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Alja [10]
2 years ago
12

What is the risk when a pwc passes too closely behind another boat?

Physics
1 answer:
Lemur [1.5K]2 years ago
4 0
The risk when a PWC (Personal Water Craft) passes too closely behind another boat is creating a blind spot. Blind spot can create a collision. 
The boat will block the view of the PWC of oncoming boats, as well as the oncoming boat's view of the PWC.So, that's why it is very important to maintain a proper lookout while turning the PWC and beware of your blind spots.
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A person on a cruise ship is doing laps on the promenade deck. on one portion of the track the person is moving north with a spe
Dmitry [639]
The resultant motion is given by pithagoras, since the two components (north and east) are perpendicular to each other.
They are asking you about the direction so you have to use trigonometry, finding that the direction is Ф=arctan(3.8/12)=17.57° north of east.
4 0
2 years ago
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Which of the following characterizes Erik Erikson's identity versus role confusion stage?
andrew11 [14]
Identity vs. Role confusion is the 5th stage of Erik Erikson's theory of psychosocial development.

During this stage, adolescents try to explore their independence. They begin to get interested in social interaction and relationships. This is also the time they begin to feel confused and insecure about themselves trying to figure out how to fit in the group. This is the reason why they engage themselves in different roles and activities. 
5 0
2 years ago
You toss a rock of mass m vertically upward. Air resistance can be neglected. The rock reaches a maximum height h above your han
VladimirAG [237]

Answer with Explanation:

We are given that

Mass of rock=m

Maximum height=h

a.At maximum height, velocity,v=0

We know that

v^2=u^2-2gh

0+2gh=u^2

u^2=2gh

Height,h=h/4

Again,v'^2=u^2-2g\times \frac{h}{4}

v'^2=2gh-\frac{gh}{2}=\frac{4gh-gh}{2}=\frac{3gh}{2}

v'=\sqrt{\frac{3gh}{2}}=\sqrt{\frac{3\times 9.8 h}{2}}=3.83\sqrt h

Where g=9.8 m/s^2

b.When height,h=3h/4

v'^2=u^2-2gh

v'^2=2gh-2g\times \frac{3h}{4}=2gh-\frac{3gh}{2}=\frac{4gh-3gh}{2}=\frac{gh}{2}

v'=\sqrt{\frac{9.8h}{2}}=2.2\sqrt h

v'=2.2\sqrt h

4 0
2 years ago
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If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?
Sav [38]
<h2>Complete Question:</h2>

You are on an aluminum ladder that is standing on the ground, trying to fix an electrical connection with a metal screwdriver having a metal handle. Your body is wet because you are sweating from the exertion; therefore, it has a resistance of 1.60 kΩ .

(a) If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?

(b) How much electrical power is delivered to your body?

<h2>Answer:</h2>

(a) 0.075A

(b) 9W

<h2>Explanation:</h2>

The voltage (V) passing across or supplied to a body is directly proportional to the current (I) flowing through the body as stated by Ohm's law. i.e

V ∝ I

=> V = I x R                 ----------------------(i)

Where;

R = constant of proportionality called resistance of the body

(a) As stated in the question;

The body is wet and thus will conduct electricity and has the following;

V = voltage supplied = 120V

R = resistance of the wet body = 1.60kΩ = 1.6 x 1000Ω = 1600Ω

Substitute these values into equation(i) as follows;

120 = I x 1600

Solve for I;

I = \frac{120}{1600}

I = 0.075A

Therefore the amount of current that will pass through your body is 0.075A

(b) Electrical power(P), which is commonly measured in Watts(W), delivered to a body is the product of the current(I) and voltage (V) supplied to the body. i.e

P = I x V           ---------------------(ii)

Where;

I = 0.075A   [as calculated above]

V = 120V     [given in the question]

Substitute these values into equation (ii) as follows;

P = 0.075 x 120

P = 9W

Therefore, the electric power delivered to your body is 9W

7 0
2 years ago
An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, a=6\times 10^{15}\ m/s^2

Charge on electron, q=1.6\times 10^{-19}\ C    

Mass of electron, m=9.1\times 10^{-31}\ kg    

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, F=qvB\ sin\theta

Here, \theta=90

So, ma=qvB

Where

v is the velocity of the electron

B is the magnetic field

v=\dfrac{ma}{qB}

v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}

v = 341250  m/s

or

v=3.41\times 10^5\ m/s

So, the speed of the electron is 3.41\times 10^5\ m/s

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0
2 years ago
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