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PilotLPTM [1.2K]
1 year ago
12

. a boat can travel in still water. (a) if the boat points directly across a stream whose current is what is the velocity (magni

tude and direction) of the boat relative to the shore? (b) what will be the position of the boat, relative to its point of origin, after 3.00 s?
Physics
1 answer:
mixer [17]1 year ago
8 0

a) The velocity of the boat relative to the shore is 3.40 m/s and b) The position of the boat relative to its point of origin after 3s is 10.20m.

Here it is given that the speed of the boat (x) = 2.20m/s

The speed of the stream current (y) = 1.20m/s

a) We have to find the velocity of the boat relative to the shore.

The speed of the boat = x + y

                                    = 2.20 + 1.20

                                    = 3.40m/s

b) Now we have to find the position of the boat after 3s

The formula for speed:

Speed = Distance/ Time

distance = speed × time

speed = 3.40m/s

Time = 3s

distance = 3.40 × 3

              = 10.20 m

Therefore we get a) speed as 3.40m/s and b) distance as 10.20m.

To know more about the boat and stream refer to the link given below:

brainly.com/question/382952

#SPJ4

 

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Calculate the electric field intensity at a point 3 cm away from point charge of 3 x 10^-9 C.
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Answer:

The electric field intensity is <u>30000 N/C.</u>

Explanation:

Given:

Magnitude of the point charge is, q=3\times 10^{-9}\ C

Distance of the given point from the point charge is, d=3\ cm=0.03\ m

Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.

Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:

E=\frac{kq}{d^2}

Plug in k=9\times 10^9\ N\cdot m^2/C^2, q=3\times 10^{-9}\ C, d=0.03\ m. Solve for 'E'.

E=\frac{(9\times 10^9\ N\cdot m^2/C^2)(3\times 10^{-9}\ C)}{(0.03\ m)^2}\\\\E=\frac{27}{0.0009}\ N/C\\\\E=30000\ N/C

Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.

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What type of friction occurs when you are trying to move an object, but the object isnt moving?
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9. A football punter attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in th
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Answer:

Angle is 55.52°

and Initial Speed is v=26.48 m/s

Explanation:

Given data

x_{o}=0m\\ y_{o}=1.23m\\a_{oy}=a_{1y}=g=-9.8m/s^{2} \\x_{1}=67.0m\\y_{1}=0m\\t_{o}=0\\a_{ox}=m/s^{2} \\t_{1}=4.50s

Applying the kinematics equations for motion with uniform acceleration in x and y direction

So

x_{1}=x_{o}+v_{ox}t_{1}=67.0m\\0+4.50v_{o}Cos\alpha =67.0m\\v_{o}Cos\alpha =14.99\\v_{o}=14.99/Cos\alpha.....(1) \\and\\y_{1}=y_{o}+v_{oy}t_{1}+(1/2)a_{oy}t_{1}^{2} =0m\\ 1+4.50v_{o}Sin\alpha+(-9.8/2)(4.5)^{2}=0\\  v_{o}Sin\alpha=21.828.....(2)

Put the value of v₀ from equation (1) to equation (2)

So

\frac{14.99}{Cos\alpha }(Sin\alpha ) =21.828\\as\\tan\alpha =Sin\alpha /Cos\alpha \\So\\14.99tan\alpha =21.828\\tan\alpha =21.828/14.99\\\alpha =tan^{-1}(21.828/14.99) \\\alpha =55.52^{o}

Put that angle in equation (1) or equation (2) to find the initial velocity

So from equation (1)

v_{o}=(\frac{14.99}{Cos\alpha } ) \\v_{o}=(\frac{14.99}{Cos(55.52) } ) \\v_{o}=26.48m/s

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