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2 years ago

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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.

1 answer:

hoa [83]2 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 wheel with radius 36 cm is rotating at a rate of 19 rev/s.(a) What is the angular speed in radians per second? rad/s(b) In a t

Sedaia [141]

(a) 119.3 rad/s

The angular speed of the wheel is

$\omega= 19 rev/s$

we need to convert it into radiands per second. We know that

$1 rev = 2 \pi rad$

Therefore, we just need to multiply the angular speed of the wheel by this factor, to get the angular speed in rad/s:

$\omega = 19 rev/s \cdot (2\pi rad/rev))=119.3 rad/s$

(b) 596.5 rad

The angular displacement of the wheel in a time interval t is given by

$\theta= \omega t$

where

$\omega=119.3 rad$

and

t = 5 s is the time interval

Substituting numbers into the equation, we find

$\theta=(119.3 rad/s)(5 s)=596.5 rad$

(c) 127.3 rad/s

At t=10 s, the angular speed begins to increase with an angular acceleration of

$\alpha = 1.6 rad/s^2$

So the final angular speed will be given by

$\omega_f = \omega_i + \alpha \Delta t$

where

$\omega_i = 119.3 rad/s$ is the initial angular speed

$\alpha = 1.6 rad/s^2$ is the angular acceleration

$\Delta t = 15 s - 10 s = 5 s$ is the time interval

Solving the equation,

$\omega_f = (119.3 rad/s) + (1.6 rad/s^2)(5 s)=127.3 rad/s$

(d) 616.5 rad

The angle through which the wheel has rotated during this time interval is given by

$\theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2$

Substituting the numbers into the equation, we find

$\theta = (119.3 rad/s)(5 s) + \frac{1}{2} (1.6 rad/s^2) (5 s)^2=616.5 rad$

(e) 222 m

The instantaneous speed of the center of the wheel is given by

$v_{CM} = \omega R$ (1)

where

$\omega$ is the average angular velocity of the wheel during the time t=10 s and t=15 s, and it is given by

$\omega=\frac{\omega_i + \omega_f}{2}=\frac{127.3 rad/s+119.3 rad/s}{2}=123.3 rad/s$

and

R = 36 cm = 0.36 m is the radius of the wheel

Substituting into (1),

$v_{CM}=(123.3 rad/s)(0.36 m)=44.4 m/s$

And so the displacement of the center of the wheel will be

$d=v_{CM} t = (44.4 m/s)(5 s)=222 m$

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2 years ago

Which are standard functions that families are responsible for providing?

Savatey [412]

C: life skills, stability, resources, affection

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3 years ago

Read 2 more answers

1) A net force of 75.5 N is applied horizontally to slide a 225 kg crate across the floor.

dimulka [17.4K]

Answer:

The acceleration of the crate is $0.3356\,\frac{m}{s^2}$

Explanation:

Recall the formula that relates force,mass and acceleration from newton's second law;

$F=m\,a$

Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:

$F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}$

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2 years ago

A long distance runner sees the finish line and accelerates at a rate in 1.2 m/s2 for

Nuetrik [128]

Answer: he did travel 15 meters.

Explanation:

We have the data:

Acceleration = a = 1.2 m/s^2

Time lapes = 3 seconds

Initial speed = 3.2 m/s.

Then we start writing the acceleration:

a(t) = 1.2 m/s^2

now for the velocity, we integrate over time:

v(t) = (1.2 m/s^2)*t + v0

with v0 = 3.2 m/s

v(t) = (1.2 m/s^2)*t + 3.2 m/s

For the position, we integrate again.

p(t) = (1/2)*(1.2 m/s^2)*t^2 + 3.2m/s*t + p0

Because we want to know the displacementin those 3 seconds ( p(3s) - p(0s)) we can use p0 = 0m

Then the displacement at t = 3s will be equal to p(3s).

p(3s) = (1/2)*(1.2 m/s^2)*(3s)^2 + 3.2m/s*3s = 15m

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2 years ago

Gauss's Law states that the net electric flux, , through any closed surface is proportional to the charge enclosed: . The analog

Natali [406]

The analogous formula for magnetic fields is the Ampere's law.

To find the answer, we need to know about the Ampere's law of magnetism.

<h3>What's Ampere's law of magnetism?</h3>

Ampere's law states that the close line integral of magnetic field around a current carrying loop is directly proportional to the current enclosed within it.

<h3>What's is the mathematical expression of Ampere's law?</h3>

Mathematically, Ampere's law is

B•dl= μ₀I

Thus, we can conclude that the analogous formula for gauss law is the Ampere's law in magnetism.

Learn more about the Ampere's law here:

brainly.com/question/17070619

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1 year ago