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3 years ago

Choose the correct statement below about the interaction between magnets, magnetic materials and charges.

Physics

1 answer:

bekas [8.4K]3 years ago

5 0

Answer: B. The south magnetic pole attracts the north pole of another magnet.

Explanation:

The correct statement below about the interaction between magnets, magnetic materials and charges is that the south magnetic pole attracts the north pole of another magnet.

We should note that a magnet typically consists of two poles which are the North Pole and the South pole. The opposite poles doesn't repel each other but attracts each other while same pole repels one another.

In this case, the North Pole will attract the South Pole while same magnetic piles repel e.g the North Pole will repel the North Pole.

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F. If the shuttle's period is synchronized with that of Earth's rotation, what is the height of the shuttle? (1 day = 8.64x104s,

oee [108]

Answer:

1.324 × 10⁷ m

Explanation:

The centripetal acceleration, a at that height above the earth equal the acceleration due to gravity, g' at that height, h.

Let R be the radius of the orbit where R = RE + h, RE = radius of earth = 6.4 × 10⁶ m.

We know a = Rω² and g' = GME/R² where ω = angular speed = 2π/T where T = period of rotation = 1 day = 8.64 × 10⁴s (since the shuttle's period is synchronized with that of the Earth's rotation), G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², ME = mass of earth = 6 × 10²⁴ kg. Since a = g', we have

Rω² = GME/R²

R(2π/T)² = GME/R²

R³ = GME(T/2π)²

R = ∛(GME)(T/2π)²

RE + h = ∛(GMET²/4π²)

h = ∛(GMET²/4π²) - RE

substituting the values of the variables, we have

h = ∛(6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg × (8.64 × 10⁴s)²/4π²) - 6.4 × 10⁶ m

h = ∛(2,987,477 × 10²⁰/4π² Nm²s²/kg) - 6.4 × 10⁶ m

h = ∛75.67 × 10²⁰ m³ - 6.4 × 10⁶ m

h = ∛(7567 × 10¹⁸ m³) - 6.4 × 10⁶ m

h = 19.64 × 10⁶ m - 6.4 × 10⁶ m

h = 13.24 × 10⁶ m

h = 1.324 × 10⁷ m

3 0

3 years ago

A force of 8 N is used to drag a chair 2.5 metres across a room. Calculate the work done to move the chair.

NNADVOKAT [17]

Answer:

Explanation:

GIVEN

Force (F) = 8 N

Distance (d) = 2.5 metres

Work done = ?

WE know we have the formula

work done = F * d

Work done = 8 * 2.5

= 20 Joule

Hope it helps :)

8 0

4 years ago

Read 2 more answers

A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2ft/s to v1 =

Ratling [72]

Answer:

Question 1)

a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.

$v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2$

This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,

$v = \omega R\\a = \alpha R$

where α is the angular acceleration.

In order to continue this question, the radius of the drums should be given.

Let us denote the radius of the drums as R, the angular acceleration of drum B is

α = 0.5/R.

b) The distance travelled by the drums can be found by the following kinematics formula:

$v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft$

One revolution is equal to the circumference of the drum. So, total number of revolutions is

$x / (2\pi R) = 6/(\pi R)$

Question 2)

a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:

$a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2$

b) $a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2$

5 0

4 years ago

if a free falling rock were equipped with a speedometer, by how much would it's speed readings increase with each second IF it w

Kaylis [27]

Accelleration(a) is changing of velocity in second.
free fall a = g
speed increase = a = g = 20 (m/s) / s

7 0

3 years ago

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent ho

faltersainse [42]

Answer:

Explanation:

Let mass of cylinder be M

Moment of inertia of cylinder

= 1/2 M R² r is radius of cylinder

If radius of equivalent hoop be k

Mk² = 1/2 x MR²

k = R / √2

1.2 / 1.414

Radius of gyration = 0.848 m

b )

moment of inertia of spherical shell

= 2 / 3 M R²

Moment of inertia of equivalent hoop

Mk²

Mk² = 2 / 3 M R²

k = √2/3 x R

= .816 X 1.2

Radius of gyration = .98 m

c )

Moment of inertia of solid sphere

= 2/5 M R²

Moment of inertia of equivalent hoop

= Mk²

Mk² = 2/5 M R²

k √ 2/5 R

Radius of gyration = .63 R

6 0

4 years ago