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love history [14]
3 years ago
10

An airplane takes off from Dallas Texas to fly to new york city traveling ne for 2,760 km the Same plane returns that day to Dal

las the airplane spent 2 hours on the trip to new york city and the return trip took 2 1/2 hours what is the airplanes speed for the round trip
1,104
1,227
1,300​
Physics
2 answers:
Furkat [3]3 years ago
6 0

<u>Answer:</u> The speed of airplane is 1227 km/hr

<u>Explanation:</u>

Average speed is defined as the ratio of total distance traveled to the total time taken.

To calculate the average speed of the airplane, we use the equation:

\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}

We are given:

Total distance traveled = (2760 + 2760) km = 5520 km  (Round trip)

Total time taken = (2 + 2.5) hr = 4.5 hr

Putting values in above equation, we get:

\text{Average speed of airplane}=\frac{5520km}{4.5hr}=1227km/hr

Hence, the speed of airplane is 1227 km/hr

Kipish [7]3 years ago
3 0
<h3>Answer</h3>

1104 km/hour

<h3>Explanation</h3>

Distance between Dallas Texas to New York = 2760 km

Time the plane took from Dallas to New York = 2 hours

Time the plane took from New York back to Dallas = 2.5 hours

Formula to use

<h3>distance = speed x time </h3>

Speed the plane took from Dallas to New York

2760 = 2 x speed

speed = 2760 / 2

          = 1380 km/hour

Speed the plane took from New York to Dallas (ROUND TRIP)

2760 = 2.5 x speed

speed = 2760 / 2.5

           = 1104 km/hour

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A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th
ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

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Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

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5 0
3 years ago
A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75
Gnom [1K]

Answer:

Explanation:

Hello,

Let's get the data for this question before proceeding to solve the problems.

Mass of flywheel = 40kg

Speed of flywheel = 590rpm

Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

Time = 30s = 0.5 min

During the power off, the flywheel made 230 complete revolutions.

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + ω₂) / 2] × 0.5

But ∇θ = 230 revolutions

∇θ/t = (530 + ω₂) / 2

230 / 0.5 = (530 + ω₂) / 2

Solve for ω₂

460 = 295 + 0.5ω₂

ω₂ = 330rpm

a)

ω₂ = ω₁ + αt

but α = ?

α = (ω₂ - ω₁) / t

α = (330 - 590) / 0.5

α = -260 / 0.5

α = -520rev/min

b)

ω₂ = ω₁ + αt

0 = 590 +(-520)t

520t = 590

solve for t

t = 590 / 520

t = 1.13min

60 seconds = 1min

X seconds = 1.13min

x = (60 × 1.13) / 1

x = 68seconds

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + 0) / 2] × 1.13

∇θ = 333.35 rev/min

8 0
3 years ago
A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The
USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

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4 0
4 years ago
An archer pulls her bowstring back 0.376 m by exerting a force that increases uniformly from zero to 251 N. (a) What is the equi
Amanda [17]

Answer:

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(B)

Explanation:

(A) Let the spring constant be k.

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(B)

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W = 0.5 × kx^2

W = 0.5 × 667.5 × 0.376 × 0.376

W = 47.2 J

8 0
3 years ago
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