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Elodia [21]
3 years ago
10

Is better to run or walk when it is raining?

Physics
1 answer:
lawyer [7]3 years ago
5 0

Answer:

run

Explanation:

run - less wet

walking - more wet

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A 2300 kg sailboat is moving west at 5.5 m/s when an eastward wind
pogonyaev

The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration <em>a</em> of

950 N = (2300 kg) <em>a</em>

<em>a</em> = (950 N) / (2300 kg)

<em>a</em> ≈ 0.413 m/s²

Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of

<em>v</em> = -5.5 m/s + <em>a</em> (11.5 s)

<em>v</em> = -0.75 m/s

which means the wind slows the boat down to a velocity of 0.75 m/s westward.

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3 years ago
the amplitude of the wave is which of the following A. Resonance B. Reflection C. Energy D. Magnitude
OLga [1]
D.Magnitude is the answer
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3 years ago
A girl is floating in a freshwater lake with her head just above the water. If she weighs 610 N, what is the volume of the subme
Elden [556K]

Answer:

The volume of the submerged part of her body is 0.0622m^{3}

Explanation:

Let's define the buoyant force acting on a submerged object.

In a submerged object acts a buoyant force which can be calculated as :

B=ρ.V.g

Where ''B'' is the buoyant force

Where ''ρ'' is the density of the fluid

Where ''V'' is the submerged volume of the object

Where ''g'' is the acceleration due to gravity

Because the girl is floating we can state that the weight of the girl is equal to the buoyant force.

We can write :

W_{girl}=B (I)

Where ''W'' is weight

⇒ If we consider ρ = 1000\frac{kg}{m^{3}} (water density) and g=9.81\frac{m}{s^{2}} and replacing this values in the equation (I) ⇒

B=W_{girl}

B=610N

ρ.V.g = 610N

1000\frac{kg}{m^{3}}.V.(9.81\frac{m}{s^{2}})=610N (II)

The force unit ''N'' (Newton) is defined as

N=kg.\frac{m}{s^{2}}

Using this in the equation (II) :

(9810\frac{N}{m^{3}}).V =610N

V=\frac{610N}{9810\frac{N}{m^{3}}}

V=0.0622m^{3}

We find that the volume of the submerged part of her body is 0.0622m^{3}

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Answer:

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