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3 years ago

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An airplane is pushed 22 degrees eastward off its northward course by a jet stream traveling at a speed of 136.73 km/hr. The new

speed is 365 km/hr. Find the plane's original velocity.

1 answer:

uranmaximum [27]3 years ago

3 0

As the speed of airplane is change due to jet stream

So the net speed is given as

$v_{net} = v_{plane} + v_{stream}$

now we can rearrange it as

$v_{plane} = v_{net} - v_{stream}$

now by the formula of vector difference we have

$v_{plane} = \sqrt{v_{net}^2 + v_{stream}^2 - 2v_{net}v_{stream}cos\theta}$

now plug in all values

$v_{plane} = \sqrt{365^2 + 136.73^2 - 2* 365* 136.73*cos22}[tex]v_{plane} = 243.7 km/hr$

so above is the speed of the plane

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The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.

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Answer:

$d=691.71km$

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

$t=\frac{d}{v_t}-\frac{d}{v_l}$

Here d is the distance at which the earthquake take place and $v_t, v_l$ is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

$t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km$

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A ruler of length 0.30m is pivoted at its centre. Equal and opposite forces of magnitude 2.0N are applied to the ends of the rul

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Answer:

0.3858 Nm

Explanation:

The torque of the couple is the dot product of the force vector and the couple vector from 1 end of the ruler to the center. This equals to the product of their magnitude times the cosine() of the angle made by their direction:

$T = \vec{F} \cdot \vec{s} = Fscos(50^0) = 2 * 0.3 * 0.643 = 0.3858 Nm$

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A stationary bomb explodes in space breaking into a number of small fragments. At the location of the explosion, the net force d

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Answer:

.D)The Vector sum of the linear momenta of the fragments must be zero.

Explanation:

.D)The Vector sum of the linear momenta of the fragments must be zero.

This statement is true. This is so because no external force is acting on the masses. The motion is created by internal force so momentum of fragments will be conserved.

A) this statement is false because kinetic energy was zero in the beginning ( the bomb was stationary in the beginning )

B ) This statement is false because it violates the law of conservation of momentum .( it does not violates only when all the fragments have equal mass )

C ) This statement is zero because kinetic energy is not a vector quantity so two kinetic energy when added can not sum up to zero.

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3 years ago

A beam of light strikes a sheet of glass at an angle of 57.0° with the normal in air. You observe that red light makes an angle

Contact [7]

<h2>Answers: </h2>

1) 1.359, 1.403

2) $2.207(10)^{8}m/s$ , $2.138(10)^{8}m/s$

Explanation:

The described situation is known as Refraction.

Refraction is a phenomenon in which a wave (the light in this case) bends or changes it direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index $n$ is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum ( $c=3(10)^{8}m/s$ ) and the speed of light $v$ in the second medium:

$n=\frac{c}{v}$ (1)

On the other hand we have the Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (2)

Where:

$n_{1}$ is the first medium refractive index . We are told is the air, hence $n_{1}\approx 1$

$n_{2}$ is the second medium refractive index

$\theta_{1}$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

Knowing this, let's begin with the answers:

<h2><u>1) Indexes of refraction for red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=38.1\º$ :

$(1)sin(57.0\º)=n_{2}sin(38.1\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(38.1\º)}$

$n_{2}=1.359$ (3)>>>Index of Refraction for red light

<h2>1b) Violet light</h2>

Again, using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=36.7\º$ :

$(1)sin(57.0\º)=n_{2}sin(36.7\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(36.7\º)}$

$n_{2}=1.403$ (4) >>>Index of Refraction for violet light

<h2><u>2) Speeds of red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Here we are going to use equation (1):

$n_{red}=\frac{c}{v_{red}}$

$v_{red}=\frac{c}{n_{red}}$

Substituting (3) in this equation:

$v_{red}=\frac{3(10)^{8}m/s}{1.359}$

$v_{red}=2.207(10)^{8}m/s$ >>>>Speed of red light

<h2>1a) Violet light</h2>

Using again equation (1):

$n_{violet}=\frac{c}{v_{violet}}$

$v_{violet}=\frac{c}{n_{violet}}$

Substituting (4) in this equation:

$v_{violet}=\frac{3(10)^{8}m/s}{1.403}$

$v_{red}=2.138(10)^{8}m/s$ >>>>Speed of violet light

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3 years ago

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Alexus [3.1K]

Answer:

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Explanation:

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3 years ago