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3 years ago

What precaution should you take while towing a trailered boat?

2 answers:

6 0

There are many precautions that one must take while towing a trailered boat:

1) Always reduce the speed, when<span> you turn it.
2) Always use extra weight when you need to stop.
3) Allow proper space while cornering
4) Keep in mind this point, as you want to stop it, keep the margin of few miles as it does not come to an end at one. </span>

Leokris [45]3 years ago

5 0

The precaution you must take while towing a trailered boat includes:

Get the right gear
Securely set up the trailer
Take a short test drive
Load the boat and check safety chains
Take it slow while you are driving

<h2>Further Explanation</h2>

Get the right gear

Ensure you look for a trailer that will fit your tow vehicle. Don’t overload the trailer and don’t use a boat that doesn’t fit the trailer. Before you move, ensure you read the towing ability and pay utmost attention to your tow rating; your vehicle’s curb rate and some other important things.

Securely set up the trailer

Once you get a trailer that fit in, ensure your truck hitch is securely bolted to the frame and ensure you inspect it carefully before you hook up the trailer. Inspect the all tires to be sure they are all in good conditions.

Soon as you hook up the trailer, ensure you use the safety chains. It is important that you hook up the trailer lights so that cars behind you can see you clearly while applying the brake or turning.

Take a short test drive

It is important to drive your trailer a short stretch so that you can be able to confirm that if trailer brakes work properly, and the lights work correctly.

Load the boat and check safety chains

Take a second look at your trailer and boat to check for safety chains, latches and light plugs, and once you have carefully gotten through the steps, then you can load your boat, tie it properly and start driving.

Take it slow while you are driving

Don’t be in haste while towing your boat. Take it slow. Use your inner and outside mirror to observe what going on behind you and apply the brakes where necessary. When you finally gets to your destination, ensure you check your boat and the trailer to make sure everything held up during transport.

LEARN MORE:

towing an outboard boat brainly.com/question/4080255

KEYWORDS:

trailered boat
boat
precaution
vehicle
mirror

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What must the diver's minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, wh

Dafna1 [17]

Answer:

1.08 m/s

Explanation:

This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.

Time taken to fall 9.5 m

vertical acceleration = a = 9.8 m/s^2.

vertical velocity = 0, (since there is only horizontal component for velocity, )

distance traveled s = 9.5 m.

Substituting these values in the equation

$s= u \timest+0.5at^{2}$

$t= \sqrt{\frac{2s}{g} }$

$t=\sqrt{\frac{2\times9.5}{9.8} }$

⇒ t= 1.392 sec

Velocity needed

We know the time taken (1.392 s) to travel 1.5 m,

So velocity = 1.5 m / 1.392 s = 1.08 m/s

hence velocity of the diver must be at least 1.08 m/s

4 0

3 years ago

According to newton's second law of motion,the acceleration of an object equals the net force acting on the object divided by th

Shtirlitz [24]

<span>According to newton's second law of motion,the acceleration of an object equals the net force acting on the object divided by the objects "Mass"

Hope this helps!</span>

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3 years ago

I just need number 2

il63 [147K]

We will apply the conservation of linear momentum to answer this question.

Whenever there is an interaction between any number of objects, the total momentum before is the same as the total momentum after. For simplicity's sake we mostly use this equation to keep track of the momenta of two objects before and after a collision:

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Note that v₁ and v₁' is the velocity of m₁ before and after the collision.

Let's choose m₁ and v₁ to represent the bullet's mass and velocity.

m₂ and v₂ represents the wood block's mass and velocity.

The bullet and wood will stick together after the collision, so their final velocities will be the same. v₁' = v₂'. We can simplify the equation by replacing these terms with a single term v'

m₁v₁ + m₂v₂ = m₁v' + m₂v'

m₁v₁ + m₂v₂ = (m₁+m₂)v'

Let's assume the wood block is initially at rest, so v₂ is 0. We can use this to further simplify the equation.

m₁v₁ = (m₁+m₂)v'

Here are the given values:

m₁ = 0.005kg

v₁ = 500m/s

m₂ = 5kg

Plug in the values and solve for v'

0.005×500 = (0.005+5)v'

v' = 0.4995m/s

v' ≅ 0.5m/s

4 0

3 years ago

Two point charges of equal magnitude are 8.0 cm apart. At the midpoint of the line connecting them, their combined electric fiel

bagirrra123 [75]

Answer:

r = 8/2 = 4cm = 0.04m

k = 9×10^9

Enet = 51 N/C

Enet = E1 + E2

since E1 = E2

E1 = Enet/2 = 51/2

E/2 = kq/r²

q = Er²/2k

q = (51 × 0.04²)/(2×9×10^9)

q = 4.5×10^-12 C

q1 = q2 = 4.5 pC

Explanation:

The electric field is a region around a

charge in which it exerts electrostatic force

on another charges. While the strength of

electric field at any point in space is called

electric field intensity. It is a vector

quantity. Its unit is NC¯¹.

According to coulomb’s law ,if a unit

positive charge q (call it a test charge) is

brought near a charge q (call a field

charge) placed in space,the charge q will

experience a force. The value of this force

depends upon the distance between the

two charges. If the charge q is moved

away from q ,this force would decrease till

at a certain distance the force would be

practically reduced to zero. The charge q

is then out of the influence of charge q.

The region of space surrounding the charge

q in which it exerts a force on the charge

q is known as E.F of the charge

q. Mathematically it is expressed as:

E =F/q

The direction of the vector E is the same

as the direction of F,because q is a

positive scalar. Dimensionally,the E.F is

force per unit charge,and its SI unit is the

newton/coulomb (N/C).

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3 years ago

Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt

Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

$= \frac{4}{3}\pi(6400+ 50)^3 - \frac{4}{3}\pi (6400)
^3\\\\= \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3$

Hence the volume of atmosphere is $2.6\times 10^{19} m^3$

(b)

Write the ideal gas equation as foll ows:

$PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles$

$no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules
$

Hence the required molecules is $13.3\times10^{43} molecules
$

(c)

Write the ideal gas equation as follows:

$PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles$

$no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules$

Hence the required molecules in Caesar breath is $1.2\times 10^{23} molecules$

(d)

Volume fraction in Caesar last breath is as follows:

$Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}$

(e)

Since the volume capacity of the human body is 500 mL.

$Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath$

5 0

3 years ago