Answers
Answer:
Let R be the total distance
R²=10.4²+(8.7+2.1)²
R²=10.4²+10.8²
R²=224.8
You square root both sides
R=14.99km
If an airplane takes off, it travels 10.4 km west, 8.7 km north, and 2.1 km up, then using the Pythagorean theorem it is 14.99km from its take-off point.
What is Pythagoras' theorem?Pythagoras' theorem is a fundamental theorem in mathematics that relates to the sides of a right-angled triangle. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
In equation form, the theorem can be written as:
c² = a² + b²
Where
c =the length of the hypotenuse,
a and b = the lengths of the other two sides.
The theorem is named after the ancient Greek mathematician Pythagoras, who is credited with his discovery and proof. Pythagoras' theorem is widely used in mathematics and has many practical applications in fields such as architecture, engineering, and physics.
Here in the question
We can use the Pythagorean theorem to find the distance from the takeoff point:
Distance = √(West)² + (North+Up)²
Plugging in the given values, we get:
Distance = √(10.4)² +( 8.7 + 2.1)²
Distance ≈ 14.99km
Therefore, the airplane is approximately 14.99km away from the takeoff point.
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Related Questions
A flat loop of wire consisting of a single turn of cross-sectional area 8.80 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.80 T in 1.10 s. What is the resulting induced current if the loop has a resistance of 2.20
Answers
Answer:
The magnitude of the induced current is 4.73 x 10⁻³ A.
Explanation:
Given;
number of turns, N = 1
cross sectional area of the loop, A = 8.8 cm² = 8.8 x 10⁻⁴ m²
change in magnetic field strength, ΔB = 1.8 T - 0.5 T = 1.3 T
change in time, Δt = 1.10 s
resistance of the loop, R = 2.2 ohm
The magnitude of the induced emf is calculated as;
[tex]emf = \frac{NA \Delta B}{\Delta t} \\\\emf = \frac{1 \times 8.8\times 10^{-4} \times 1.3}{1.10} \\\\emf = 1.04 \times 10^{-3} \ V[/tex]
The induced current in the loop is calculated as;
[tex]I = \frac{emf}{R} \\\\I = \frac{1.04 \times 10^{-3}}{2.2} \\\\I= 4.73 \times 10^{-4} \ A[/tex]
Therefore, the magnitude of the induced current is 4.73 x 10⁻³ A
Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for slinging supplies to affected areas, similar to the electromagnetic lift catapults used to launch planes from aircraft carriers. This catapult is located at a fixed point 400 meters away and 50 meters below the target site. The catapult is capable of launching the payload at 67 meters per second and an initial launch angle of 50 degrees. Using your knowledge of kinematics equations, determine whether this would be sufficient to deliver the payload to the drop site.
Answers
Answer:
Please see below as the answer is self-explanatory.
Explanation:
We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).Both movements are independent each other, due to they are perpendicular.In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:[tex]v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)[/tex]
Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:[tex]t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)[/tex]
In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:[tex]\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)[/tex]
In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:[tex]v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)[/tex]
Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:[tex]\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)[/tex]
Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.why type of volcano is built almost entirely from ejected lava fragments
Answers
Answer:
Shield volcanoes
Explanation:
what is the name for a force of gravity acting on an object
Answers
Answer:
Weight
Explanation:
We call it "Ralph" or "the object's weight".
I want answer plzz??
Answers
Answer:
a=0.5g ms^-2
Explanation:
Let tension be T
acceleration be a
g be gravitational acceleration 9.81ms^-2
They share same T and a
by force diagram the net force on m:
T-2.1g=2.1a
net force on M:
6.3g-T=6.3a
solve:
(T-2.1g)/(6.3g-T)=2.1a/6.3a
3T+T=12.6g
T=3.15g newton
3.15g-2.1g=2.1a
a=0.5g ms^-2
As dancers twirl faster and faster around their partners, they are demonstrating what type of energy?
A. Binding energy,
B. Nuclear energy,
C. Kinetic energy,
Answers
What factors determine electric potential?
A. Mass and distance
B. Charge and density
C. Charge and distance
O D. Mass and charge
Answers
Answer: C. Charge and distance
A scientist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The balloon is spherical, with a radius of 2.10 m, and filled with hydrogen. The total mass of the balloon (including the hydrogen within it) and the instruments it carries is 24.0 kg. The density of air at ground level is 1.29 kg/m3. (a) What is the magnitude of the buoyant force (in N) acting on the balloon, just after it is released from ground level
Answers
Answer:
Fb = 490.4 N
Explanation:
According to Archimedes' principle, any object submerged in a fluid, receives a push upward (which we call buoyant force) equal to the weight of the volume of the fluid removed by the object.We can express this force (Fb), in terms of the density and the volume of the fluid, as follows:[tex]F_{b} = \rho * V * g (1)[/tex]
The volume removed from the fluid by the balloon is just the volume of the balloon, assuming that it is a perfect sphere, as follows:[tex]V = \frac{4}{3}*\pi * R^{3} = \frac{4}{3} *\pi *(2.1m)^{3} = 38.8 m3 (2)[/tex]
Replacing by the givens and (2) in (1), we get:[tex]F_{b} = \rho * V * g = 1.29 kg/m3* 38.8m3*9.8m/s2 = 490.4 N (3)[/tex]
what are three physical properties of gases
Answers
Answer: Gases have three characteristic properties: (1) they are easy to compress, (2) they expand to fill their containers, and (3) they occupy far more space than the liquids or solids from which they form. An internal combustion engine provides a good example of the ease with which gases can be compressed.
