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3 years ago

I NEED A 100% ACCURATE ANSWER FOR THIS QUESTION ASAP NO LINKS !!!

Physics

2 answers:

marysya [2.9K]3 years ago

7 0

Answer:

3. Higher in some places and lower in other places

MrRa [10]3 years ago

6 0

Answer:

3) higher in some places nd lower In other places

Explanation:

The pressure in a fluid is given by the law

p = p0 + pgh

where

p0 is the atmospheric pressure

p is the fluid's density

g is the acceleration of gravity

h is the depth at which the pressure is calculated

As we see, the pressure depends on the value of h (depth): therefore, points which are located at more depth experience a larger pressure than points located near the fluid's surface.

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Find the value of F1 + F2 + F3.

Dovator [93]

Answer:

F = 0.78[N]

Explanation:

The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.

For F₁

$F_{y}=2[N]$

For F₂

$F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]$

For F₃

$F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]$

Now we can sum each one of the forces in the given axes:

$F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]$

Now using the Pythagorean theorem we can find the total force.

$F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]$

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3 years ago

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Arada [10]

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How does mass affect acceleration during free fall?

nirvana33 [79]

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3 years ago

3. When encountering low visibility from rain or fog, you

zimovet [89]

Answer:

c. low beams and fog lights

Explanation:

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i just got the answer wrong and the drivers ed gave me this explanation !!

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3 years ago

Read 2 more answers

A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040

Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force ( $F_{B}$ ), as follows: