All categories

2 years ago

Each element has an atomic number state which is meant by atomic number

1 answer:

3 0

Answer: The atomic number is the number of protons in an atom, and isotopes have the same atomic number but differ in the number of neutrons. The number of protons in an atom is called its atomic number. This number is very important because it is unique for atoms of a given element. All atoms of an element have the same number of protons, and every element has a different number of protons in its atoms.

Explanation:

You might be interested in

How many mL of water would be displaced by 408 g of lead

Natasha2012 [34]

For this, we must know the density of lead. This is 11.36 grams per cubic centimeter. The mass present is 408 grams. Therefore, the volume occupied will be:
408 / 11.36 = 35.9 cubic centimeters. This is the volume of water that will be displaced.
Now, we know that one cubic centimeter is equivalent to one mL. Therefore,
35.9 ml of water will be displaced.

6 0

3 years ago

A ball is thrown at an angle of 38° to the horizontal. What happens to the

finlep [7]

Answer:

Vy = V0 sin 38 where Vy is the initial vertical velocity

The ball will accelerate downwards (until it lands)

Note the signs involved if Vy is positive then g must be negative

The acceleration is constant until the ball lands

t (upwards) = (0 - Vy) / -g = Vy / g final velocity = 0

t(downwards = (-Vy - 0) / -g = Vy / g final velocity = -Vy

time upwards = time downwards (conservation laws)

8 0

2 years ago

What are the names of the 4 types of fronts? How are they created?

jeka57 [31]

Answer:

Stationary Front, warm front, cold front, Occluded Front.

Explanation:

Stationary Front. When the surface position of a front does not change (when two air masses are unable to push against each other; a draw), a stationary front is formed.

cold front is the leading edge of a cooler mass of air at ground level that replaces a warmer mass of air and lies within a pronounced surface trough of low pressure. It often forms behind an extratropical cyclone (to the west in the Northern Hemisphere, to the east in the Southern), at the leading edge of its cold air advection pattern—known as the cyclone's dry "conveyor belt" flow. Temperature differences across the boundary can exceed 30 °C (86 °F) from one side to the other. When enough moisture is present, rain can occur along the boundary. If there is significant instability along the boundary, a narrow line of thunderstorms can form along the frontal zone. If instability is weak, a broad shield of rain can move in behind the front, and evaporative cooling of the rain can increase the temperature difference across the front. Cold fronts are stronger in the fall and spring transition seasons and weakest during the summer.

A warm front is a density discontinuity located at the leading edge of a homogeneous warm air mass, and is typically located on the equator-facing edge of an isotherm gradient. Warm fronts lie within broader troughs of low pressure than cold fronts, and move more slowly than the cold fronts which usually follow because cold air is denser and less easy to remove from the Earth's surface. This also forces temperature differences across warm fronts to be broader in scale. Clouds ahead of the warm front are mostly stratiform, and rainfall gradually increases as the front approaches. Fog can also occur preceding a warm frontal passage. Clearing and warming is usually rapid after frontal passage. If the warm air mass is unstable, thunderstorms may be embedded among the stratiform clouds ahead of the front, and after frontal passage thundershowers may continue. On weather maps, the surface location of a warm front is marked with a red line of semicircles pointing in the direction of travel.

In meteorology, an occluded front is a weather front formed during the process of cyclogenesis. The classical view of an occluded front is that they are formed when a cold front overtakes a warm front, such that the warm air is separated (occluded) from the cyclone center at the surface. The point where the warm front becomes the occluded front is called the triple point; a new area of low-pressure that develops at this point is called a triple-point low. A more modern view of the formation process suggests that occluded fronts form directly during the wrap-up of the baroclinic zone during cyclogenesis, and then lengthen due to flow deformation and rotation around the cyclone.

3 0

2 years ago

Read 2 more answers

1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/

mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

8 0

2 years ago

1. Astronomical observatories have been available since ancient times, and many cultures set aside special sites for astronomica

nydimaria [60]

Answer:

e telescopes

Explanation:

may i be marked brainliest?

4 0

3 years ago