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3 years ago

An airplane flies 12 m/s due north with a velocity of 35.11 m/s. how far east does it fly?

2 answers:

4 0

<span>The question says: The airplane flew due north for the given duration (i interpret m/s to be milliseconds) and at the given velocity.

The statement mentions no change in direction or angle, therefore the airplane does not fly east (or the airplane travels 0.0m east).</span>

goblinko [34]3 years ago

3 0

The plane's velocity of 35.11 m/s is actually due in a north-eastward direction. The 12 m/s velocity is the vertical component of the plane's velocity, hence it is pointing northwards. We will use the formula:

Vy = Vsin∅

To determine the angle ∅ at which the plane is flying. This is:

12 = 35.11 * sin∅
∅ = 20.0 degrees

The eastward velocity is:

Vx = Vcos∅
Vx = 35.11 * cos(20)
Vx = 33.0 m/s

The plane's eastward velocity is 33.0 m/s

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2 years ago

During 4 hours one winter afternoon, when the outside temperature was 11° C, a house heated by electricity was kept at 24° C wit

Masteriza [31]

Answer:

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2 years ago

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Two point charges of +2.50 x 10^-5 C and -2.50 x 10^-5 C are separated by 0.50m. Which of the following describes the force betw

mr Goodwill [35]

Answer:

Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?

Using Coulomb's law:

F = 1/(4πE) x Q1 x Q2/ r^2

Where

k= 1/(4πE) = 9 x 10^9Nm2/C2

Therefore,

F = 9x 10^9 x 2.50 x 10^-5 x2.50 x

10^-5/. ( 0.5)^2

F= 5.625/ 0.25

F= 22.5N approximately

F= 23N.

To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.

Hence To = 23N, attractive. C ans.

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2 years ago

Choose all correct sentences Group of answer choices The power is maximum when the value of the impedance is greater than the va

Illusion [34]

Answer:

True b and c

Explanation:

In an RLC circuit the impedance is

$Z = \sqrt{[R^{2} + ( (wL)^{2} + (\frac{1}{wC})^{2} ] }$

examine the different phrases..

a) False. The maximum impedance is the value of the resistance

b) True. Resonance occurs when

(wL)² + (1 / wC)² = 0

w² = 1 / LC

c) True. In resonance the impedance is the resistive part and the power is maximum

d) False. In resonance the inductive and capacitive part cancel each other out

e) False. The impedance is always greater outside of resonance, but at the resonance point they are equal

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2 years ago

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago