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3 years ago

You endanger a swimmer with the wake from your boat. under minnesota boating laws, what have you done?

Physics

1 answer:

mixas84 [53]3 years ago

8 0

The following enumeration is the list of actions which are considered to be illegal when operating your boat:

1 Operating the boat in a harsh or inconsiderate manner

2 Operating the boat in the absence of safety equipment

3 Actual weight exceeds the carrying capacity of the boat

4 Not prohibiting the riders to sit on gunwales, transom, stern, sides, bow or decking over the bow sides when the boat is underway, except when sufficient guards or railings are provided

5 Operating the boat or giving permission to other people to operate the boat while drunk or under illegal substances

6 Operating the boat and the wash or wake endanger any property or person

7 Operating the boat in an area designated for swimming

8 Operating the boat beyond a slow, no wake speed in locations marked as no wake zones

We can see that in this case, the rule # 6 is being trespassed. Therefore under Minnesota Boating Laws, we have committed an Illegal Operating Practice.

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A person pushes horizontally with a force of 220. N on a 61.0 kg crate to move it across a level floor. The coefficient of kinet

Ede4ka [16]

Answer:

(a) 161.57 N

(b) 0.958 m/s^2

Explanation:

Force applied, F = 220 N

mass of crate, m = 61 kg

μ = 0.27

(a) The magnitude of the frictional force,

f = μ N

where, N is the normal reaction

N = m x g = 61 x 9.81 = 598.41 N

So, the frictional force, f = 0.27 x 598.41

f = 161.57 N

(b) Let a be the acceleration of the crate.

Fnet = F - f = 220 - 161.57

Fnet = 58.43 N

According to newton's second law

Fnet = mass x acceleration

58.43 = 61 x a

a = 0.958 m/s^2

Thus, the acceleration of the crate is 0.958 m/s^2.

7 0

3 years ago

The mass of the earth is 5.96/10kg^24. The radius of the earth is approximately 6.37x10^6 calculate the force of gravity

n200080 [17]

<h2>Answer:g=9.79 $ms^{-2}$ ,A object of mass $m$ at the surface of earth experiences a force $mg$ </h2>

Explanation:

Let $M$ be the mass of earth.

Let $R$ be the radius of earth.

Let $G$ be the universal gravitational constant.

Given,

$M=5.96\times 10^{24}Kg$

$R=6.37\times 10^{6}m$

$G=6.67259 \times 10^{-11}$ $Nm^{2}Kg^{-2}$

Let $g$ be the acceleration due to gravity.

Then, $g=\dfrac{GM}{R^{2}}$

$g=\frac{6.67259 \times 10^{-11}\times 5.96\times 10^{24}}{(6.37\times 10^{6})^{2}}$

$g=9.79ms^{-2}$

A object of mass $m$ at the surface of earth experiences a force $mg$

3 0

3 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth’s center. Satellite A is to o

Vera_Pavlovna [14]

Answer:

Explanation:

Orbital radius of satellite A , Ra = 6370 + 6370 = 12740 km

Orbital radius of satellite B , Rb = 6370 + 19110 = 25480 km

Orbital potential energy of a satellite = - GMm / r where G is gravitational constant , M is mass of the earth and m is mass of the satellite

Orbital potential energy of a satellite A = - GMm / Ra

Orbital potential energy of a satellite B = - GMm / Rb

PE of satellite B /PE of satellite A

= Ra / Rb

= 12740 / 25480

= 1 / 2

b ) Kinetic energy of a satellite is half the potential energy with positive value , so ratio of their kinetic energy will also be same

KE of satellite B /KE of satellite A

= 1 / 2

c ) Total energy will be as follows

Total energy = - PE + KE

- P E + PE/2

= - PE /2

Total energy of satellite B / Total energy of A

= 1 / 2

Satellite B will have greater total energy because its negative value is less.

5 0

3 years ago

Are there any exceptions to the rule that planets rotate with small axis tilts and in the same direction as they orbit the sun?

Alona [7]

<span>Venus, Uranus, and Pluto are exceptions</span>

4 0

3 years ago

A car travelling at a constant speed of 70km/h passes a stationary police car. The police car immediately goes on the chase acce

Virty [35]

Answer:

18.24 seconds

Explanation:

First you convert the km/h to m/s, 70km/h=(175/9)m/s,85km/h=(425/18)m/s.

You know it took 10 seconds for the police to reach 85 km/h. Calculate the distance that the car is ahead of the police (175/9)*10=1750/9m. Then by divide 1750/9 with 425/18, you will get the value 8.24. Add the 10 seconds with the 8.24 you will get 18.24 sec which is the total time.

5 0

3 years ago