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Talja [164]
2 years ago
9

A man wishes to travel due north in order to cross a river 5 kilometers wide flowing due East at 3 kilometers per hour . if he c

an row at 10 kilometers per hour In still water, find either by scale drawing or calculation
1. the direction in which he must heads his boat in order to get to his destination directly opposite his starting point​
Physics
1 answer:
Liono4ka [1.6K]2 years ago
7 0

Answer:

17.46° W of N

Explanation:

His prow must point westward so that the westward component of his total velocity is equal to the eastward river speed.

3 = 10sinθ

sinθ = 3/10

θ = 17.4576... or about 17.46° W of N

it will take him 5 / 10cos17.46 = 0.52414... hrs. or about 31 min 27 s to cross.

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In what direction does centripetal force act on an object that is traveling in a circular path?
Setler [38]

According to Newton's first law of motion, it is the natural tendency of all moving objects to continue in motion in the same direction that they are moving ... unless some form of unbalanced force acts upon the object to deviate its motion from its straight-line path.

Hope this helped, have a great day!
8 0
3 years ago
"Giant Swing", the seat is connected to two cables as shown in the figure (Figure 1) , one of which is horizontal. The seat swin
Bingel [31]
The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N. 

<span>Fx = [(233 + 840)/g]*v²/7.5 </span>

<span>v = 32.3*2*π*7.5/60 m/s = 25.37 m/s </span>

<span>The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º </span>

<span>Thh horizontal and vertical forces must balance each other. First the vertical components: </span>

<span>233 + 840 = Ti*cos40º </span>

<span>solve for Ti. (This is the answer to the part b) </span>

<span>Horizontally </span>

<span>[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º </span>

<span>Solve for Th </span>

<span>Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º </span>

<span>using v and Ti computed above.</span>
3 0
3 years ago
Tara is an electrician. Which field of science does Tara need to know the most about in order to do her job?
katovenus [111]
Chemistry and physics
3 0
3 years ago
You are looking down on a N = 9 turn coil in a magnetic field B = 0.5 T which points directly down into the screen. If the diame
Novay_Z [31]

Answer:

The magnitude of the voltage is 2.27\times10^{-4}\ V and the direction of the current is clockwise.

Explanation:

Given that,

Number of turns = 9

Magnetic field = 0.5 T

Diameter = 3 cm

Time t = 0.14 s

We need to calculate the flux

Using formula of flux

\phi=NAB

Put the value into the formula

\phi=9\times\pi\times(1.5\times10^{-2})^2\times0.5

\phi=0.003180

We need to calculate the emf

Using formula of emf

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{0.003180}{0.14}

\epsilon =-0.000227\ V

\epsilon=-2.27\times10^{-4}\ V

Negative sign shows the direction of current.

Hence, The magnitude of the voltage is 2.27\times10^{-4}\ V and the direction of the current is clockwise.

8 0
3 years ago
The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental
pav-90 [236]

A) 5.0\cdot 10^{-11} m

The energy of an x-ray photon used for single dental x-rays is

E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J

The energy of a photon is related to its wavelength by the equation

E=\frac{hc}{\lambda}

where

h=6.63\cdot 10^{-34}Js is the Planck constant

c=3\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

Re-arranging the equation for the wavelength, we find

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m

B) 2.0\cdot 10^{-11} m

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m

4 0
2 years ago
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