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inessss [21]
2 years ago
14

A plane initially traveling at 200 m/s due west experiences a 10 m/s head wind coming from the opposite direction. A). What will

be the relative velocity of the plane from the frame of reference of a bystander in the ground looking at the plane when it experiences the head wind current? B) Will the plane be late, on time or early after the wind current?
Physics
1 answer:
Hitman42 [59]2 years ago
8 0
Ok so it would be late and the relative velocity would be 190 m/s because 200 m/s - 10 m/s is 190 m/s. Hope this helps.
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A water-skier with weight Fg = mg moves to the right with acceleration a. A horizontal tension force T is exerted on the skier b
Degger [83]

Answer:

The correct relationships are T-fg=ma and L-fg=0.

(A) and (C) is correct option.

Explanation:

Given that,

Weight Fg = mg

Acceleration = a

Tension = T

Drag force = Fa

Vertical force = L

We need to find the correct relationships

Using balance equation

In horizontally,

The acceleration is a

T-Fd=ma...(I)

In vertically,

No acceleration

w=L

mg-L=0

Put the value of mg

L-fg=0....(II)

Hence,  The correct relationships are T-fg=ma and L-fg=0.

(A) and (C) is correct option.

3 0
2 years ago
A long, straight wire carrying a current of 3.45 A moves with a constant speed v to the right. A 5-turn circular coil of diamete
d1i1m1o1n [39]

Answer:

I = 69.3  μA

Explanation:

Current through the straight wire, I = 3.45 A

Number of turns, N = 5 turns

Diameter of the coil, D = 1.25 cm

Resistance of the coil, R = 3.25 \mu ohms

Distance of the wire from the center of the coil, d = 20 cm = 0.2 m

The magnetic field, B₁, when the wire is at a distance, d, from the center of the coil.

B_{1} = \frac{\mu_{0}I }{2\pi d}

B_{1} = \frac{4\pi* 10^{-7}  *3.45 }{2\pi *0.2}\\B_{1} =0.00000345 T

Magnetic field B₂ when the wire is at a distance, 2d from the center of the coil

B_{2} = \frac{\mu_{0}I }{2\pi(2d)) } \\B_{2} = \frac{\mu_{0}I }{4\pi d } \\

B_{2} = \frac{4\pi* 10^{-7}  *3.45 }{2\pi *2*0.2}\\B_{2} = 0.000001725 T

Change in the magnetic field, ΔB = B₂ - B₁ = 0.00001725 - 0.0000345

ΔB = -0.000001725

Induced current, I = \frac{E}{R}

E = -N (Δ∅)/Δt

Δ∅ = A ΔB

Area, A = πr²

diameter, d = 0.0125 m

Radius, r = 0.00625 m

A = π* 0.00625²

A = 0.0001227 m²

Δ∅ =  -0.000001725 * 0.0001227

Δ∅ = -211.6575 * 10⁻¹²

E = -N (Δ∅)/Δt

E = -5\frac{-211.6575 * 10^{-12} }{4.70} \\E = 225.17 * 10^{-12} V

Resistance, R = 3.25 μ ohms = 3.25 * 10⁻⁶ ohms

I = E/R

I = \frac{225.17 * 10^{-12} }{3.25 * 10^{-6} }

I = 0.0000693 A

I = 69 .3 * 10⁻⁶A

I = 69.3  μA

3 0
2 years ago
Rachel is helping her younger brother replace a broken part in his toy ambulance. This part is responsible for converting electr
Studentka2010 [4]
The light bulb, it takes electrical energy and turns it into l<span>ight energy!</span>
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2 years ago
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Which of the following sketches represents a possible configuration for this problem?
garri49 [273]
Where are the following sketches?
7 0
2 years ago
An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular accele
leva [86]

To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.

Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.

The average angular acceleration

\alpha = \frac{\omega_f - \omega_0}{t}

Here

\alpha = Angular acceleration

\omega_{f,i} = Initial and final angular velocity

There is not initial angular velocity,then

\alpha = \frac{\omega_f}{t}

We know that the relation between the tangential velocity with the angular velocity is given by,

v = r\omega

Here,

r = Radius

\omega = Angular velocity,

Rearranging to find the angular velocity

\omega = \frac{v}{r}}

\omega = \frac{30}{0.20} \rightarrow Remember that the radius is half te diameter.

Now replacing this expression at the first equation we have,

\alpha = \frac{30}{0.20*6}

\alpha = 25 rad /s^2

Therefore teh average angular acceleration of each wheel is 25rad/s^2

3 0
1 year ago
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