Answer:
Explanation:
A) 215 + 45.0 = 260 km/h West
B) 215 - 45.0 = 170 km/h West
C) √(215² + 45.0²) = 219.6588... 220 km/h
θ = arctan(45.0/215) = 11.8214... 11.8° S of W
The airplane's ground velocity is westward at 170 km/h regardless of wind direction. The south wind does not affect the plane's westward velocity in component C.
Vector addition will determine each scenario's plane velocity relative to the ground. We'll simplify by assuming a consistent horizontal breeze.
Let's divide the airplane's velocity and wind velocity into westward (negative) and eastward (positive) components.
Data: Plane airspeed = 215 km/h.
45 km/h wind.
A. West wind:
The airplane's westward (negative) ground velocity:
West aeroplane velocity = -215 km/h
-45 km/h west wind.
Find Vg when the wind blows west:
Vg (west) = Aeroplane velocity + Wind velocity (west)
Vg(west) = -215 km/h - (-45 km/h)
Vg (west)=-170 km/h
B. East wind: 2. Westward (negative) aeroplane velocity relative to the ground:
West aeroplane velocity = -215 km/h
Wind velocity (east) = 45 km/h (because the wind is blowing eastward)
Find Vg when the wind blows east:
Vg (west) = Aeroplane velocity (west) + Wind velocity (east).
Vg (west) = -215 + 45 km/h
Vg (west)=-170 km/h
C. Southerly wind: 3. Westward (negative) aeroplane velocity:
West aeroplane velocity = -215 km/h
Wind velocity (south) = 0 km/h (no effect since wind is not moving westward)
Find Vg when the wind blows south:
Vg (west) = Aeroplane velocity (west) + Wind velocity (south).
Vg (west) = -215 + 0 km/h.
Vg (west)=-215 km/h
All resultant velocities are westward and 170 km/h.
Part C's plane's westward velocity is unaffected by the south wind. The plane's ground velocity remains -215 km/h (westward). Since the heading is westward, no angle is needed.
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You are standing 30 m due east of a 50 kg person who is running at a speed of 20 m/s due west. What is the magnitude of that person's angular momentum about you (in units of )
Hi there!
We can use the following equation for angular momentum:
[tex]\large\boxed{L = mrv}[/tex]
m = mass (kg)
r = distance from reference point (m)
v = velocity (m/s)
We can simply plug-and-chug the values provided in the question.
L = (50)(30)(20) = 30000 kgm²/s
A guitar string 63.6 cm long vibrates with a standing wave that has five antinodes. Which harmonic is this
Answer:
fifth harmonic
Explanation:
Two steel guitar strings have the same length. String A has a diameter of 0.489 mm and is under 410 N of tension. String B has a diameter of 1.27 mm and is under a tension of 809 N. Calculate the ratio of the wave speeds, vA/vB, in these two strings.
Answer:
Explanation:
vA / vB = √(TA/(m/L)) / √(TB/(m/L))
The lengths are the same, so the L divides out to 1
The material is identical so the mass will be directly proportional to the cross sectional area of the string
vA / vB = √(TA/(πdA²/4)) / √(TB/(πdB²/4))
π/4 is common so divides out to 1
vA / vB = √(TA/dA²) / √(TB/dB²)
vA / vB = √(410/0.489²) / √(809/1.27²)
vA / vB = 41.407 / 23.396
vA / vB = 1.8488961...
vA / vB = 1.85
Define faith in your own words.
Answer:
Faith, to me is almost like hope but for faith you need to believe. Some people associate the word faith with religion. But that's not it. Faith is really just a stronger term than believing someone. Like you can believe in someone but not have faith in that same person. You can have faith in someone but to do that you have to believe and trust them.
Explanation:
I need help with us history
Answer:
English is the language
Answer: im most likely wrong but i think its A
Explanation:
explain the different conditions that can result in hot and cold lahars, and explain how lahars change the earth's surface?
I need help with this answer I believe it's a democracy
Answer:
a. democracy
Explanation:
beacouse the government control of their members
Need help ASAP, 1 MC
Answer:
The first one is the only one that is true all the time
Explanation:
The second one may be true if friction is high enough.
The other three are false all the time
As a truck rounds a curve, a box in the bed of the truck slides to the side farthest from the center of the curve. This movement of the box is a result of
Answer:
inertia .
because yes
Một chất điểm khối lượng m=200g chuyển
động chậm dần với vận tốc biến đổi theo qui luật
v=30 – 0,4t2 (SI). Lực hãm tác dụng vào chất điểm
lúc t = 5 giây là
A. 8 N B. 0,8 N C. 4 N D. 0,4 N
Which statement is true for a series circuit
Answer: they have one path to flow
Explanation: share the same current
What are the characteristics of high energy wave?
A. Low frequencies and short wavelengths.
B. High frequencies and long wavelengths.
C. Low frequencies and long wavelengths.
D. High frequencies and short wavelengths
Answer:
D. High frequency and short wavelengths.
Explanation:
If a wave is high in energy it will have a higher frequency.
High frequency = short wavelengths
4. Protons and neutrons are held together to form this _________
Answer:
strong nuclear force.
Explanation:
hope this helps you!!
The maximum speed with which an 1000 kg car makes a 180-degree turn is 10 m/s. The radius of the circle through which the car is turning is 24 m. Determine the force of friction and the coefficient of friction acting upon
the car.
a rocket ship is moving through space at 1000 m/s. It accelerates in the same direction at 4m/s/s. What is its speed after 100 seconds
Answer:
Acceleration = (final velocity - starting velocity) / time
4 = (x-1000) / 100
<br/>x = 1400 m/s
Explanation:
The final velocity of the rocket ship which is moving with an initial velocity of 1000 m/s and acceleration of 4 m/s² after 100 seconds is 1400 m/s.
What is velocity?Velocity of a moving body is the rate of its speed. Mathematically velocity is the ratio of distance travelled to the time taken with a unit of m/s. Acceleration is the rate of change in velocity of moving body. The unit of acceleration is 4 m/s² .
Thus acceleration can be determined from the change in velocity with respect to the change in time. Now, the relation between initial velocity, acceleration, a and time, t with the final velocity is written in the equation below:
v = u + at.
Where, v is the final velocity and u be the initial velocity.
Given here the initial velocity is 1000 m/s. Acceleration of the rocket is 4 m/s² . Thus the velocity after 100 seconds is calculated as follows:
v = 1000 m/s + ( 4 m/s² × 100 s )
= 1400 m/s.
Hence, the speed of the rocket after 100 seconds will be 1400 m/s.
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A box of mass 7.7 - kg is accelerated from rest across a floor at a rate of 2.6 m/s2 for 18.5 s. Find the net work done on the box
Answer:
Explanation:
The net work will change the kinetic energy
W = ½mv² = ½m(at)² = ½ma²t²
W = ½(7.7)2.6²(18.5²) = 8907.3985 = 89 kJ
A woman skis from rest down a hill 20 m high. If friction is negligible, what is her speed at the bottom of the slope? Select one: O a. 20 m/s O b. 12 m/s O c. 400 m/s O d. 6 m/s
Hi there!
We can use the work-energy theorem to solve.
Recall:
[tex]\large\boxed{E_i = E_f}}[/tex]
Initial energy = final energy
The initial energy is purely potential (she starts from rest), and, if we assign the bottom of the slope as the 0 line, her energy at the bottom is purely kinetic.
PE = mgh
KE = 1/2mv²
We can begin by setting the two equal:
mgh = 1/2mv²
Cancel out the mass and rearrange to solve for velocity:
2gh = v²
v = √2gh
Plug in given values and use g ≈ 10 m/s²:
v = √2(10)(20) = 20 m/s
Two Blocks are connected by a massless rope over a massless,
frictionless pulley, as shown in the figure. The mass of block 2
is m2 = 10.1 kg, and the coefficient of kinetic friction
between block 2 and the incline is Mk = 0.200. The angle 0 of
the incline is 27.5º. If block 2 is moving up the incline at
constant speed, what is the mass mi of block 1?
The mass of block 1 will be 1.99 kg.The tension force is applied along the whole length of the wire, pulling energy equally on both ends.
What is tension force?The tension force is described as the force transferred through a rope, string, or wire as it is pulled by opposing forces.
