A shell is launched with a velocity of 100 m/s at an angle of 30.0° above horizontal from a point on a cliff 50.0 m above a leve

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

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:

529.15 m/s

Explanation:

h = Highest point = 70000 m

g = Gravitational acceleration = 2 m/s²

m = Sulfur's mass

Since both potential and kinetic energies are conserved

The velocity at which the liquid sulfur exited the volcano is 529.15 m/s

The question lacks details. Here is the full question.

The accompanying image was captured with a camera capable of shooting between one and two frames per second. A series of photos was merged into this single image, meaning the vehicles depicted are actually the same car, documented at different intervals.

Assuming the camera produced 1.3 frames per second for this image and that the length of the car is approximately 5.3 meters, based on this information and the photo, how fast was the car moving?

Answer: v = 6.5 m/s

Explanation: The problem requires calculating the car's velocity. Velocity can be computed using:

Since the camera captured 7 images of the car and its length is noted as 5.3, the car's displacement is:

Δx = 7(5.3)

Δx = 37.1 m

The camera operates at 1.3 frames per second and recorded 7 images, thus the time driven by the car is:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

v = 6.87 m/s

Explanation:

The diverse structures of carbon-based compounds are influenced by several factors:

1. the capacity of bonds to rotate freely,

2. the ability of carbon to create four covalent bonds,

3. the spatial arrangement of bonds resembling a tetrahedron.