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IgorC [24]
3 years ago
7

jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh

at is the Cessna's speed relative to the 747? A 634 mph B 578 mph C 590 mph D 444 mph E 610 mph

Physics
1 answer:
Greeley [361]3 years ago
8 0

Answer:

C. 590 mph

\vert v_{cj}\vert=589.49\ mph

Explanation:

Given:

  • velocity of jet, v_j=500\ mph
  • direction of velocity of jet, east relative to the ground
  • velocity of Cessna, v_c=150\ mph
  • direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

\vec v_j=500\ \hat i\ mph

&

\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)

\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph

Now the vector of relative velocity of Cessna with respect to jet:

\vec v_{cj}=\vec v_j-\vec v_c

\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )

\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph

Now the magnitude of this velocity:

\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}

\vert v_{cj}\vert=589.49\ mph is the relative velocity of Cessna with respect to the jet.

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D is the correct answer hope this helps!!!

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3 years ago
A string of length L is under tension, and the speed of a wave in the string is v. What will be the speed of a wave in the strin
lawyer [7]

Answer:

The speed increases by square root 2 times.

Explanation:

Let the wave velocity initially be 'v' with tension in the string as 'T' of mass 'm'.

Given:

Initial length of the string (L₁) = L

Final length of the string (L₂) = 2L

Wave velocity in a stretched string of length 'L' is given as:

v=\sqrt{\frac{LT}{m}}

From the above equation we can conclude that if tension 'T' and mass 'm' remains the same, the wave velocity is directly proportional to the square root of length of the stretched string. Therefore,

v=k\sqrt L\\\\\frac{v_1}{\sqrt{L_1}}=\frac{v_2}{\sqrt{L_2}}

Now, plug in the given values and solve for v₂ in terms of v₁. This gives,

v_2=\frac{v}{\sqrt{L}}\times \sqrt{2L}\\\\v_2=\sqrt2 v

Therefore, the speed of the wave is increased by square root 2 times.

3 0
3 years ago
Two objects, A and B, have the same size and shape. Object A is twice as massive as B. The objects are simultaneously dropped fr
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Answer:

acceleration

Explanation:

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5 0
3 years ago
A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq
MAVERICK [17]

Answer:

Approximately 18 volts when the magnetic field strength increases from \rm 20\; mT to \rm 80\;mT at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF \epsilon that a changing magnetic flux induces in a coil is:

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt},

where

  • N is the number of turns in the coil, and
  • \displaystyle \frac{d\phi}{dt} is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux \phi is equal to

\phi = B \cdot A\cdot \cos{\theta},

where

  • B is the magnetic field strength at the coil, and
  • A\cdot \cos{\theta} is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so \theta = 0 and A\cdot \cos{\theta} = A. The area of this circular coil is equal to \pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}.

A\cdot \cos{\theta} = A doesn't change, so the rate of change in the magnetic flux \phi through the coil depends only on the rate of change in the magnetic field strength B. The size of the magnetic field at the instant that B = \rm 50\; mT will not matter as long as the rate of change in B is constant.

\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}.

As a result,

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V.

9 0
3 years ago
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F = gravitational force = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

m₁ = mass of object 1 = 235 kg

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r = distance between objects = 2.59 x 10⁴ m

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4 0
3 years ago
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