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A boy can swim 3.0 meter a second in still water while trying to swim directly across a river from west to east, he is pulled by

lana66690 [7]

Answer:

Angle: $48.19^o$

Explanation:

<u>Two-Dimension Motion</u>

When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.

Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.

To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

$v_b\ cos\alpha$

where $v_b$ is the speed of the boy in still water and $\alpha$ is the angle respect to the shoreline. If the river flows at speed $v_s$ , we now set

$v_b\ cos\alpha=v_s$

$\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}$

$\alpha=48.19^o$

8 0

2 years ago

A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i

nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

$U=mgh$

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J$

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

$U=mgh$

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

$U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J$

3. 0 J

As before, the gravitational potential energy of the ball is given by

$U=mgh$

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J$

4 0

3 years ago

How do I do number 8, plz respond quick, I have a big unit test on this and I’m rocking a 65 science average rn. Thank you for t

xxTIMURxx [149]

Answer:

3 is the correct answer for number 8

6 0

3 years ago

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Bromine, a liquid at room temperature, has a boiling point

lukranit [14]

Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.

3 0

2 years ago

The index of refraction is the ratio of the speed of light in a vacuum to the speed of light in a medium. Which true statement i

Alika [10]

Answer:

A. Materials with a low index of refraction cause light to refract very little.

7 0

2 years ago

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