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Dominik [7]
2 years ago
13

if 130N centripetal force is needed to keep a 0.45kg ball that is attached to a string that is 0.7m long to complete 5 full rota

tions every 10 seconds. what is the centripetal acceleration and velocity felt by the ball?
Physics
1 answer:
andrezito [222]2 years ago
4 0

Answer:

centripetal acceleration of the ball is 6.9 m/s/s

tangential speed of the ball is 2.2 m/s

Explanation:

As we know that ball complete 5 rotations in 10 seconds

so frequency of rotation of ball is given as

f = \frac{5}{10} = 0.5 Hz

now we know that angular frequency is given as

\omega = 2\pi f

\omega = 2(\pi)(0.5)

\omega = \pi rad/s

Now centripetal acceleration is given as

a_c = \omega^2 R

a_c = \pi^2 (0.7)

a_c = 6.9 m/s^2

now the velocity of the ball at this angular frequency is given as

v = R\omega

v = 0.7 (\pi)

v = 2.2 m/s

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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    

Finally, we can calculate the time from equation (1):

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!

4 0
2 years ago
Why is the Mid-Atlantic Ridge an ideal place for SWARM to collect electrical conductivity data?
ElenaW [278]

Answer: Mid-ocean ridges are geologically important because they occur along the kind of plate boundary where new ocean floor is created as the plates spread apart. Thus the mid-ocean ridge is also known as a "spreading center" or a "divergent plate boundary." The plates spread apart at rates of 1 cm to 20 cm per year.

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2 years ago
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Lightwaves travel from the air into a lens made of glass. Their velocity decreases as they enter the glass. How does this affect
Papessa [141]
The waves become longer but slower
4 0
3 years ago
a runner starts from rest and has an acceleration of 3 m/s^2. How fast is she running after 1.1 seconds
grigory [225]
Acceleration x time = velocity 

Since you're given acceleration and time, just plug the values into the equation. 

3\frac{m}{s^{2}} x 1.1 s = ? 

Solve that equation, and remember your velocity should be in m/s.
8 0
2 years ago
In a nuclear fusion reaction, the mass of the products is _____ the mass of the reactants.
JulijaS [17]

Answer:

more than

Explanation:

In a nuclear fusion reaction, the mass of the products is more than the mass of the reactants.

7 0
2 years ago
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