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Ugo [173]
3 years ago
15

A submarine is stranded on the bottom of the ocean with its hatch 25 m below the surface. In this problem, assume the density of

sea water is 1.03 × 103 kg/m3. show answer No Attempt Calculate the force, in newtons, needed to open the hatch from the inside, given it is circular and 0.25 m in diameter. The air pressure inside the submarine is 1.00 atm.
Physics
1 answer:
topjm [15]3 years ago
4 0

Answer:

F = 49.55 x 10 ³ N

Explanation:

Given :

ρ = 1.03 x 10³ kg / m³ ,  h= 25 m , g = 9.8 m / s² , r = 0.25 m

So solve to determine force

A = π * r² = π * (0.25 m)²

A = 196.34 x 10 ⁻³ m²

The pressure of the water

P = F / A

Determine the force using the pressure into the area

F = P * A

P = ρ * g * h

Replacing numeric

P = 1.03 x 10³ kg / m³ * 9.8 m / s² * 25 m

P = 252.35 x 10³ N / m²

F = P * A  ⇒ F = 252.35 x 10³ N / m²  * 196.34 x 10 ⁻³ m²  

F = 49.55 x 10 ³ N

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100 points!!! Please help!!!
Greeley [361]

Answer:

41°

Explanation:

Kinetic energy at bottom = potential energy at top

½ mv² = mgh

½ v² = gh

h = v²/(2g)

h = (2.4 m/s)² / (2 × 9.8 m/s²)

h = 0.294 m

The pendulum rises to a height of  above the bottom.  To determine the angle, we need to use trigonometry (see attached diagram).

L − h = L cos θ

cos θ = (L − h) / L

cos θ = (1.2 − 0.294) / 1.2

θ = 41.0°

Rounded to two significant figures, the pendulum makes a maximum angle of 41° with the vertical.

7 0
3 years ago
Match the type of heat transfer with its description
VikaD [51]

Answer:

1. Convection

2. Radiation

3. Conduction

Hope this helps!

Explanation:

5 0
3 years ago
A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5​​​​m​​/s2 for 4s. After t
Ber [7]

Answer:

+7.0 m/s

Explanation:

Let's take rightward as positive direction.

So in this problem we have:

a = -2.5 m/s^2 acceleration due to the wind (negative because it is leftward)

t = 4 s time interval

v = -3.0 m/s is the final velocity (negative because it is leftward)

We can use the following equation:

v = u + at

Where u is the initial velocity

We want to find u, so if we rearrange the equation we find:

u = v - at = (-3.0 m/s) - (-2.5 m/s^2)(4 s)=+7.0 m/s

and the positive sign means the initial direction was rightward.

6 0
3 years ago
Ben walks 500 meters from his house to the corner store. He then walks back toward his house but stops to talk to a neighbor whe
spin [16.1K]
Average velocity =

      (displacement) / (time for the displacement)
and
      (direction of the displacement) .

Displacement =

      (distance from the start-point to the end-point)
and
      (direction from the start-point to the end-point)   .

When Ben is 200 meters from the corner store,
he is (500 - 200) = 300 meters from his house.

His displacement is

         300 meters in the direction
                             from his house to the neighbor .


His average velocity is

         (300/910) =  0.33 meters per second, in the
                               direction from his house to the neighbor .


7 0
3 years ago
Read 2 more answers
A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

6 0
3 years ago
Read 2 more answers
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