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vredina [299]
2 years ago
7

A child on a sled has a combined mass of 54 kg. At the top of a 2.7 meter hill, the sled has a

Physics
1 answer:
leonid [27]2 years ago
5 0

Answer:

8.78?m/s

Explanation:

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Chemical equations should be balanced so that they demonstrate the law of conservation of mass. Which of the following statement
faust18 [17]

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The products must contain the same numbers and types of atoms

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2 years ago
What type of friction occurs when you are trying to move an object, but the object isnt moving?
Tasya [4]

the answer is static friction


7 0
3 years ago
A ball dropped from a bridge takes three seconds to reach the water below how far is the bridge above the water?
tatuchka [14]

<u>Given that:</u>

Ball dropped from a bridge at the rate of 3 seconds

Determine the height of fall (S) = ?

      As we know that, S = ut + 1/2 ×a.t²

                          u =initial velocity = 0

                          a= g =9.81 m/s  (since free fall)

                            S = 0+ 1/2 × 9.81 × 3²

                          <em> S = 44.145 m</em>

<em>44.145 m far is the bridge from water</em>

6 0
3 years ago
What are alkaline earth metals used for?
Tomtit [17]

Answer:

C firework

Explanation:

from Quizlet

6 0
3 years ago
Read 2 more answers
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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