When a submarine dives to a depth of 5.0 × 102 m, how much pressure, in pa, must its hull be able to withstand? how many times l

Answer:

σ₁ = C/m²

σ₂ =  C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface  ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

Charge (Q₁) = 572.8 C

Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.

V = constant

∴

  = C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = = =  C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  = 0.0314 m².

σ₂ = = = C/m²

In the study of physics, Hooke's law can be expressed as:

F = kx

This law indicates that the spring force F is proportional to the extension x, with k being the spring constant.

In experiments, this is often examined using the setup illustrated in the included figure. The spring is tested, and a known weight is applied underneath it. This weight exerts a gravitational pull, essentially its weight, on the spring. While the spring elongates, the displacement can be measured using a ruler.

Several potential errors can arise during this experiment. Firstly, the person's measurement reading may be faulty. Digital scales offer greater accuracy as they reduce human error, while ruler readings can be subjective, especially if not viewed at eye level. Additionally, the object's weight may be inaccurately measured if the scale is untrustworthy. Lastly, the measuring equipment may not be correctly calibrated.

Answer:

Explanation:

a )

Each blade resembles a rod with its axis positioned near one end.

The moment of inertia for one blade is:

= 1/3 x m l²

where m stands for the mass of the blade

l represents the length of each blade.

Total moment of inertia for 3 blades is:

= 3 x  x m l²

ml²

2 )

Details provided include:

m = 5500 kg

l = 45 m

Substituting these values produces:

moment of inertia of one blade:

= 1/3 x 5500 x 45 x 45

= 37.125 x 10⁵ kg.m²

Moment of inertia for 3 blades:

= 3 x 37.125 x 10⁵ kg.m²

= 111.375 x 10⁵ kg.m²

c )

Angular momentum

= I x ω

I denotes the moment of inertia of the turbine

ω symbolizes angular velocity

ω = 2π f

f indicates the rotational frequency of the blades

d )

We have I = 111.375 x 10⁵ kg.m² (Calculated)

f = 11 rpm (revolutions per minute)

= 11 / 60 revolutions per second

ω = 2π f

=  2π x  11 / 60 rad / s

Calculating angular momentum yields

= I x ω

111.375 x 10⁵ kg.m² x  2π x  11 / 60 rad / s

= 128.23 x 10⁵  kgm² s⁻¹.

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

1 hour = 3,600 seconds
1 km = 1,000 meters

75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s

The mean speed during the deceleration is

                                   (1/2)(20-5/6 + 0) = 10-5/12 m/s.

Traveling at this average speed for 21 seconds,
the bus covers

                        (10-5/12) × (21) = 218.75 meters.