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3 years ago

If gravity did not affect the path of a horizontally thrown ball the ball would ____

1 answer:

5 0

If gravity did not affect the path of a horizontally thrown ball the ball would eventually slow due to air resistance. If the ball were somehow thrown in a vacuum devoid of gravity and air, it would technically continue in a straight line forever.

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Which property of table salt is also a property of other ionic compounds

Jet001 [13]

s alluded to in the other answers, salt refers to any ionic compound that doesn't have “oxides” in it. Table salt is sodium chloride. Going down the periodic table, the first column contains lithium, sodium, potassium, rubidium, cesium, and francium. This group (alkali metals) of atoms (and their corresponding positive ions) gets larger in the order shown above. Therefore, their ionic bonds with chloride (or any nonmetal) gets smaller. The trend of their corresponding compounds is a decreasing hardness, decreasing melting point, decreasing boiling point, and decreasing thermal stability. These are the major periodic trends of these corresponding compounds. Other metal ions generally have higher positive charges on them. This makes the ionic bonds considerably larger and you can probably surmise most of their corresponding properties listed above. However, the details of their lattice structures may cause the overall trend to vary.

3 0

3 years ago

Read 2 more answers

Is this statement true or false concerning squall line thunderstorm development? These often form ahead of the advancing front b

Pavlova-9 [17]

Answer: The following statement is true about squall line thunderstorm development: <em><u>These often form ahead of the advancing front but rarely behind it because lifting of warm, humid air and the generation of a squall line usually occur in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.</u></em>

<em>An upper-level wave, accountable for the fabrication of a squall line, extend in front of and backside a cold front, the air backside the front is cold, steady and settling while the air ahead of the front is hot and co-seismic.</em>

3 0

2 years ago

Given: G = 6.67259 × 10−11 N m2 /kg2

8_murik_8 [283]

Answer:

Work done W 2.938*10^9 J

Explanation:

given data:

mass m = 944 Kg

Mass of moon M = 7*10^22 Kg

Radius of the moon R = 1.5*10^6 m

gravitational constant G = 6.67*10^{-11} Nm^2/Kg^2

we know that work done is given as

Work done $W = \frac{GMm}{R}$

$= \frac{6.67*10^{-11}*7*10^{22}*944}{1.5*10^6 m}$

= 2.938*10^9 J

8 0

3 years ago

Calculate a. The heat that must be supplied to a 500.0 g copper kettle containing 400.0 g of water to raise its temperature from

uysha [10]

Explanation:

<h2>For Copper</h2>

dH copper = mCdT copper

<em>(Specific Heat of copper=0.385 J/g C ) </em>

dH = 500 g (0.385 J/g C) (78 C rise)

dH = 15,015 Joules

<h2> For Water</h2>

dH water = m C dT water

<em>(Specific Heat of copper=</em>4.184 J/g-C<em>) </em>

<em />

dH = 400 g (4.184 J/g-C) (78 C rise)

dH = 130,540 Joules

total heat = 15,015 + 130,540 = 145,555 Joules

<h2>Percentage for Water </h2>

(130,540 Joules / 145,555 Joules) x 100 = 89.7 %

If we consider that we have 3 Significant Figures,

then, the answers become ,

15.0 KJ must be added for Copper

130.5 KJ must be added for Water

and the total of 145 KJ must be added in the kettle with the water

89.7 % of heat goes to the Water

7 0

2 years ago

Consider a long, closely wound solenoid with 10,000 turns per meter. What current, in amperes, is needed in the solenoid to prod

Makovka662 [10]

Answer:

0.4344A

Explanation:

From Ampere's law, it can be shown that the magnetic field B inside a long solenoid is

$B= \mu_0NI$

Where

B= Magnetic field strenght at distance d

I= current

$\mu_0 =$ Permeability of free space ( $4\pi*10^{-7} Tm/A$ )

N= Number of loops

Our values are defined as follow,

$N=10000$

$B=5.25*10^{-5}$ T

$B'=5.25*10^{-5} * 104 = 5.46*10^{-3}T$

As a current required to become 104 times the Earth's magnetic field is required, we use B '

$B'= \mu_0NI$

$5.46*10^{-3}=4\pi*10^{-7}*10000*I$

$I=\frac{5.46*10^{-3}}{4\pi*10^{-7}*10000}$

$I=0.4344A$

<em>Therefore is needed 0.4344A in the solenoid to produce a magnetic field inside the solenoid, near its center, that is 104 times the Earth's magnetic field.</em>

4 0

2 years ago