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3 years ago

How many atoms are present in 3 molecules of chromium

1 answer:

5 0

Answer:

1,8066 x 1024 atoms

Explanation:

1 mole of chromium contains 6,022 x 1023 atoms

so

3 moles of chromium contains 3 times as many atoms

3 x 6,022 x 1023 = 1,8 066 x 1024

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If you see lightening bolt and count for 4 seconds before you hear the thunder how far away was the lightening strike ? It was n

rjkz [21]

well, you divide 4 by 5, so .8

.8 of a mile is 4224 feet.

i really hope this helps :)

4 0

3 years ago

If the speed of an object is increasing, then the forces acting on that object must be ________.

In-s [12.5K]

Answer:

Unbalanced.

Explanation:

Usually, unbalanced forces cause acceleration, or increased movement.

5 0

3 years ago

A 70- kg bicycle rides his 9.8- kg bicycle with a speed of 16 m/ s. What is the magnitude of the braking force of the bicycle co

Rus_ich [418]

Answer:

F = -319.2 N

Explanation:

Given that,

The mass of a bicyclist, m = 70 kg

Mass of the bicycle = 9.8 kg

The speed of a bicycle, v = 16 m/s

We need to find the magnitude of the braking force of the bicycle come to rest in 4.0 m.

The braking force is given by :

$F=ma\\\\=\dfrac{m(v-u)}{t}\\\\=\dfrac{(70+9.8)(0-16)}{4}\\\\=-319.2\ N$

So, the required force is 319.2 N.

3 0

2 years ago

A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm.

SashulF [63]

Answer:

Part a)

$F = 0$

Part b)

$F = 0.25 N$

Part c)

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

Explanation:

As we know that force on a current carrying wire is given as

$\vec F = i(\vec L \times \vec B)$

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

$F = i(\vec L \times \vec B)$

here we know that L and B is parallel to each other so

$F = 0$

Part b)

For 68.1 cm length wire we have

$F = iLB sin\theta$

here we know that

$cos\theta = \frac{68.1}{166}$

$\theta = 65.8$

so we have

$F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8$

$F = 0.25 N$

Part c)

For 151 cm length wire we have

$F = iLB sin\phi$

here we know that

$cos\phi = \frac{151}{166}$

$\theta = 24.5$

so we have

$F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5$

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

4 0

2 years ago

the density of aluminum is 2700 kg/m3. if transverse waves travel at in an aluminum wire of diameter what is the tension on the

Sunny_sXe [5.5K]

The tension on the wire is 52.02 N.

From the question, we have

Density of aluminum = 2700 kg/m3

Area,

A = πd²/4

A = π x (4.6 x 10⁻³)²/4

A = 1.66 x 10⁻⁵ m²

μ = Mass per unit length of the wire

μ = ρA

μ = 2700 kg/m³ x 1.66 x 10⁻⁵ m²

μ = 0.045 kg/m

Tension on the wire = √T/μ

34 = √T/0.045

34² = T/0.045

T = 52.02 N

The tension on the wire is 52.02 N.

Complete question:

The density of aluminum is 2700 kg/m3. If transverse waves propagate at 34 m/s in a 4.6-mm diameter aluminum wire, what is the tension on the wire.

To learn more about tension visit: brainly.com/question/14336853

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5 0

1 year ago