Horizontal displacement?

What is the density of a block of marble that occupies 236 cm3 and has a mass of 824 g? answer in units of g/cm3 ?

REY [17]

Since the formula is mass/volume the density is 3.4915......g/cm^3

4 0

3 years ago

Graph are blank a relationship

Maru [420]

I have no clue sorry maybe try applying the answers and questions then I'll answer for u

7 0

3 years ago

Two tougboats are toeing a ship each exerts a force of 6000N and the angle between the two ropes is 60 calculate the resultant f

Arlecino [84]

Answer:

10392.30N

Explanation:

We proceed by computing the individual force exerted by the boats

For the first boat

The angle is 30 degree to the vertical

Hence

Force = F cos θ

F=6000 cos 30

F=6000*0.866

F=5196.15 N

Since the boats are two and also at the same angle and also exerting the same force

The Net force = 2*5196.15

Net force=10392.30N

7 0

2 years ago

How far from the surface of Earth is the magnitude of Earth's gravitational field equal to 7.86 N/kg?

AfilCa [17]

Answer:

7.48 x 10⁵ m

Explanation:

g = 7.86 N/kg

M = 5.97 x 10²⁴ kg, R = 6.37 x 10⁶ m.

Find height h

g = GM/(R + h)²

(R + h)² = GM/g = 6.67 x 10⁺¹¹ x 5.97 x 10²⁴ /7.86 = 5.066 x 10¹³

R + h = 7.12 x 10⁶ m

so

h = 7.12 x 10⁶ - 6.37 x 10⁶ = 7.48 x 10⁵ m

4 0

2 years ago

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0

2 years ago