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maks197457 [2]
2 years ago
12

What is the gravitational acceleration on the board of the Deepsea Challenger when it is in the Mariana Trench, 11 km below the

surface of the Pacific Ocean?
a. It is slightly less than 9.81 m/s2.
b. It is very large due to the high pressure.
c. It is zero.
d. It is significantly less than 9.81 m/s2.
Physics
1 answer:
kap26 [50]2 years ago
8 0

The gravitational acceleration on the board of the Deepsea Challenger when it is in the Mariana Trench, 11 km below the surface of the Pacific Ocean is very large due to the high pressure. That is option B

<h3>What is gravitational acceleration?</h3>

The gravitational acceleration is the acceleration of an object that is under free fall. The acceleration due to gravity is 9.81 m/s2. above the sea level.

One of the factors that affect gravitational acceleration is depth. This is because as the depth increases so will pressure increase.

Increase in pressure leads to increase in gravitational acceleration.

Therefore, the gravitational acceleration on the board of the Deepsea Challenger when it is in the Mariana Trench, 11 km below the surface of the Pacific Ocean is very large due to the high pressure.

Learn more about acceleration here:

brainly.com/question/605631

#SPJ1

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A copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)
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<h2>The increase in length = 1.87 x 10⁻²</h2>

Explanation:

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The increase in length can be found by the relation

L = L₀ ( 1 + α ΔT )

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α  is the coefficient of linear expansion and ΔT is the increase in temperature .

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Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

cos \theta = \frac {V_{river}}{V_{boat}}

When solving for theta, we get that:

\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})

so now we can substitute the corresponding values:

\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})

Which yields:

\theta = 60^{o}

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

V_{y}=5.60km/hr*cos(210^{o})

V_{y}=-4.85km/hr

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

t=\frac{y}{V_{y}}

t=\frac{5.60km}{4.85km/hr}=1.155hr

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

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V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr

and now up stream:

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Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

t_{ds}=\frac{x}{v_{ds}}

t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr

and the time up stream:

t_{us}=\frac{x}{v_{us}}

t_{us}=\frac{2.80km}{2,80km/hr}=1hr

so the total time will be:

t_{ds}+t_{us}=0.333hr+1hr=1.333hr

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

t=\frac{d}{v_{boat}}

t=\frac{5.60km}{5.60km/hr}

so

t= 1hr

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