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3 years ago

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Which of the following equations can be used to directly calculate an object's impulse?

1 answer:

Iteru [2.4K]3 years ago

7 0

I believe the answer is B

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Argon, atomic number 18, has an atomic weight of 39.9. Potassium, atomic number 19, has an atomic weight of 39.1. Which of these

Rufina [12.5K]

Answer: D

Explanation:

Atomic weight is measured by adding the number of protons and neutrons in an atom's nucleus. Argon's atomic number is 18 while potassium's is 19. This means that Argon will always have 18 protons while potassium will always have 19 protons.

To make the numbers easier to work with, round each atomic weight. We'll say the atomic weight of potassium is 39 and the atomic weight of argon is 40. To see how many neutrons each one has, I can set up a simple equation for each using the following equation:

Atomic weight = protons + neutrons

Potassium:

39 = 19 + N --> N = 20

Argon:

40 = 18 + N --> N = 22

An atom is defined by the number of protons it has, but the number of neutrons can vary. We call these isotopes, or atoms with the same number of protons but a different number of neutrons. As the math shows, argon typically has more neutrons per atom than potassium does.

6 0

3 years ago

Read 2 more answers

You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.

valentinak56 [21]

Answer:

$F=6\times 10^{-7}\ N$

Explanation:

Given:

quantity of point charge, $q=-4\times 10^{-9}\ C$

radial distance from the linear charge, $r=0.09\ m$

linear charge density, $\lambda=3\times 10^{-9}\ C.m^{-1}$

<u>We know that the electric field by the linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.\epsilon_0.r}$

$E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}$

$E=150\ N.C^{-1}$

<u>Now the force on the given charge can be given as:</u>

$F=E.q$

$F=150\times 4\times 10^{-9}$

$F=6\times 10^{-7}\ N$

3 0

3 years ago

A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

Zielflug [23.3K]

Answer: $F_{v} =\mu_{k} mg$

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to start crate moving is $F_{s} =\mu_{s} mg$ .

but once crate starts moving the force of friction is reduced $F_{v} =\mu_{k} mg$ .

Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie $F_{v} =\mu_{k} mg$ .

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.

Magnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

4 0

3 years ago

If the period of a wave is 0.05 s, then what is its frequency?

yKpoI14uk [10]

Frequency = 1/time period = 1/0.05 = 20s^-1.

8 0

3 years ago

Read 2 more answers

A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density

Amiraneli [1.4K]

Answer:

a) K = 2/3 π G m ρ R₁³ / R₂ , b) U = - G m M / r

Explanation:

The law of universal gravitation is

F = G m M / r²

Part A

Let's use Newton's second law

F = m a

The acceleration is centripetal

a = v² / R₂

G m M / R₂² = m v² / R₂

v² = G M / R₂

They give us the density of the planet

ρ = M / V

V = 4/3 π R₁³

M = ρ V

M = ρ 4/3 π R₁³

v² = 4/3 π G ρ R₁³ / R₂

K = ½ m v²

K = ½ m (4/3 π G ρ R₁³ / R₂)

K = 2/3 π G m ρ R₁³ / R₂

Part B

Potential energy and strength are related

F = - dU / dr

∫ dU = - ∫ F. dr

The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1

U- U₀ = G m M ∫ dr / r²

U - U₀ = G m M (- r⁻¹)

We evaluate for

U - U₀ = -G m M (1 / $r_{f}$ - 1 / $r_{i}$ )

They indicate that for ri = ∞ U₀ = 0

U = - G m M / r

6 0

3 years ago