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2 years ago

Which of these is most likely to create a boating hazard around river bridges?

Physics

1 answer:

OlgaM077 [116]2 years ago

6 0

c) debris collecting around pilings.

Boating Hazards:

Dams, submerged items, freezing water, rapidly changing weather, sunstroke, and current are just a few of the dangers that boaters may encounter. It's not always easy to see these risks. These risks must be recognized by boaters, and they must always be prepared to prevent hazards.

Operator negligence is the most frequent reason for boating accidents, according to US Coast Guard (USCG) recreational boating statistics from 2019. Inattentiveness on the part of the operator can result in crashes, people falling overboard, and slip-and-fall incidents on board, all of which can result in life-threatening injuries.

The greatest places to find more about any potential local risks are marinas and local boaters. Check any nearby marine charts as well to learn about potential dangers and how to avoid them.

Learn more about hazards here:

brainly.com/question/11491168

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A 62.0-kg skier is moving at 6.10 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.10

7nadin3 [17]

b) 747.3 J

a) 7.88 m/s

Explanation:

We start by solving part b) first.

The internal energy generated when crossing the rough patch is equal (in magnitude) to the work done by the friction force on the skier.

The magnitude of the friction force is:

$F_f=\mu mg$

where:

$\mu=0.300$ is the coefficient of friction

m = 62.0 kg is the mass of the skier

$g=9.8 m/s^2$ is the acceleration due to gravity

Therefore, the work done by friction is:

$W=-F_f d =-\mu mg d$

where

d = 4.10 m is the length of the rough patch

The negative sign is due to the fact that the friction force is opposite to the direction of motion of the skier

Substituting,

$W=-(0.300)(62.0)(9.8)(4.10)=-747.3 J$

So, the internal energy generated in crossing the rough patch is 747.3 J.

If we take the bottom of the hill as reference level, the initial mechanical energy of the skier is sum of his kinetic energy + potential energy:

$E=K_i +U_i = \frac{1}{2}mu^2 + mgh$

where

m = 62.0 kg is the mass

u = 6.10 m/s is the initial speed

h = 2.50 m is the height of the hill

After crossing the rough patch, the new mechanical energy is

$E'=E+W$

where

W = -747.3 J is the work done by friction

At the bottom of the hill, the final energy is just kinetic energy,

$E' = K_f = \frac{1}{2}mv^2$

where v is the final speed.

According to the law of conservation of energy, we can write:

$E+W=E'$

So we find v:

$\frac{1}{2}mu^2 +mgh+W = \frac{1}{2}mv^2\\v=\sqrt{u^2+2gh+\frac{2W}{m}}=\sqrt{6.10^2+2(9.8)(2.50)+\frac{2(-747.3)}{62.0}}=7.88 m/s$

8 0

3 years ago

Find the initial velocity

Ivenika [448]

a = dv/t = (330-x)/1.5 =50

x= 330-75 = 255 m/s

Intial velocity 255 m/s

4 0

3 years ago

A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. H

lukranit [14]

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

$a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2$

Let d is the distance moved in 2.25 s. Using second equation of motion,

$d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m$

So, it will move 6.32 m from rest in 2.25 seconds.

4 0

3 years ago

Which statement about the mass and the weight of an object is correct?

Fantom [35]

Answer:

HOPE IT HEPLED U !!!!

Explanation:

i think it's (C) They have different units ....

8 0

3 years ago

Answer the following using equations, number substitution and keep units. 1. What is the speed of an object that travels 5m in 1

DIA [1.3K]

Explanation:

1. Distance, d = 5 m

Time, t = 10 s

Speed = distance/time

$v=\dfrac{5}{10}=0.5\ m/s$

2. Mass, m = 10 kg

Acceleration, a = 3 m/s³

Force, F = mass (m) × acceleration (a)

F = 10 × 3

= 20 N

3. Mass, m = 7 kg

Height, h = 4 m

Potential energy, E = mgh

E = 7 × 9.8 × 4

E = 274.4 J

Hence, this is the required solution.

6 0

3 years ago