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devlian [24]
3 years ago
5

A golf ball is released from rest from the top of a very tall building. Choose a coordinate system whose origin is at the starti

ng point of the ball, and whose y axis points vertically upward. Neglecting air resistance, calculate the ve- locity of the ball after 3.04 s. The acceleration of gravity is 9.8 m/s2 . Answer in units of m/s.
Physics
2 answers:
Alinara [238K]3 years ago
7 0

Answer: v=29.792m.s^{-1}

Explanation:

<u>Here given that:</u>

  • initial velocity of the golf ball, u =0m.s^{-1}
  • a coordinate system pointing upwards with origin at the point of release.
  • time, t = 3.04 seconds
  • acceleration due to gravity, g = 9.8 m.s^{-2}

Now, we need to find the velocity (v) after the time 3.04 s.

From the given and required data we choose the suitable equation of motion: v = u+at

Putting the given values in the above equation:

v= 0 - gt ∵the given frame of reference points in the upward directoin so we have negative on the downward side.

v=0-9.8\times 3.04

v=29.792m.s^{-1}

Zigmanuir [339]3 years ago
5 0

Answer:

Velocity of the ball after 3.04 (s) = 29.79 (m/s)

Explanation:

From the free fall movement we have the following formulas: Vf^{2} = Vo^{2} - 2gh and h=Vo*t - \frac{g*t^{2} }{2}, First we need to find the height to time iqual to 3.04 s using the formula: h=Vo*t - \frac{g*t^{2} }{2} and remember that golf ball was released from the rest (Vo= 0 (m/s)) so h= (0 (m/s))*(3.04 (s)) - \frac{9.8 (m/s^2)*(3.04 (s))^{2} }{2}, we get: h = -45.28 (m) with the height that we have got, now the velocity of the ball is calculate using Vf^{2} = Vo^{2} - 2gh solving for Vf, we get: Vf = \sqrt{Vo^{2}-2*g*h } replacing the values given Vf = \sqrt{(0 m/s)^{2}-2*(9.8 m/s^2)*(-45.28 m) }, so we get: Vf = 29.79 (m/s).

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[dV/dt] = [dV/dr] * [dr/dt]

[dV/dt] = [4π r^2] * [ dr/ dt]

r = 60 mm, [dr / dt] = 4 mm/s

[dV / dt ] = [4π(60mm)^2] * 4mm/s = 180,955.7 mm/s 


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The wall in this case provides the opposite reactive force.

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39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
3 years ago
5N to the left and 20 N to the right. If the couch has a mass of 45 Kg determine its acceleration
Alexus [3.1K]

Answer:

D) 0.33 m/s² to the right

Explanation:

Apply Newton's second law.  Take right to be positive and left to be negative.

∑F = ma

20 N − 5 N = (45 kg) a

a = 0.33 m/s²

The couch accelerates at 0.33 m/s² to the right.

5 0
3 years ago
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