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3 years ago

15

If someone drops a cup, it falls to the ground. Why doesn't the gravitational force between the person's hand and the cup keep t

he cup from falling?

1 answer:

Genrish500 [490]3 years ago

5 0

Answer:

The gravity from the person's hand is weaker than the gravity from the pull of the earth

Explanation:

The gravity from the person's hand is weaker than the gravity from the pull of the earth

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How do oxygen and beryllium atoms transform into oxygen ion O2- and Be2 beryllium ion Be2? tysm

Hoochie [10]

On the whole, the metals burn in oxygen to form a simple metal oxide. Beryllium is reluctant to burn unless it is in the form of dust or powder. Beryllium has a very strong (but very thin) layer of beryllium oxide on its surface, and this prevents any new oxygen getting at the underlying beryllium to react with it.

7 0

3 years ago

Read 2 more answers

How does the current model of the atom differ from<br> j j thomas model

bixtya [17]

Jj thomas' model contained electrons.

The current model contains a nucleus with electrons orbiting around it. :)

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3 years ago

At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi

Amanda [17]

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

3 0

3 years ago

The bonds of oxygen molecules are broken by sunlight. The minimum energy required to break the oxygen-oxygen bond is 495 kJ/mol.

makvit [3.9K]

Answer:

The wavelength of sunlight that can cause this bond breakage is 242 nm

Explanation:

The minimum energy of the sunlight that'll break Oxygen-oxygen bond must match 495 KJ/mol

But 1 mole of any molecule contains 6.02 × 10²³ molecules/mol

Each molecule of Oxygen will require (495 × 10³)/(6.02 × 10²³) = 8.22 × 10⁻¹⁹ J

E = hf

v = fλ

f = v/λ

f = frequency of the sunlight

λ = wavelength of the sunlight

v = speed of light = 3.0 × 10⁸ m/s

E = hv/λ

λ = hv/E

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

λ = (6.63 × 10⁻³⁴)(3 × 10⁸)/(8.22 × 10⁻¹⁹)

λ = 2.42 × 10⁻⁷ m = 242 nm.

5 0

3 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago