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stiv31 [10]
3 years ago
10

A forcue of 186 N acts on a 7.3-kg bowling ball for 0.40 s. what is the chanhe in velocity of the bowling ball

Physics
2 answers:
baherus [9]3 years ago
3 0

As we know that Force = mass * acceleration.   
Here a force of 186 N acting on 7.3 kg ball will create acceleration = 186/7.3 = 25.48 m/s2  
Also, change in velocity can be found as, acceleration multiplied by time it is acting on object. 
Therefore, change in velocity = 24.48 * 0.4 = 10.19 m/s
motikmotik3 years ago
3 0
Given are: Force = 186 N; Mass = 7.3 Kg; time = 0.40 s Initial Velocity Vi = 0
Final Velocity Vf = ?
Unknown: Change in velocity or acceleration 
F =ma;  but a = Vf - Vi/t  Solve for Vf or Final velocity
therefor  F = m(Vf-Vi/t) Ft = m(Vf - Vi/t)   But Vi = 0
Vf = (186 N)(0.40 S)/7.3 Kg  answer Vf = 10.2 m/s
Solving for acceleration or change in velocity
a = Vf - Vi/t
a = 10.2 m/s - 0/0.4s  therefore change in velocity a = 25.5 m/s²



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Batman (95kg) is standing on top of a 50m high building looking out over the city of Gotham. Given that he uses the potential en
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Answer:

47 kJoules (kJ)

Explanation:

Potential enegy on Earth is given by the relationship:

P.E. = mgh, where m is mass, g is the acceleration due to Earth's gravity, and h is height. Since we are given metric values, we will look for an answer that is consistent with Joules, the metric measure of energy. 1 Joule is defined as 1 kg*m^2/s^2, so we wnat units of kg, m, and sec.

We are given:

m = 95kg

h = 50 meters

Earth's gravity, g is 9.8 m/s^2

Enter the data:

P.E. = mgh

P.E. = (95kg)(9.8m/s^2)(50m)

P.E. = 46550 kg*m^2/s^2 or 46550 Joules(J)

Since we only have 2 sig figs, and since 1kJ =- 1000J

We can state the potential energy is 47kJ.

Spiderman has 47kJ of potential energy for the start of any dive back to Earth. [He needed that same amount of energy to reach that height, but we don't know from where it came. A jump, helicopter, beamed up by Scotty, or tossed up by Doctor Octopus.]

3 0
1 year ago
A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

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t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

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Answer:

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Answer

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 \tau = 3.63 \times 10^{-3} N.m

8 0
2 years ago
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