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zloy xaker [14]
4 years ago
9

Give two examples of spatial interference which can be easily observed. Small explanation of each object please, just so I can u

nderstand.
Physics
1 answer:
MissTica4 years ago
7 0

Answer:

Spatial interference describes the correlation between waves at different points in space.

The two example of spatial interference are as following:

  • alphanumeric characters can be easily identified in isolation but difficult to be identified when these characters are surrounded by optotypes because visual acuity will be reduced by crowded conditions that limit resolution at several stages of visual processing.
  • Birds are able to identify visual target slowly that are presented at the top than to  that same visual target presented at the bottom.

Hence, different waves in a space correlate result into Spatial interference.

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A 12.0 g bullet was fired horizontally into a 1 kg block of wood. The bullet initially had a speed of 250 m/s. The block of wood
LuckyWell [14K]

Answer:

The rise in height of combined block/bullet from its original position is 0.45m

Explanation:

Given;

mass of bullet, m₁ = 12 g = 0.012 kg

mass of block of wood, m₂ = 1 kg

initial speed of bullet, u₁ = 250 m/s.

initial speed of block of wood, u₂ = 0

From the principle of conservation of linear momentum, calculate the final speed of the combined block/bullet system.

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final speed of the combined block/bullet system.

0.012 x 250 + 0 = v (0.012 + 1)

3 = v (1.012)

v = 3/1.012

v = 2.96 m/s

From the principle of conservation of energy, calculate the rise in height of the block/bullet combined from its original position.

¹/₂mv² = mgh

¹/₂v² = gh

¹/₂ (2.96)² = (9.8)h

4.3808 = 9.8h

h = 4.3808/9.8

h = 0.45 m

Therefore, the rise in height of combined block/bullet from its original position is 0.45m

7 0
3 years ago
Read 2 more answers
State a Newtowns second law​
astraxan [27]

Answer:

the second law states that the force F is the product of an object's mass and its acceleration a: F = m * a. For an external applied force, the change in velocity depends on the mass of the object.

5 0
3 years ago
I really need help with this
katen-ka-za [31]

<u>C</u> is the correct answer, because energy cannot be created neither destroy. The energy is changing from chemical to from electric to light, and from light to heat.  

8 0
3 years ago
What is the easiest way to increase the magnetic force acting on the rotor in an induction motor?
Schach [20]

Answer:

Explanation:

Magnets are of two major forms namely the permanent magnet and the temporary magnets. Temporary magnets magnetizes and demagnetize easily while permanent magnets does not magnetizes and demagnetize easily.

This permanents magnets are applicable in loudspeakers, generators, induction motor etc.

To increase the

The following will tend to increase the magnetic force acting on the rotor in an induction motor.

1. Increasing the strength of the bar magnet. Increase in strength of the magnet will lead to increase in the magnetic force acting on the rotor.

2. Increase in the magnetic line of force also known as the magnetic flux around the magnet will also increase the magnetic force acting on the rotor.

6 0
4 years ago
The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion.
lesya [120]

Answer:

(a) m = 1.6 x 10²¹ kg

(b) K.E = 2.536 x 10¹¹ J

(c) v = 7.12 x 10⁵ m/s

Explanation:

(a)

First we find the volume of the continent:

V = L*W*H

where,

V = Volume  of Slab = ?

L = Length of Slab = 4450 km = 4.45 x 10⁶ m

W = Width of Slab = 4450 km = 4.45 x 10⁶ m

H = Height of Slab = 31 km = 3.1 x 10⁴ m

Therefore,

V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)

V = 6.138 x 10¹⁷ m³

Now, we find the mass:

m = density*V

m = (2620 kg/m³)(6.138 x 10¹⁷ m³)

<u>m = 1.6 x 10²¹ kg</u>

<u></u>

(b)

The kinetic energy will be:

K.E = (1/2)mv²

where,

v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)

v = 3.17 x 10⁻¹⁰ m/s

Therefore,

K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²

<u>K.E = 2.536 x 10¹¹ J</u>

<u></u>

(c)

For the same kinetic energy but mass = 77 kg:

K.E = (1/2)mv²

2.536 x 10¹¹ J = (1/2)(77 kg)v²

v = √(2)(2.536 x 10¹¹ J)

<u>v = 7.12 x 10⁵ m/s</u>

7 0
3 years ago
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