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3 years ago

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1 answer:

5 0

1) the molar mass of an element is found on the periodic table. It is often referred to as the mass number.
2) the molar mass of a compound is the sum of the molar masses of its elements. Don't forget that if there's two of an element in the compound, you must multiply that element's molar mass by two.
3) You can only find the amount of protons and electrons given atomic number. They are both equal to the atomic number. To find the amount of neutrons, you need the mass number as well. The amount of neutrons = mass number - atomic number.
4) I'm not sure about kinetic theory. I think it might say that everything's always moving. Particles in solids move slower than those in liquids which move slower than those in gasses. Not sure what else.

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Abox has a mass of 14.4 kg. The length is 4 m longthe width is 1 m and the height is 5 m. What is the density of the box?

valkas [14]

Answer:

0.72 kg per cubic m

Explanation:

Mass = 14.4 kg

Volume = lbh = 4*1*5 = 20 cubic m

$\because \: density \\ \\ = \frac{mass}{volume} \\ \\ = \frac{14.4}{20} \\ \\ = 0.72 \: kg {m}^{ - 3}$

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3 years ago

Which visible quality of smoke shows that it contains a

kvasek [131]

Answer:

C: Smoke rises out of very hot substances

Explanation:

Edge

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3 years ago

A positive test charge q is released from rest at distance r away from a charge of +Q and a distance 2r away from a charge of +2

Svetradugi [14.3K]

Answer:

B. to the right

Explanation:

Given:

test charge +q

distance of the test charge from +Q, r

distance of test charge from +2Q, 2r

Force on the test charge due to +Q:

$F_1=k.\frac{Q.q}{r^2}$

Force on the test charge due to +Q:

$F_2=k.\frac{2Q.q}{(2r)^2}$

$F_2=k.\frac{Q.q}{2r^2}$

Since all the charges are positive here, so they will try to repel the test charge away. And the force due to charge +Q will be greater so initially the test charge will move rightwards away from the +Q charge.

3 0

3 years ago

Read 2 more answers

on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker

DIA [1.3K]

Momentum
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.

Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum

Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum

Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude

!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v [solve for v]
v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s

If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]

The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.

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3 years ago

Indicate all ways in which the graph used in question 4 would change if the Normal Force applied to the object was increased. Be

Maksim231197 [3]

Answer:

Answer:

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# CARRY ON LEARNINGAnswer:

PLEASE BRAINLESS ME INEED

# CARRY ON LEARNING

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3 years ago