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3 years ago

What at the bottom of the ocean? take that! detailed answer!!!

2 answers:

4 0

Answer: Essentially, no one is certain of the things that lie at the bottom of the sea. The oceans deepest darkest deep depths have proven more challenging to reach than traveling to the moon. In all of human history, 12 people have reached the moon, while only three have reached the bottom of the deepest part of the ocean: the Challenger Deep. The deepest point- located within the Marina Trench, reaches nearly 10,000 meters deep- beyond the depths which light can reach.

Explanation:

Alik [6]3 years ago

3 0

Answer:

Unlike mapping the land, we can't measure the landscape of the sea floor directly from satellites using radar, because sea water blocks those radio waves. But satellites can use radar to measure the height of the sea's surface very accurately.

Explanation:

94 percent of the ocean is still unexplored so thats it

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A rifle has a mass of 7-kg and the bullet has a mass of 0.7-kg. If the velocity of the bullet is 350-m/s after the rifle is fire

nevsk [136]

Answer:

-35 m/s

Explanation:

Momentum is conserved.

Momentum before firing = momentum after firing

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Before the bullet is fired, the bullet and rifle have no velocity, so u₁ and u₂ are 0.

0 = m₁v₁ + m₂v₂

Given m₁ = 0.7 kg, v₁ = 350 m/s, and m₂ = 7 kg:

0 = (0.7 kg) (350 m/s) + (7 kg) v

v = -35 m/s

The rifle recoils at 35 m/s in the opposite direction.

8 0

3 years ago

Two charges are separted by a distance d and exert mutual attractive forces of f on each other. If the charges are separated by

ASHA 777 [7]

The new force between the charges when the distance become 2d will be F'=4F

<h3>What is electrostatic force?</h3>

When two charged particles are separated by the distance d then the force of attraction or repultion acts on the charged particle depending upon the nature of the charge whether it is positive or negative.

The formula for the electrostatic force is given by

$F=\dfrac{kq_1q_2}{d^2}$

Now if the value of d becomes 2d then the formula will become:

$F'=\dfrac{kq_1q_2}{(2d)^2}=\dfrac{kq_1q_2}{4d}$

$\dfrac{F'}{F}=\dfrac{1}{4}$

$F=4F'$

Hence the new force between the charges when the distance become 2d will be F'=4F

To know more about electrostatic force follow

brainly.com/question/17692887

#SPJ4

3 0

2 years ago

(1 point) find the half-life (in hours) of a radioactive substance that is reduced by 1010 percent in 9595 hours.

Bad White [126]

The decay function is of the form
$N(t) = N_{0} \, e^{-kt}$
where
N₀ = initial amount
k = decay constant
t = hours

The material decays by 10% in 95 hours. Therefore
$0.9N_{0} = N_{0} \, e^{-95k} \\\\ -95k = ln(0.9) \\\\ k= \frac{ln(0.9)}{-95}=0.001109$

The time for the half life is given by
$0.5N_{0} = N_{0} \, e^{-0.001109t} \\\\ -0.0001109t = ln(0.5) \\\\ t = \frac{ln(0.5)}{-0.001109} = 625 \, h$

Answer: The half life is 625 hours

8 0

3 years ago

2. Who were the first philosophers to propose that atoms existed? What time period was<br> this?

Rus_ich [418]

Answer:

"Greek Origins" were the first philosophers to propose that atoms existed.

Explanation:

The idea that all matter is made up of tiny, indivisible particles, or atoms, is believed to have originated with the Greek philosopher Leucippus of Miletus and his student Democritus of Abdera in the 5th century B.C. (The word atom comes from the Greek word atomos, which means “indivisible.”)

7 0

3 years ago

The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire

kakasveta [241]

Answer:

$F_{AB} = 172.1356\\F_{AC} = 258.2033\\F_{AD} = 368.8004$

Explanation:

Using the diagram (see attachment) we extract the following position vectors:

$Vector (OA) = 6i + 0j + 0k \\Vector (OB) = 0i + 3j + 2k \\Vector (OC) = 0i - 2j + 3k$

Next step is to find unit vectors $u_{AB} ,u_{AC}, u_{AD}, u_{AE}$ as follows:

$u_{AB} = \frac{vector(AB)}{magnitude(AB)} \\= \frac{OB - OA}{magnitude({vector(OB - OA))} }\\=\frac{-6i +3j+2k}{\sqrt{6^2 + 3^2+2^2} } \\\\=-0.857 i +0.429j+0.286k\\\\u_{AC} = \frac{vector(AC)}{magnitude(AC)} \\= \frac{OC - OA}{magnitude({vector(OC - OA))} }\\=\frac{-6i -2j-3k}{\sqrt{6^2 + 2^2+3^2} } \\\\=-0.857 i -0.286j+0.429k\\\\u_{AD} = +1i\\u_{AC} = -1k$

Using the diagram we find the corresponding vectors Forces:

$F_{AB} = F_{AB} i + F_{AB}j +F_{AB}k\\F_{AC} = F_{AC} i + F_{AC}j +F_{AC}k\\F_{AD} = F_{AD} i + F_{AD}j +F_{AD}k\\W = -160 k$

Equation of Equilibrium:

$Sum of forces = 0\\F_{AB}. u_{AB} + F_{AC}.u_{AC} + F_{AD}.u_{AD} + W = 0\\(-0.857F_{AB}i + 0.429F_{AB}j +0.286F_{AB}k) + (-0.857F_{AC}i - 0.286F_{AC}j +0.429F_{AC}k) + (+1F_{AD} i) + (-160k) = 0$

Comparing i , j and k components as follows:

$-0.857F_{AB} -0.857F_{AC} +1F_{AD} = 0\\+ 0.429F_{AB} - 0.286F_{AC} = 0\\+0.286F_{AB} +0.429F_{AC} = 160$

Solving Above equation simultaneously we get:

$F_{AB} = 172.1356\\F_{AC} = 258.2033\\F_{AD} = 368.8004$

3 0

3 years ago