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ANTONII [103]
1 year ago
8

When there's a hazard ahead, it's almost always quicker for you to _________ than to come to a full stop.

Physics
1 answer:
TEA [102]1 year ago
5 0

When there's a hazard ahead, it's almost always quicker for you to steer away than to come to a full stop.

<h3>What is an hazard?</h3>

Hazard refers to any obstacle or other feature which causes risk or danger.

Living organisms respond to hazards via the production of adrenaline hormone. This hormone causes a flight response away from the hazard.

Therefore, when there's a hazard ahead, it's almost always quicker for you to steer away than to come to a full stop.

Learn more about hazards at: brainly.com/question/5338299

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A mass of 30.0 grams hangs at rest from the lower end of a long vertical spring. You add different amounts of additional mass ΔM
ra1l [238]

Answer:

K = 373.13 N/m

Explanation:

The force of the spring is equals to:

Fe - m*g = 0     =>    Fe = m*g

Using Hook's law:

K*X = m*g    Solving for K:

K = m/X * g

In this equation, m/X is the inverse of the given slope. So, using this value we can calculate the spring's constant:

K = 10 / 0.0268 = 373.13N/m

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When did galileo discover projectile motion?
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Between 1589-1592 when he discovered projecctile motion
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A student attaches a block of mass M to a vertical spring so that the block-spring system will oscillate if the block-spring sys
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Consider the data collected in science class. Different masses were thrown with varied amounts of
lesya [120]

Answer:

A: In all cases, the acceleration was the same.

Explanation:

I know this because its a clear obvious answer not only that it was one of my USA TESTPREP questions and it was right.

All you mainly have to do is the math - F=ma , In each case , the acceleration is 5 m/s squared

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31. If you threw a baseball straight out at 45 m/s from a height of 1.5 meters (A) how long would it be in the air? B) How far o
coldgirl [10]

Answer:

A) t = 0.55 s

B) x = 24.8 m

Explanation:

A) We can find the time at which the ball will be in the air using the following equation:

y_{f} = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}    

Where:

y_{f} is the final height= 0  

y_{0} is the initial height= 1.5 m

v_{0y} is the component of the initial speed in the vertical direction = 0 m/s        

t: is the time =?      

g: is the gravity = 9.81 m/s²

0 = 1.5 m - \frac{1}{2}9.81 m/s^{2}t^{2}

By solving the above equation for t we have:

t = \sqrt{\frac{2*1.5 m}{9.81 m/s^{2}}} = 0.55 s  

Hence, the ball will stay 0.55 seconds in the air.

                             

B) We can find the distance traveled by the ball as follows:

x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}

Where:  

a: is the acceleration in the horizontal direction = 0  

x_{f} is the final position =?  

x_{0} is the initial position = 0      

v_{0x} is the component of the initial speed in the horizontal direction = 45 m/s                                                                                            

x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}

x_{f} = 0 + 45 m/s*0.55 s + 0 = 24.8 m

Therefore, the ball will travel 24.8 meters.

I hope it helps you!

3 0
2 years ago
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