Is recommend attaching the answer choices; Meters, Liters, Grams are three basic ones
Answer:
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Explanation:
is obviously the correct answer they are both so different yet so alike
The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

We can apply the first Newton's law in x and y-direction.
If we do a free body diagram of the system we will have:
x-direction
All the forces acting in this direction are:
(1)
Where:
- T(1) is the tension due to the rope 1
- T(2) is the tension due to the rope 2
Here we just conclude that T(1) = T(2)
y-direction
The forces in this direction are:
(2)
Here W is the weight of the steel beam.
We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.
Knowing that T(1) = T(2) and W = mg, we have:



T(1) must be equal to 5479 N, so we have:


Therefore, the maximum angle allowed is θ = 37.01°.
You can learn more about tension here:
brainly.com/question/12797227
I hope it helps you!
The yellow star will live longer as it has less mass
Answer:

Explanation:
First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.
Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:
