The light from Star 1 reaches Star 3 in 138 years and Star 2 in 63 years. Which of these conclusions about the stars is correct?

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 2.07 Hz. On this tray is an empty cu

Anettt [7]

Answer:

Explanation:

Given

Frequency of SHM is $f=2.07\ Hz$

Amplitude of SHM is $A=3.13\ cm$

Cup begins to slip when it overcomes the friction force

Friction force $F_s=\mu mg$

Applied force $F=ma$

$ma=\mu mg$

$a=\mu g$

and maximum acceleration during SHM is

$a=A\omega ^2$

$a=A(2\pi f)^2$

$a=3.13\times 10^{-2}\times (2\pi 2.07)^2$

$a=5.296\ m/s^2$

$\mu =\frac{a}{g}$

$\mu =\frac{5.296}{9.8}=0.54$

6 0

3 years ago

Read 2 more answers

Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of

Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = $\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}$

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = $\sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}$

(|r₂₁|)² = 148.25

$F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }$

= 0.050938(0.19107i + 0.54933j) N

= (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0

3 years ago

How do mass and speed affect kinetic energy?

Lunna [17]

The mass affects the kinetic energy because the more the mass the more energy is given to the object and the speed<span> affects by making it go faster and longer, so whenever speed goes up so does energy.</span>

6 0

3 years ago

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How do the dark lines of an atom''s absorption spectrum relate to the bright lines of its emission spectrum?

tangare [24]

Wouldn't it be neat if an electron falling closer to the nucleus ... emitting a
photon ... actually gave out more energy than it needed to climb to its original
energy level by absorbing a photon ! If there were some miraculous substance
that could do that, we'd have it made.

All we'd need is a pile of it in our basement, with a bright light bulb over the pile,
connected to a tiny hand-crank generator.

Whenever we wanted some energy, like for cooking or heating the house, we'd
switch the light bulb on, point it towards the pile, and give the little generator a
little shove. It wouldn't take much to git 'er going.

The atoms in the pile would absorb some photons, raising their electrons to higher
energy levels. Then the electrons would fall back down to lower energy levels,
releasing more energy than they needed to climb up. We could take that energy,
use some of it to keep the light bulb shining on the pile, and use the extra to heat
the house or run the dishwasher.

The energy an electron absorbs when it climbs to a higher energy level (forming
the atom's absorption spectrum) is precisely identical to the energy it emits when
it falls back to its original level (creating the atom's emission spectrum).

Energy that wasn't either there in the atom to begin with or else pumped
into it from somewhere can't be created there.

You get what you pay for, or, as my grandfather used to say, "For nothing
you get nothing."

3 0

3 years ago

A girl rides her bike 5.4 km due east. While riding she experiences a resistive force from the air that has a magnitude of 3.1 N

kotegsom [21]

Answer:

(A) = 34,560 J

(B) Work done during the round trip is not zero, hence the resistive force is not a conservative force.

Explanation:

This question is incomplete, here is the complete question:

A girl rides her bike 5.4 km due east. While riding she experiences a resistive force from the air that has a magnitude of 3.1 N and points due west. She then turns around and rides due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 3.3 N and points due east.? (a) Find the work done by the resistive force during the round trip. (b) Based on your answer to part (a), is the resistive force a conservative force? Explain.

SOLUTION:

distance covered by the girl due east (De) = 5.4 km = 5400 m

magnitude of resistive force by the air (Fw) = 3.1 N due west

distance covered by the girl due west (Dw) = 5.4 km = 5400 m

magnitude of resistive force by the air (Fe) = 3.3 N due east

(A) to find the work done by the resistive force during the round trip we have to get the work done by the resistive force due west and add it to the work done by the resistive force due east

work done by the resistive force due west = (Fw.cosθ) x De

where θ is the angle between the displacement and the force. The

displacement is due east while the force is due west, hence θ = 180°

work done (W1) = (3.1 x cos 180) x 5400 = -16,740 j

work done by the resistive force due east = (Fe.cosθ) x Dw

where θ is the angle between the displacement and the force. The

displacement is due west while the force is due east, hence θ = 180°

work done (W2) = (3.3 x cos 180) x 5400 = -17,820 j

Hence work done during the round trip = W1 + W2 = (-16,740 ) + (-17,820)

= 34,560 J

(B) A conservative force is a force in which the work done by the force in moving an object around a closed path is zero. From part (A) above the work done during the round trip is not zero, hence the resistive force is not a conservative force.

3 0

3 years ago