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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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A sinusoidal sound wave moves through a medium and is described by the displacement wave function
labwork [276]

By applying the wave equation we know that the displacement on the y-axis is 1.999 micrometers.

We need to know about wave equations to solve this problem. The displacement of the wave on the y-axis can be explained by the wave equation

y = A cos (kx - ωt)

where y is y-axis displacement, A is amplitude, k is wave number, x is x-axis displacement, ω is angular speed and t is time.

the wavenumber and angular speed of the wave equation can be determined respectively by

k = 2π / λ

ω = 2πf

where k is the wavenumber, λ is wavelength and f is frequency.

From the question above, we know that:

y = 2.00cos (15.7x - 858t)

(x in meters, t in second, y in micrometers)

x = 0.05 m

t = 3 ms

Convert time to second

t = 3ms = 0.003 s

By applying the wave equation, we get

y = 2.00cos (15.7x - 858t)

y = 2.00cos (15.7(0.05) - 858(0.003))

y  = 2 cos(-1.789)

y = 1.999 micrometers

For more on wave equation on: brainly.com/question/25699025

#SPJ4

7 0
2 years ago
A snake slithered across the sidewalk in between two sign posts. At the first sign post his speed was
masha68 [24]

This is an example of Uniform Speed

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5 0
3 years ago
What damage a tornado can cause
Ksivusya [100]
Alrighty, so, houses can be destroyed, people can be killed, animals can be killed, habitats can be destroyed, and trees can be uprooted. Tornadoes can create massive damage.

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5 0
3 years ago
A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

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