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3 years ago

Describe all the ways that newtons laws can apply in a car crash

1 answer:

7 0

Newtons first law - Objects in the car at rest (The human) will remain at rest unless affected by an unbalanced force. Well the unbalanced force would be the crash and this would set the human in motion and they would ether fly out the car if not wearing a seat belt or if wearing one they would get bad whip lash

Newtons second law - With more mass requires more force, so since the human is pretty light or even if heavy in a big crash there will be so much more from it that this will send the human flying.

Newtons 3rd law - Objects A puts force onto objects b and object b excretes the same amount of force back onto object a, so in a crash the human would hit the car hard and the car would excrete the same amount of force back on the human which would really damage him/her

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What does MA mean for you in real life?

andrezito [222]

To me "Ma" mean in real life is either mama or Massachusttes.

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3 years ago

A ball is thrown straight upward and rises to a maximum

Leviafan [203]

Answer:

Explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

$v_f^2 - v_i^2 = 2 a d$

here we know that

$v_f = 0$

also we have

$a = -9.81 m/s^2$

so we have

$0 - v_i^2 = 2(-9.81)(16)$

$v_i = 17.72 m/s$

Now we need to find the height where its speed becomes half of initial value

so we have

$v_f = 0.5 v_i$

now we have

$v_f^2 - v_i^2 = 2 a d$

$(0.5v_i)^2 - v_i^2 = 2(-9.81)h$

$-0.75v_i^2 = -19.62 h$

$0.75(17.72)^2 = 19.62 h$

$h = 12 m$

3 0

3 years ago

A boat with a weight of 547 newtons is floating in a harbor. What is the buoyant force on the boat?

Veronika [31]

Answer:

Explanation:

its correct

3 0

3 years ago

A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th

Evgen [1.6K]

Answer:

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Explanation:

Given that

$\dfrac{dx}{dt}= -1\ m/s$

From the diagram

$tan\theta=\dfrac{21}{x}$

By differentiating with time t

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

When x= 10 m

$tan\theta=\dfrac{21}{10}$

θ = 64.53°

Now by putting the value in equation

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

$sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)$

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Therefore rate of change in the angle is 0.038\ rad/s

8 0

3 years ago

A object was fired at 17ms at 40 degree angle horizontal , if object traveled 29meters how long was it in the air

Oksi-84 [34.3K]

First you need to find the initial x component which is 13.02. After that you plug it into the 3rd formula on the AP physics formula sheet. Then rearrange it into the quadratic formula and solve for time. The answer should be 4.1 seconds.

8 0

2 years ago