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hjlf
3 years ago
13

A surfer is moving at a constant velocity of 5.2 m/s north relative to a wave which is pushing him west at a constant velocity o

f 8.6 m/s toward the beach. The surfer begins at a distance of 41 meters away from the beach. How far north does the surfer make it (in meters) before becoming beached? Round answer to 1 decimal place.
Physics
1 answer:
Vika [28.1K]3 years ago
5 0
The speed and distances are directly proportional. Use ratios to solve for vertical y-distance. The ratio of x-distance west to y-distance north equals the x-velocity to y-velocity.

x/y = vx/vy
41/y = 8.6/5.2
41/y = 1.65
41/1.65 = y
24.8 m = y

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A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th
GREYUIT [131]

Answer:

1.26\cdot 10^7 m/s

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

F=qvB

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

qvB=m\frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

v=\frac{qBr}{m}

where in this problem we have:

q=1.6\cdot 10^{-19}C is the magnitude of the charge of the electron

B=208 G=208\cdot 10^{-4}T is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

r=3.45 mm = 3.45\cdot 10^{-3} m

m=9.11\cdot 10^{-31} kg is the mass of the electron

So, the electron's speed is

v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s

6 0
3 years ago
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4 0
3 years ago
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Nostrana [21]

Answer:

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7 0
2 years ago
A 87 arrow is fired from a bow whose string exerts an average force of 105 on the arrow over a distance of 75 .
timofeeve [1]

The solution would be like this for this specific problem:

 

V^2 = 2AS = 2FS/M

V = sqrt(2FS/M) = sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps

So the speed of the arrow as it leaves the bow is 42.5 mps.

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

6 0
3 years ago
A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sph
arsen [322]

Answer:

option E

Explanation:

given,

I is moment of inertia about an axis tangent to its surface.

moment of inertia about the center of mass

I_{CM} = \dfrac{2}{5}mR^2.....(1)

now, moment of inertia about tangent

I= \dfrac{2}{5}mR^2 + mR^2

I= \dfrac{7}{5}mR^2...........(2)

dividing equation (1)/(2)

\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}

\dfrac{I_{CM}}{I}=\dfrac{2}{7}

I_{CM}=\dfrac{2}{7}I

the correct answer is option E

4 0
3 years ago
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