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VARVARA [1.3K]
3 years ago
9

A plane flies 1,480 miles in 4 hours, with the wind. On its return trip, it took 5 hours flying against the same wind to go 1,15

0 miles. How fast would the plane have flown without any wind? Type in your numerical answer only; do not type any words or letters with your answer.
Physics
1 answer:
Katarina [22]3 years ago
3 0

Answer:300 mi/hr

Explanation:

Given

Plane Flies 1480 miles in 4 hr with the wind

While returning it took 5 hr to cover 1150 miles against the wind

Let v and v_0 be the absolute speed of Plane and wind

Thus during with the flow of wind

time=\frac{Distance}{Speed}

\frac{1480}{v+v_0}=4

370=v+v_0   ------------1

During return trip wind speed oppose Plane thus

\frac{1150}{v-v_0}=5

230=v-v_0    ----------2

adding 1 & 2 we get

600=2v+v_0-v_0

v=300 mph

Thus Speed of plane without wind is 300 mi/hr

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X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
4 years ago
The decibel level of an orchestra is 90 db, and the violin section achieves a level of 80 dB. How does the sound intensity from
skad [1K]

Answer:

The difference in the decibel corresponses to a constant difference in the loudness perceived.

The refore the sound intensity from the orchestra is like 100 times that of the violin.

Explanation:

4 0
3 years ago
A 30 g horizontal metal bar, 13 cm long, is free to slide up and down between two tall, vertical metal rods that are 13 cm apart
natita [175]

Answer:

Terminal speed, v = 6901.07 m/s

Explanation:

It is given that,

Mass of the horizontal bar, m = 30 g = 0.03 kg

Length of the bar, l = 13 cm = 0.13 m

Magnetic field, B=5.5\times 10^{-2}\ T

Resistance, R = 1.2 ohms

We need to find the terminal speed oat which the bar falls. When terminal speed is reached,  

Force of gravity = magnetic force

mg=ilB..................(1)

i is the current flowing

l is the length of the rod

Due to the motion in rods, an emf is induced in the coil which is given by :

E=Blv, v is the speed of the bar

iR=Blv

i=\dfrac{Blv}{R}

Equation (1) becomes,

mg=\dfrac{B^2l^2v}{R}

v=\dfrac{mgR}{B^2l^2}

v=\dfrac{0.03\times 9.8\times 1.2}{(5.5\times 10^{-2})^2(0.13)^2}

v = 6901.07 m/s

So, the terminal speed at which the bar falls is 6901.07 m/s. Hence, this is the required solution.

5 0
3 years ago
Does anyone know what the answers are?
baherus [9]

Answer:

Option (C) is the answer

Explanation:

may be it is possible if that we stand so far

4 0
3 years ago
Which of the following is an example of how pictures of a planet's surface can provide evidence about the planet's natural resou
Volgvan
I think it’s c but I’m not for sure
5 0
3 years ago
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