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harkovskaia [24]
3 years ago
13

Please help With the question! Ill mark brainliest for whoever answers First.

Physics
2 answers:
Galina-37 [17]3 years ago
7 0
Meteor maybe?

I’m pretty sure it’s meteor but if not sorry
rewona [7]3 years ago
5 0

Answer:

I believe its an Asteroid. Hope that helps

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Aaron's normal response time to apply the car brakes is 0.7 seconds. Aaron's response time doubles when he is tired. How far wil
Likurg_2 [28]

Aaron's car is moving at speed of 30 m/s

His reaction time is given as 0.7 s

but when he is tired the reaction time is doubled

Now we need to find the distance covered by his car when he is tired during the time when he react to apply brakes

So here since during this time speed is given as constant so we can say that distance covered can be product of speed and time

So here we can use

d = v*t

d = 30 * 1.4

d = 42 m

So the car will move to 42 m during the time when he apply brakes

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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:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

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}

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3 years ago
Trong thí nghiệm Y-âng về giao thoa ánh sáng : khoảng cách giữa hai khe là 0,5 mm ; khoảng cách từ mặt phẳng chứa hai khe đến mà
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không có câu hỏi, trả lời như nào nhỉ?

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