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AVprozaik [17]
3 years ago
8

BA-9 What piece of safety equipment is required on every canoe and kayak?

Physics
2 answers:
kirza4 [7]3 years ago
5 0

The answer is a lifevest

olya-2409 [2.1K]3 years ago
5 0

Answer:

canoes and kayaks of any length must have a properly fitting PFD (personal flotation device.) (Type I, II, III or V in flotation rating) for each person on board.

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A trait of an organism that helps it survive in its environment is calles
kompoz [17]
Answer is adaptation. An organism develops a trait over time to help survive in its environment called an adaptation. You could take a giraffe for example. A long time ago giraffes actually had short necks, but now since their food is higher they soon developed a longer neck and this is what we now see in the present. This goes for any artic animal. Polar bears and seals have a white fur adaptation to help them blend in with their environment. A chameleon changes colors in order to hide from predators and sneak up on prey. These are all adaptations
7 0
3 years ago
Read 2 more answers
A billiards ball B rests on a horizontal surface and is struck by another billiards ball A of the same mass m = 0.2 kg. Ball A i
SOVA2 [1]

Answer:

v1 = 15.90 m/s

v2 = 8.46 m/s

mechanical energy before collision = 32.4 J

mechanical energy after collision = 32.433 J

Explanation:

given data

mass m = 0.2 kg

speed = 18 m/s

angle =  28°

to find out

final velocity and  mechanical energy both before and after the collision

solution

we know that conservation of momentum remain same so in x direction

mv = mv1 cosθ + mv2cosθ

put here value

0.2(18) = 0.2 v1 cos(28) + 0.2 v2 cos(90-28)

3.6 =  0.1765 V1 + 0.09389 v2    ................1

and

in y axis

mv = mv1 sinθ - mv2sinθ

0 = 0.2 v1 sin28 - 0.2 v2 sin(90-28)

0 = 0.09389 v1 - 0.1768 v2   .......................2

from equation 1 and 2

v1 = 15.90 m/s

v2 = 8.46 m/s

so

mechanical energy  before collision = 1/2 mv1² + 1/2 mv2²

mechanical energy before collision = 1/2 (0.2)(18)² + 0

mechanical energy before collision = 32.4 J

and

mechanical energy after collision = 1/2 (0.2)(15.90)² + 1/2 (0.2)(8.46)²

mechanical energy after collision = 32.433 J

7 0
3 years ago
What type of stress is caused by two tectonic plates sliding past one another
prisoha [69]
Extensional stress. is your answer.
5 0
3 years ago
Read 2 more answers
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

  = 4.59\times 10^{-19} \ J

or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

  = 2.87 \ eV

(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

5 0
3 years ago
A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at
Readme [11.4K]

Answer:

 KE = 1.75 J

Explanation:

given,

mass of ball, m₁ = 300 g = 0.3 Kg

mass of ball 2, m₂ = 600 g = 0.6 Kg

length of the rod = 40 cm = 0.4 m

Angular speed = 100 rpm= 100\times \dfrac{2\pi}{60}

                         =10.47\ rad/s

now, finding the position of center of mass of the system

    r₁ + r₂ = 0.4 m.....(1)

 equating momentum about center of mass

  m₁r₁ = m₂ r₂

   0.3 x r₁ = 0.6 r₂

   r₁ = 2 r₂

Putting value in equation 1

2 r₂ + r₂ = 0.4

 r₂ = 0.4/3

 r₁ = 0.8/3

now, calculation of rotational energy

KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2

KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)

KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)

KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)

 KE = 1.75 J

the rotational kinetic energy is equal to 1.75 J

7 0
3 years ago
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