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mojhsa [17]
3 years ago
7

A paper airplane is thrown to the north from the top of a cliff with a speed of 2 km/h. The paper airplane lands 15 minutes (0.2

5 hours) later. How far from the cliff did the paper airplane land?
A.0.25 km to the north
B.0.5 km to the north
C.0.7 km to the north
D.1.5 km to the north
Physics
1 answer:
Luden [163]3 years ago
5 0
If you neglect air resistance, then you can solve for the horizontal distance using the formula:

dx=vixt

Where:
dx = horizontal distance
vix = initial horizontal velocity
t = time in flight

Now you can see that you have all the given you need. 
vix= 2km/h
t = 0.25 hours

We use hours because you need the units to be the same. So just plug that in your equation and you will have your answer:

dx = vixt
     =(2km/h)(0.25h)
     =0.5km

This means that the answer to your question is B.
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Here we know that

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now from above all values we have

\omega_f = (-8.4 rad/s) + (-2.8 rad/s^2)(1.5)

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\omega_f = -12.6 rad/s

so final angular speed is -12.6 rad/s

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A constant force of 8N acting on an object displaces it through a distance of 3.0 m in the direction of force. Calculate work-do
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\blue{\bold{\underline{\underline{Answer:}}}}

  • \green{\tt{Work\:done=24\:J}}

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}

\green{\underline{\bold{Given :}}} \\  \tt: \implies Constant \: force(F) = 8 \: N \\  \\ \tt: \implies Displacement(s) = 3 \: m \\  \\ \red{\underline{\bold{To \: Find : }}} \\  \tt:  \implies Work \: Done(W.D) = ?

• <u>According to given question</u> :

\green{ \star} \tt \:  \theta \:  = 0 \degree \:  \:  \:  \: (Angle \: between \: force \: and \: displacement) \\  \\  \bold{As \: we \: know \: that} \\  \tt:  \implies Work \: Done = FS \: cos  \: \theta \\  \\  \tt:  \implies Work \: Done = 8 \times 3 \times cos \:0 \degree \\  \\ \green{ \circ} \tt \: cos  \: 0 \degree = 1  \\  \\  \tt:  \implies Work \: Done =24 \times 1 \\  \\   \green{\tt:  \implies Work \: Done =24 \: J}

5 0
3 years ago
A thin soap bubble of index of refraction 1.33 is viewed with light of wavelength 550.0nm and appears very bright. Predict a pos
faust18 [17]

Answer:

The possible thickness of the soap bubble = 1.034\times 10^{-7}\ m.

Explanation:

<u>Given:</u>

  • Refractive index of the soap bubble, \mu=1.33.
  • Wavelength of the light taken, \lambda = 550.0\ nm = 550.0\times 10^{-9}\ m.

Let the thickness of the soap bubble be t.

It is given that the soap bubble appears very bright, it means, there is a constructive interference takes place.

For the constructive interference of light through a thin film ( soap bubble), the condition of constructive interference is given as:

2\mu t=\left ( m+\dfrac 12 \right )\lambda.

where m is the order of constructive interference.

Since the soap bubble is appearing very bright, the order should be 0, as 0^{th} order interference has maximum intensity.

Thus,

2\mu t=\left (0+\dfrac 12\right )\lambda\\t=\dfrac{\lambda}{4\mu}\\\ \ = \dfrac{550\times 10^{-9}}{4\times 1.33}\\\ \ = 1.034\times 10^{-7}\ m.

It is the possible thickness of the soap bubble.

6 0
3 years ago
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