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3 years ago

13

I need some help with this one :)

1 answer:

SVEN [57.7K]3 years ago

3 0

Explanation:

I think so c is answer

good morning

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How does energy from an ocean wave power stuff in a house?

Gala2k [10]

<span>The density of the energy that is transported under the waves under the ocean surface is about five times higher compared to the wind energy 20 meter (about 65 feet) above. In other words, the amount of energy in a single wave is very high. However it turns out that when they tried to make the high power of the waves to power things in a house it only made it past the first few stages there were very few that made it to some stages. This started taking place in around 2001. hope this helped if not let me know and i could explain more:)

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6 0

4 years ago

A quantity of a gas has an absolute pressure of 400 kPa and an absolute temperature of 110 degrees kelvin. When the temperature

KiRa [710]

Answer:

A) 854.46 kPa

Explanation:

P₁ = initial pressure of the gas = 400 kPa

P₂ = final pressure of the gas = ?

T₁ = initial temperature of the gas = 110 K

T₂ = final temperature of the gas = 235 K

Using the equation

$\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$

Inserting the values

$\frac{400}{110}=\frac{P_{2}}{235}$

P₂ = 854.46 kPa

3 0

3 years ago

The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point loca

zlopas [31]

Answer:

The value is $I = 0.0932 ML ^2$

Explanation:

From the question we are told that

The rotational inertia about one end is $I_R = \frac{1}{3} ML^2$

The location of the axis of rotation considered is $d = 0.4 L$

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is $0.4 M$

Generally the length of the rod from the its beginning to the axis of rotation consider is

$k = 1 - 0.4 L = 0.6L$

Generally the mass of the portion of the rod from the its beginning to the axis of rotation consider is

$m = 1- 0.4 M = 0.6 M$

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

$I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2$

$I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2$

Generally the rotational inertia about the axis of rotation consider for the second portion of the rod is

$I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2$

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

$I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2$

=> $I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]$

=> $I = 0.0932 ML ^2$

8 0

3 years ago

PLZZZZ HELPPPPPPP MEHHHHHHHH ASAP!!! 20 POINTS!!<br> I am desperate.... <br> :'(

Katarina [22]

Answer:

i think it is d

Explanation:

8 0

3 years ago

Which statement about a right triangle is NOT true?

Alex787 [66]

Answer:

The square of the hypotenuse is equal to the sum of the squares of the other two lengths.

Explanation:

only one that made sense

6 0

3 years ago

Read 2 more answers