Given that,
Mass of block 1=? kg
The coefficient of the kinetic friction,μ=0.200
Now consider the weight component in the uphill direction.The weight is found as;
[tex]\rm W=m_1gsin \theta[/tex]
The force is balanced in the vertical direction as;
[tex]\rm T=F_f-W[/tex]
When the force of friction is;
[tex]\rm F_F=\mu_k N[/tex]
[tex]\rm F_f=(m_1 gcos \theta)[/tex]
Substitute the value in the vertical balanced equation;
[tex]\rm T=m_1gsin(27.5)^0-\mu_kmgcos27.5^0[/tex]
[tex]\rm T-m_2g=0\\\\T=m_2g[/tex]
[tex]\rm (10.1) g=m_1g(0.699-0.2 \times(-0.714) ) \\\\ (10.1) g=m_1g (0.699+0.1428) \\\\\ (10.1) g= m_1 \times 0.8418 \\\\ m_1 =11.99 \ kg[/tex]
Hence the mass of block 1 will be 1.99 kg.
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A 4.0 kilogram projectile is fired horizontally from a 500 kilogram cannon initially at rest. The momentum of the projectile after being fired is 600 kilogram-meters per second to the north
(neglecting friction).
What is the speed of the cannon after firing?
0.83 m/s
1.2 m/s
o
3.3 m/s
150 m/s
The speed of the cannon after firing is 1.2 m/s
This can be solved using the law of conservation of momentum.
From the law of conservation of momentum,
⇒ The momentum of the projectile is equal to the momentum of the cannon.
MV = P................ Equation 1⇒ Where :
M = mass of the cannon,V = velocity of the cannonP = momentum of the projectile.⇒ make V the subject of the equation
V = P/M.................. Equation 2From the question,
⇒ Given:
P = 600 kgm/sM = 500 kg⇒ Substitute these values into equation 2
V = 600/500V = 1.2 m/sHence, The speed of the cannon after firing is 1.2 m/s
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How large is the acceleration of a 25 kg mass with a net force of 75 N applied horizontally to it?
Answer:
Explanation:
F = ma
a = F/m
a = 75/25
a = 3 m/s²
What happens to the gravitational force between two objects if the distance between them triples?
A. The force increases by a factor of 9
B. The force decreases by a factor of 9
C. The force decreases by a factor of 3
D. The force increases by a factor of 3
Three people are trying to move a box. Which set of forces will result in a net
force on the box of 20 N to the left?
Answer: push force gravity tension and reverse force
Explanation: sense they are pushing the box there is a push force gravity because this is likely on earth tension because it is the reverse of gravity and the reverse force because you have to have the reverse of push
a car moves at a speed of 30m/s to the west of 3hr, what is its displacement of the car in km?
Answer:
Explanation:
30 m/s • 3 hr •3600 s/hr / 1000 m/km = 324 km west
Someone please help with this question. From my knowledge the answer I believe to be correct is 4Em but I’m still not so sure. Please explain!
Answer choices:
1/2 Em
Em
2Em
4Em
Answer:
Explanation:
For an ideal spring over a frictionless horizontal surface, stored energy is only a function of the spring constant k and the distance of compression. The mass of the block doing the compressing is irrelevant
Energy stored in the first example is
Em = ½kd²
Energy stored in the second example is
E₂m = ½k(2d)² = 4(½kd²) = 4Em
So the second situation has four times as much stored spring potential energy as the first situation
4 Em is correct
Good job!
Determine the rTo understand the concept of nodes of a standing wave.
The nodes of a standing wave are points where the displacement of the wave is zero at all times. Nodes are important for matching boundary conditions, for example that the point at which a string is tied to a support has zero displacement at all times (i.e., the point of attachment does not move).
Consider a standing wave, where y represents the transverse displacement of a string that extends along the x direction. Here is a common mathematical form for such a wave:
y(x,t)=Acos(kx)sin(ωt),
where A is the maximum transverse displacement of the string (the amplitude of the wave), which is assumed to be nonzero, k is the wavenumber, ω is the angular frequency of the wave, and t is time.
Part A
Which one of the following statements about wave y(x,t) is correct?
adius of the 236U nucleus.
