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Lostsunrise [7]
3 years ago
12

The deepest point of the pacific ocean is 11,033 m, in the mariana trench. what is the gauge pressure in the water at that point

? the density of seawater is 1025 kg/m3.
Physics
1 answer:
velikii [3]3 years ago
7 0

Given that,

Depth of seawater, h = 11,033 m

Density seawater, p (rho) = 1025 kg/m³

Gauge Pressure , P = ??

Since, we know that:

Pressure, P = pgh

Pressure = 1025 * 9.81 * 11033

Pressure = 1109395723.3 N/m²

or

Pressure = 1.1 x 10∧8 Pascal

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In an effort to protect a rhino, volunteers are following its steps with air monitoring and ground cameras. The rhino starts on
finlep [7]

Answer:

Explanation:

We shall represent each displacement by vectors . i will represent east , -i west , j north and - j south .

Rhino walks 1.5 km west on day 1.

D₁ = - 1.5 i

The rhino then moves 0.7 km on a straight line in a direction of 15o east of north toward ground camera B

D₂ = .7 sin15 i + .7cos15 j

On the second day, the rhino moves 2.5 km directly south

D₃ = - 2.5 j

D = D₁ + D₂ + D₃

= - 1.5 i + .7 sin15 i + .7cos15 j - 2.5 j

= - 1.5 i + .181 i + .676 j - 2.5 j

= - 1.32 i - 1.824 j

magnitude  of total displacement

= √ (1.32² +1.824²

= 2.25 km

For direction we shall calculate slope with x axis

Tanθ = - 1.824 / - 1.32

= 54°

So rhino will be towards 54° south of west as both x and y coordinates are negative.

6 0
3 years ago
PLEASE HELP EMERGENCY
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Answer:

mechanical energy to heat energy to chemical energy

8 0
3 years ago
You are looking at a yellow flower growing outside in the sunshine. Why does it look yellow?
Alchen [17]

Answer:

It looks yellow because that is the only (major) color reflected.

Visible spectra is from about 4000-7000 Angstroms (10^-10 m).

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3 0
2 years ago
It is a good idea to read or exchange text messages during all of the following times EXCEPT:
Dafna11 [192]

Answer: A while driving a car. It is unsafe

8 0
3 years ago
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Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
3 years ago
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