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2 years ago

According to Newton´s second law of motion, what does an object with more mass require? Question 8 options:

2 answers:

6 0

More force to accelerate. If you push a car compared to a beach ball, which will you have to push harder for it to move? That's pretty much what the question is asking, if that helps any:)

trapecia [35]2 years ago

6 0

more for force to accelerate

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NASA launches a probe with a mass of 15,000 kg to another planet more massive than Earth. Which statement is true about the prob

vivado [14]

Answer: The weight of the probe will increase.

8 0

2 years ago

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

alexandr1967 [171]

Answer:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

$m_e$ is the mass of the electron

Solving for v, we find

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

$r=5.3\cdot 10^{-11}m$

while the electron mass is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

Substituting into the formula, we find

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

7 0

3 years ago

A spotlight on a boat is y = 2.2 m above the water, and the light strikes the water at a point that is x = 8.5 m horizontally di

slava [35]

Answer:

The answer to the question is

The distance d, which locates the point where the light strikes the bottom is 29.345 m from the spotlight.

Explanation:

To solve the question we note that Snell's law states that

The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction

n₁sinθ₁ = n₂sinθ₂

y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and

θ₁ = 14.511 °

n₁ = 1.0003 = refractive index of air

n₂ = 1.33 = refractive index of water

Therefore sinθ₂ = $\frac{n_1sin\theta_1}{n_2}$ = $\frac{1.003*0.251}{1.33}$ = 0.1885 and θ₂ = 10.86 °