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Mrac [35]
3 years ago
13

A boat crosses a river of width 161 m in which the current has a uniform speed of 1.09 m/s. The pilot maintains a bearing (i.e.,

the direction in which the boat points) perpendicular to the river and a throttle setting to give a constant speed of 2.77 m/s relative to the water. What is the magnitude of the speed of the boat relative to a stationary shore observer? Answer in units of m/s.

Physics
1 answer:
hoa [83]3 years ago
7 0

Answer:

|v|=2.98m/s

Explanation:

In order to solve this problem we must first draw a diagram that will represent the situation. (See picture attached)

In this case the boat is being dragged by the river while it has a given velocity relative to the water and perpendicular to the flow of the river. We can take these velocities as vectors so we can add them up:

|v|=\sqrt{(1.09m/s)^{2}+(2.77m/s)^{2}}

which yields:

|v|=2.98m/s

so an observer that is stationary on the shore will se the boat having a speed of 2.98m/s.

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2500 centimeters is 25 meters. 25 decimeters is 2.5 meters. 2500 centimeters is 10 times longer than 25 decimeters.
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3 years ago
The slope of a position-timw graph can be used to find the moving objects.....?​
Fofino [41]

Answer:

Velocity

Explanation:

The slope of a position-time graph gives velocity of a moving object.

8 0
3 years ago
Longer wavelengths will have ________ frequencies, and shorter wavelengths will have ________ frequencies.
MakcuM [25]
Within a single medium, the product of (wavelength) x (frequency)
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3 0
3 years ago
An object is thrown upwards with a speed of 14 m/s. How long does it take to reach a height of 5.0 m above the projection point
Mamont248 [21]

Answer:

2.43 s

Explanation:

Using newton's equation of motion.

T = (v-u)/g

Where T = time taken for the object to return to the point of projection , u = initial velocity, v = final velocity, g = acceleration due to gravity.

Given: v =-14 m/s, u = 14 m/s, g = -9.8 m/s²

T = (-14-14)/-9.81

T = 2.85 s

Note: We look for the object's speed at 5.0 m.

using

v² = u²+2gs.................................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, s = distance.

Given:  u = 14 m/s, g = -9.81 m/s², s = 5.0 m

Substitute into equation 1

v² = 14²+(-9.81×5×2)

v² = 196-98.1

v = √97.9

v = 9.89

We look for the time taken for the velocity to decrease from 14 m/s to 9.89 m/s.

using

v = u+gt

t =(v-u)/g........................... Equation 2

Where t = time taken for the object to decrease it velocity from 14 m/s to 9.89 m/s

Given: v = 9.89 m/s, u =14 m/s g = -9.81 m/s²

t = (14-9.89)/-9.81

t = -4.11/-9.81

t = 0.42 s

Thus,

Time taken to reach 5.0 m above projection point = T-t

=2.85-0.42

2.43 s

4 0
3 years ago
What is the independent variable, the dependent variable of the graph? What is the relationship between the variable based on th
sergeinik [125]

Answer:

independent” variable goes on the x-axis (the bottom, horizontal one) and the “dependent” variable goes on the y-axis (the left side, vertical one).

Explanation:

this is the answer I don't know if it's right

6 0
3 years ago
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