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3 years ago

b. Comparing and Contrasting Compare the change in atmospheric pressure with elevation to the change in water pressure

Physics

2 answers:

sleet_krkn [62]3 years ago

8 0

Atmospheric pressure will decrease with elevation Water pressure, however increases with depth

Hope this helped :3

babunello [35]3 years ago

3 0

Answer:

hfyn done hm jjhohhy u kk

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Martha was leaning out of the window on the second floor of her house and speaking to Steve. Suddenly, her glasses slipped from

DiKsa [7]

Answer:

s = 23.72 m

v = 21.56 m/s²

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

u = 0 m/s

g = 9.8 m/s²

s = 0 + 0.5 × 9.8 × 2.2²

s = 23.72 m

b) impact velocity

v = √(2gh)

v = √(2× 9.8 × 23.72)

v = √464.912

v = 21.56 m/s²

6 0

3 years ago

It has been suggested, and not facetiously, that life might have originated on Mars and been carried to Earth when a meteor hit

damaskus [11]

Answer:

$a=3125000 m/s^2\\a=3.125*10^6 m/s^2$

Acceleration, in m/s, of such a rock fragment = $3.125*10^6m/s^2$

Explanation:

According to Newton's Third Equation of motion

$V_f^2-V_i^2=2as$

Where:

$V_f$ is the final velocity

$V_i$ is the initial velocity

a is the acceleration

s is the distance

In our case:

$V_f=V_{escape}, V_i=0,s=4 m$

So Equation will become:

$V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2$

Acceleration, in m/s, of such a rock fragment = $3.125*10^6m/s^2$

5 0

2 years ago

How many Fahrenheit degrees are needed to equal one Celsius (or Kelvin) degree?

Nat2105 [25]

One Celsius degree is the same size as one Kelvin. Each of them is the size of 1.8 Fahrenheit degrees.

3 0

2 years ago

An open 1-m-diameter tank contains water at a depth of 0.5 m when at rest. As the tank is rotated about its vertical axis the ce

Morgarella [4.7K]

Answer:

Angular velocity (w) = 8.86 rad/s

Explanation:

Angular velocity (w) = $\sqrt{} 4ghi/R^{2}$

g= 9.81 m/s

R= 0.5

hi (initial depth) = 0.5m

Hence= $\sqrt4* 9.81* 0.5/0.5^{2}$ = 8.86 rad/s

3 0

2 years ago

At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the

san4es73 [151]

Answer:

189 m/s

Explanation:

The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.

So, F = W

mv²/r = mg

v² = gr

v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m

So, v = √gr

v = √(9.8 m/s² × 3.63 × 10³ m)

v = √(35.574 × 10³ m²/s²)

v = √(3.5574 × 10⁴ m²/s²)

v = 1.89 × 10² m/s

v = 189 m/s

5 0

3 years ago