape into the air as gas.
The temperature that causes this to happen is known as the boiling point.
A marble rolls off of a table that is 0.97 meters tall with a velocity of 1.87 m/s. How long
does it take the marble to hit the floor?
O 0.44 seconds
O 0.98 seconds
O 0.39 seconds
O 0.52 seconds
Waves that hit a fixed boundary return to the starting point on the____
side that it began
A.Opposite
B.same
Answer:
A: Opposite
Explanation:
From Newton's third law of motion, to every action there is an equal an opposite reaction. Thus, when the wave hits a fixed boundary, it is returned as a reflected wave to the starting point albeit on the other side.