Answer:
The nodes of a standing wave are points where the displacement of the wave is zero at all times nodes are important for matching boundary conditions for example that the point at which a string is tied to a support has zero displacement at all times ie the point of attachment does not move consider a standing
A block slides on a rough 45 degree incline. The coefficient of friction is µk what is the ratio of acceleration when the block accelerates down the incline to the acceleration when the block is projected up the incline
Answer:
[tex]\frac{a_{d}}{a_{i}} = \frac{(1 -mu)}{mu}[/tex]
= (1 - μ)/μ
Explanation:
Always draw a diagram!
Up the incline:
[tex]Fr_{max}[/tex] = maximum friction
[tex]Fr_{max}[/tex] = μk
k = R = mg.cos(45) = mg.sin(45)
Resolution of forces parallel to the slope:
F (Fp in the diagram) = force of propulsion
g = gravity
[tex]F - Fr_{max} = ma_{i}[/tex]
[tex]F -[/tex] μ.mg.cos(45) [tex]= ma_{i}[/tex]
Down the decline:
Resolution of forces:
[tex]mg.sin(45) - Fr_{max} = ma_{d}[/tex]
[tex]mg.sin(45) -[/tex] μ.mg.cos(45) [tex]= ma_{d}[/tex]
Then, find the ratio:
[tex]\frac{ma_{d}}{ma_{i}} = \frac{mg.sin(45) - mu.mg.cos(45)}{-F + mu.mg.cos(45)} \\\\ \frac{a_{d}}{a_{i}} = \frac{k - k.mu}{-F + k.mu} \\\\ = \frac{k(1 -mu)}{-F + k.mu}[/tex]
Potentially, there is no need to consider F in this situation, in which case:
[tex]\frac{a_{d}}{a_{i}} = \frac{k(1 -mu)}{k.mu} \\\\ = \frac{(1 -mu)}{mu}[/tex]
= (1 - μ)/μ
As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constants. The angular speed changes from 3.70 rad/s at t = 0 to 2.00 rad/s at t = 8.60 s.
a. Use this information to determine σ and ω0.
σ = _______s−1
ωo = ______rad/s
b. Determine the magnitude of the angular acceleration at t = 3.00 s.
______rad/s2
c. Determine the number of revolutions the wheel makes in the first 2.50 s
_______rev
d. Determine the number of revolutions it makes before coming to rest.
_______rev
Hi there!
a.
We can use the initial conditions to solve for w₀.
It is given that:
[tex]\frac{d\theta}{dt} = w_0e^{-\sigma t}[/tex]
We are given that at t = 0, ω = 3.7 rad/sec. We can plug this into the equation:
[tex]\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}[/tex]
Now, we can solve for sigma using the other given condition:
[tex]2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}[/tex]
b.
The angular acceleration is the DERIVATIVE of the angular velocity function, so:
[tex]\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}[/tex]
c.
The angular displacement is the INTEGRAL of the angular velocity function.
[tex]\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.[/tex]
[tex]\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}[/tex]
[tex]\theta = 8.471 rad[/tex]
Convert this to rev:
[tex]8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}[/tex]
d.
We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.
[tex]0 = 3.7e^{-0.0714t}\\\\t = \infty[/tex]
Evaluate the improper integral:
[tex]\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad[/tex]
Convert to rev:
[tex]51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}[/tex]
Please help me! Some people have proposed a new way to build houses in areas that are likely to experience tsunamis. In this design, a house wouldn’t have solid walls on all four sides. Instead, some of the wall areas would be replaced by substances that water can travel through quickly, as shown in the diagram. How would this design help a house survive a tsunami? What drawbacks might there be to this design?
Answer:
I think some drawbacks are that since there are no solid walls meaning it is weak and if murphy's law is in place, the water will destroy the substance. Tsunami waves also happen very quickly so even if the water can travel thru the substance quickly, it probably won't be quick enough. This design could help if the wave is smaller because less destruction would occur.
Explanation:
yeah
9. P waves move faster than S waves A. True B. False
What is the weight of a 5kg object at the surface of the earth?
A. 49N
B. 49kg
C. 5.0kg
D. 25N
Answer: The answer is A) 49 N(Newtons).