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inna [77]
3 years ago
9

You are attempting to row across a stream in your rowboat. Your paddling speed relative to still water is 3.0 m/s (i.e., if you

were to paddle in water without a current, you would move with a speed of 3.0 m/s). You head off by rowing directly north, across the stream.
Now assume that the stream flows east at 4.0 m/s. Draw the vectors \vec{v}_w, representing the velocity of the stream, and \vec{v}_{tot}, representing the velocity of your rowboat relative to the stream bank. Be sure to draw both vectors.
Draw \vec{v}_{tot} and \vec{v}_w starting at the tail and tip of \vec{v}_{still} respectively. The location, orientation and length of the vectors should be appropriate. Each vector's length is displayed in meters per second.

Physics
1 answer:
Nataliya [291]3 years ago
6 0

Answer:

Please check the attached file for the diagram

Explanation:

The velocity of the of the rowboat V_{tot}  through the river is the resultant velocity. It is obtained taking a vector sum of the velocity in still water and the velocity of the river.

There are several ways to take this vector sum, but the question makes it simple for us to use Pythagoras's theorem because the East and North directions are perpendicular to each other.

Hence;

V_{tot}^2=V_{still}^2+V_{w}^2\\V_{tot}^2=3^2+4^2

V_{tot}=\sqrt{3^2+4^2}\\ V_{tot}=\sqrt{25}=5m/s

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A mass of 240 grams oscillates on a horizontal frictionless surface at a frequency of 2.5 Hz and with amplitude of 4.5 cm.
mr_godi [17]

Answer:

(a) The spring constant is 59.23 N/m

(b) The total energy involved in the motion is 0.06 J

Explanation:

Given;

mass, m = 240 g = 0.24 kg

frequency, f = 2.5 Hz

amplitude of the oscillation, A = 4.5 cm = 0.045 m

The  angular speed is calculated as;

ω = 2πf

ω = 2 x π x 2.5

ω = 15.71 rad/s

(a) The spring constant is calculated as;

\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m

(b) The total energy involved in the motion;

E = ¹/₂kA²

E = (0.5) x (59.23) x (0.045)²

E = 0.06 J

5 0
3 years ago
You collect some more data on that horse at a later time interval, but now you are measuring thehorse’s velocity, not its positi
Monica [59]

Answer:

a)  x(t) = 10t + (2/3)*t^3

b) x*(0.1875) = 10.18 m

Explanation:

Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.

Given:

- v(t) = 10 + 2*t^2 (radar gun)

- x*(t) = 10 + 5t^2 + 3t^3  (our coordinate)

Find:

-The position x of horse as a function of time t in radar system.

-The position of the horse at x = 2m in our coordinate system

Solution:

- The position of horse according to radar gun:

                              v(t) = dx / dt = 10 + 2*t^2

- Separate variables:

                              dx = (10 + 2*t^2).dt

- Integrate over interval x = 0 @ t= 0

                             x(t) = 10t + (2/3)*t^3

- time @ x = 2 :

                              2 = 10t + (2/3)*t^3

                              0 = 10t + (2/3)*t^3 + 2

- solve for t:

                              t = 0.1875 s

- Evaluate x* at t = 0.1875 s

                              x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3

                              x*(0.1875) = 10.18 m

3 0
4 years ago
Solving a physics problem is . . .
Vsevolod [243]
C. A step-by-step process that takes time, and is essential for learning physics concepts.
3 0
3 years ago
Read 2 more answers
Question 7 (1 point)
ioda

Answer:

1150 secs

distance = speed x time

time = distance / speed

230,000 / 200 = 1150

time = 1150 seconds

7 0
3 years ago
When a voltage difference is applied to a piece of metal wire, a current flows through it. if this metal wire is now replaced wi
Nikitich [7]

The resistance of the cylindrical wire is R=\frac{\rho l}{A}.

Here R is the resistance, l is the length of the wire and A is the area of cross section. Since the wire is cylindrical A=\frac{\pi d^2}{4} .

Comparing two wires,

R_1=\frac{\rho_1 l}{A_1} \\ R_2=\frac{\rho_2 l}{A_2}

Dividing the above 2 equations,

\frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2}  \frac{A_2 }{A_1}  \\ \frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2}  \frac{d_2^2 }{d_1^2}  \\

Since d_2=2d_1

The above ratio is

\frac{R_1}{R_2}=\frac{1.68(10^{-8})  }{1.59(10^{-8}) } (4)\\ \frac{R_1}{R_2}=4.2264

We also have,

\frac{E/R_1}{E/R_2} =\frac{I_1}{I_2} \\ I_2=\frac{R_1}{R_2}I_1 \\ I_2=4.23I_1

The current through the Silver wire will be 4.23 times the current through the original wire.

8 0
3 years ago
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