Explanation:
How is it difficult to perform electrostatic experiments on humid days
Answers
Answer:
that should be the right answer
Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of m A = 17.0 kg and an initial velocity of v 0A = 8.00 m/s, due east. Object B, however has a mass of m B = 29.0 kg and an initial velocity of v 0B = 5.00 m/s, due north. Find the magnitude and direction of the total momentum of the two-object system after the collision.
Answers
Answer:
pf = 198.8 kg*m/s
θ = 46.8º N of E.
Explanation:
Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:[tex]p_{ox} = p_{oAx} + p_{oBx} (1)[/tex]
We can do exactly the same for the initial momentum along the y-axis:[tex]p_{oy} = p_{oAy} + p_{oBy} (2)[/tex]
The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:[tex]p_{fx} = (m_{A} + m_{B} ) * v_{fx} (3)[/tex]
We can repeat the process for the y-axis, as follows:[tex]p_{fy} = (m_{A} + m_{B} ) * v_{fy} (4)[/tex]
Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:[tex]v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} = 2.96 m/s (5)[/tex]
In the same way, we can find the component of the final momentum along the y-axis, as follows:[tex]v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} = 3.15 m/s (6)[/tex]
With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:[tex]v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)[/tex]
The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:[tex]p_{f} = (m_{A} + m_{B})* v_{f} = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)[/tex]
Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.We can find this angle applying the definition of tangent of an angle, as follows:[tex]tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)[/tex]
⇒ θ = tg⁻¹ (1.06) = 46.8º N of E
Physics gravity question, Please help
Answers
Answer: 2.7 x10^-4 N
Explanation: 6.674 ×10^-11 × 1000 x 1000 divided by 0.5 squared.
In a transverse wave, the distance from any two consecutive crest or any two consecutive troughs is called the
Answers
Answer:
Wavelength
Explanation:
Answer:
Wavelength if it's the horizontal distance. Wave Height if it's vertical distance.
Explanation:
PLEASE HELP WITH THIS ONE QUESTION
A bird lands on a bird feeder which is connected to a spring. The mass of the bird is exactly the same as the mass of the bird feeder. How does the added mass affect the period of oscillation of the bird feeder?
The period is less, but not halved.
The period is more, but not doubled.
The period is halved.
The period is doubled.
Answers
Answer:
The period is more, but not doubled.
Explanation:
Recall that the period of a mass on a spring is T=2πmk.
An object is moving with constant non-zero velocity. Which of thw following statements about it must be true
Answers
Answer:
The net force on the object is zero.
Explanation:
An object is moving with constant non-zero velocity. If velocity is constant, it means that the change in velocity is equal to 0. As a result, acceleration of the object is equal to 0. Net force is the product of mass and acceleration. Hence, the correct option is (d) "The net force on the object is zero".
How fast is a ball going when it hits the ground after being dropped from a
height of 16 m? The acceleration of gravity is 9.8 m/s2
A. 22.5 m/s
B. 28.1 m/s
O C. 17.7 m/s
D. 19.3 m/s
Answers
Hi there!
[tex]\large\boxed{\text{C. 17.7m/s}}[/tex]
Use the following kinematic equation to solve:
vf² = vi² + 2(ad)
Since the initial velocity is 0 m/s because it started at rest, we can eliminate this part of the equation:
vf² = 2ad
Plug in the given acceleration and distance:
vf² = 2(9.8)(16)
vf ≈ 17.7. The correct answer is C.
How does the current in a series circuit compare
Answers
What happens when a moving object experiences no net force?
Answers
Answer:
An object with no net forces acting on it which is initially at rest will remain at rest. If it is moving, it will continue to move in a straight line with constant velocity. Forces are "pushes" or "pulls" on the object, and forces, like velocity and acceleration are vector quantities.
The course an object travels along is called DIRECTION.
A
TRUE
B
FALSE
DO NOT WRITE DOWN NONSENSE OR YOU WILL BE REPORTED!!!!!!!!
Answers
Which of the following is a contact force?
