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never [62]
2 years ago
7

17.food that is both nutrient- and calorie-dense can be a good choice in what circumstance?

Physics
1 answer:
Arlecino [84]2 years ago
6 0

Answer:

Explanation: you can choose B or C but i would choose C

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Pushing a door closed is an example of force. true or false​
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True

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How many additional valence electrons does fluorine need to have a full
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D= Seven

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How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?
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There is too much information given, it's hard to understand exactly which variables are important in this problem. 
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A wave amplitude 0.36m interferes with a second wave of amplitude 0.22m traveling in the same direction. What is the largest res
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3 years ago
A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin
Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

3 0
3 years ago
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