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Stolb23 [73]
3 years ago
6

________ occurs when an object in the outer reaches of the solar system passes between earth and a far distant star, temporarily

blocking light from the star.
Physics
1 answer:
murzikaleks [220]3 years ago
6 0
<span>Occulation occurs when an object in outer space reaches the solar system and passes between earth and a far distant star, temporarily blocking light from the star. Occulation is an astronomy term referring to when an object in the foreground blocks the view of an object in the background.</span>
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Can someone please help me ASAP!
Artist 52 [7]
I’m not 100% sure but i think it’s A because if you divide the speed by the time you get 2 and also all the other answer choices don’t make any sense!
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3 years ago
Problem: Hooke's law states that the force on a spring varies directly with the distance that it is stretched. If a spring has a
Scorpion4ik [409]

Answer:

Restoring force of the spring is 50 N.

Explanation:

Given that,

Spring constant of the spring, k = 100 N/m

Stretching in the spring, x = 0.5 m

We need to find the restoring force of the spring. It can be calculated using Hooke's law as "the force on a spring varies directly with the distance that it is stretched".

F=kx

F=100\ N/m\times 0.5\ m

F = 50 N

So, the restoring force of the spring is 50 N. Hence, this is the required solution.

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3 years ago
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

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The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
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pshichka [43]

Answer:

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1 year ago
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