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NikAS [45]
3 years ago
12

Like the moon's orbit, Earth's orbit is not a perfect circle, but an oval How do you think this might affect Earth's tides?

Physics
1 answer:
ExtremeBDS [4]3 years ago
8 0
During the parts in the orbit where the moon is farthest away from the earth the tides will be low. Whereas during the parts where the moon is closer to the earth the tides will be higher.

Good Luck!
Please correct me if i'm wrong
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A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of
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Answer:

a) 4.485 kg b) 3.94 kg

Explanation:

since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .

44 = m * 9.81m/s^2

m = 44/9.81 = 4.485kg

b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)

44 = m (a +g)

44 = m (1.37 + 9.81)

44/11.18 = m

m = 3.94 kg

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C. Alpha

Explanation:

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Which instrument should, ideally, have zero resistance?
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Calculate the kinetic energy of a dog,
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a projectile is launched from level ground with an initial speed v0 at an angle θ with the horizontal. if air resistance is negl
LekaFEV [45]

The time of flight of the launched particle will be 2v₀ sinθ / g.

<h3>What is kinematics?</h3>

The study of motion without considering the mass and the cause of the motion.

A projectile is launched from level ground with an initial speed v₀ at an angle θ with the horizontal. If air resistance is negligible.

Then the time of flight of the launched particle will be given as,

Determine how long it requires the projectile to reach its highest point in order to calculate the flight duration. Just twice as long as the maximum altitude is the travel time.

\rm v_y = v_{o_y} + a_y t

At maximum height, \rm v_y = 0.

The time to reach maximum height is given as

\rm t_{1/2} = - \dfrac{v_{o_y} }{ a_y}

The time of flight is given as,

\rm t = 2 \times t_{1/2} = - \dfrac{2v_{o_y}}{ a_y}

Substituting in \rm v_{o_y} = v_o \ \sin(\theta ) \ and \ a_y = -g, then we have

The time of flight of the launched particle will be 2v₀ sinθ / g.

More about the kinematics link is given below.

brainly.com/question/7590442

#SPJ4

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