Magnetic
Friction
Electric
Gravity
Answers
Answer:
magnetic
Explanation:
because magnetic attracted other magnetic by pulling them contact forcely that creates a strong force when they are nearly close to one another
Plsss I want answer???
Answers
Answer:
s=136.89/2g meter
s=6.98 meter (correct to 3 sig.fig. taking g=9.81ms^-2)
Explanation:
u= + 11.7 ms^-1
a= - g ms^-2
At highest point: v=0ms^-1
v^2=u^2+2as
0=11.7^2+2(-g)s
s=136.89/2g meters
In 1666 at the age of 23, what scientist
developed the theories of gravitation inspired
by an apple falling from a tree?
A. Galileo Galilei
B. Issac Newton
C. Albert Einstein
D. Nicolaus Copernicus
Answers
Answer:
B - Isaac NewtonExplanation:
He first thought of his system of gravitation which he hit upon by observing an apple fall from a tree,
The incident occurring in the late summer of 1666.
While sitting in a boat, a fisherman observes that 2 complete waves pass by his position every 4 seconds. What is the period of these waves?A)2s B)8s C) 0.5 s D) 4s
Answers
Answer:
2
Explanation:
Because 2 waves 4 secs means 1 in 2s
The period of these waves is 2s. and The right option is A)2s.
The period of a wave is the time taken by a wave to complete one cycle.
The formula of period from the question is given below.
⇒ Formula:
T = t/2.................... Equation 1⇒ Where:
T = Period of the wavet = time taken for two complete oscillationFrom the question,
⇒ Given:
t = 4 seconds.⇒ Substitute these value above into equation 1
T = 4/2T = 2 seconds.Hence, The period of these waves is 2s.
The right option is A)2s
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An object is experiencing an acceleration of 0.4 m/s^2 while traveling in a circle of 35 m. What is it’s velocity?
Answers
Answer:
v = 3.74 m/s
Explanation:
Given that,
The acceleration of the object in circular path, a = 0.4 m/s²
The radius of the circle, r = 35 m
We need to find the velocity of the object. The acceleration of an object on the circular path is given by :
[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{0.4\times35}\\\\v=3.74\ m/s[/tex]
So, the velocity of the object is equal to 3.74 m/s.
A transverse wave is represented below. 1.5 m 0.20 m What is the approximate amplitude and wavelength of the wave? amplitude = 0.20 m, wavelength = 0.60 m B. amplitude = 0.20 m, wavelength = 0.30 m C. amplitude = 0.10 m, wavelength = 0.60 m OO amplitude = 0.10 m, wavelength = 0.30 m
Answers
Answer:
C. amplitude = 0.10 m, wavelength = 0.60 m
Explanation:
The diagram shows an oscillating progressive wave, with its amplitude and wavelength.
Amplitude of a wave is the maximum distance covered either upward or downward.
So that,
amplitude of the wave, A = [tex]\frac{0.2}{2}[/tex]
= 0.1
Amplitude of the wave = 0.1 m
Wavelength in this case is the distance from crest to crest, or trough to trough of the wave.
So that,
wavelength = [tex]\frac{1.5}{2.5}[/tex]
= 0.6
wavelength of the wave = 0.6 m
Therefore, the amplitude of the wave is 0.10 m, while the wavelength is 0.60 m.
If a person walks 3 m north and 5 meters east, how would you find the displacement for that person? what would the displacement be?
Answers
Answer:
AC)=(AB)2+(BC)2−−−−−−−−−−−−√=42+32−−−−−−√
⇒displacement=16+9−−−−−√=25−−√=5m
An object is spun in a circle of radius 1.5m with a frequency of 6.0Hz. what is it’s velocity ?
Answers
Answer:
56.55 m/s
Explanation:
Frequency=1/T
1/6 = 0.1666666
V=(2*pi*1.5)/0.166666
What us a magnetic domain?
Answers
magnetic domain is a region within a magnetic material in which the magnetization is in a uniform direction. This means that the individual magnetic moments of the atoms are aligned with one another and they point in the same direction.
What is the unique geological feature found on Mercury surface?
Answers
Answer:
The surface of Mercury has landforms that indicate its crust may have contracted. They are long, sinuous cliffs called lobate scarps. These scarps appear to be the surface expression of thrust faults, where the crust is broken along an inclined plane and pushed upward.
Explanation:
I hope this helps a little bit.
I'm asking for a quick favor. I'm trying to understand an equation that has to do with Projectile motion, Bernoulli's principle, and Magnus Effect. Basically focused on understanding air resistance on a projectile. I would like to discuss this privately rather than have it on this public forum. I'll give you 100 of my points if you help.
Answers
Explanation:
Projectile motion, Bernoulli's principle, and Magnus Effect.
Sure I would be happy to discuss projectile motion!
I'll do it if you mark brainliest :) I need the points thanks