Answer:
0.60 N, towards the centre of the circle
Explanation:
The tension in the string acts as centripetal force to keep the ball in uniform circular motion. So we can write:
(1)
where
T is the tension
m = 0.015 kg is the mass of the ball
is the angular speed
r = 0.50 m is the radius of the circle
We know that the period of the ball is T = 0.70 s, so we can find the angular speed:
And by substituting into (1), we find the tension in the string:
And in an uniform circular motion, the centripetal force always points towards the centre of the circle, so in this case the tension points towards the centre of the circle.
Explanation:
Answer:
(a)T= M2 × g, (b)T= (M1 + M2)g, (c)T= M2 (a + g) and (d)T=(M1 + M2) (a + g)
Explanation:
M1 is hanged upper and M2 is lower at Rest.
(a) For M2
T2 = Weight of the Body M2= M2 × g
(b) T1 = Weight of the Body M2 + Weight of the Body M2
T1 = M1 g + M2 g = (M1 + M2)g
M1 is hanged upper and M2 is lower at accelerated upwards ( F = T - W)
(c) For M2
⇒T = M2a + M2g = M2 (a + g)
(d) For M1
T = (M1 + M2) a + (M1 + M2) g
⇒ T = (M1 + M2) (a + g)
The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s.
After one half-life, 8 g of radioactive isotope will remain in the sample.
<h3>What is radioactivity?</h3>
The act of producing radiation spontaneously is known as radioactivity. This is accomplished by an unstable atomic nucleus that want to give up some energy in order to move to a more stable form.
The following formula is used to compute the number of half lives elapsed:
![\rm N=\frac{N_0}{2^n} \\\\ N=\frac{16}{2} \\\\ N= 8 \ gram](https://tex.z-dn.net/?f=%5Crm%20N%3D%5Cfrac%7BN_0%7D%7B2%5En%7D%20%5C%5C%5C%5C%20N%3D%5Cfrac%7B16%7D%7B2%7D%20%5C%5C%5C%5C%20N%3D%208%20%5C%20gram)
Hence,8 gram of radioactive isotope remains in the sample after 1 half-life.
To learn more about the radioactivity, refer to the link;
brainly.com/question/1770619
#SPJ1
Answer:
The amount of time is 31.45 sec.
Explanation:
Given that,
Voltage = 10.0 V
Resistance ![R=2.80\times10^{6}\ \Omega](https://tex.z-dn.net/?f=R%3D2.80%5Ctimes10%5E%7B6%7D%5C%20%5COmega)
Capacitance ![C=3.50\ \mu F](https://tex.z-dn.net/?f=C%3D3.50%5C%20%5Cmu%20F)
We need to calculate the time constant
Using formula of time constant
![\tau=RC](https://tex.z-dn.net/?f=%5Ctau%3DRC)
Put the value into the formula
![\tau=2.80\times10^{6}\times3.50\times10^{-6}](https://tex.z-dn.net/?f=%5Ctau%3D2.80%5Ctimes10%5E%7B6%7D%5Ctimes3.50%5Ctimes10%5E%7B-6%7D)
![\tau=9.8\ sec](https://tex.z-dn.net/?f=%5Ctau%3D9.8%5C%20sec)
The amount of time required for the current in the circuit to decay to 4.00% of its original value.
We need to calculate the amount of time
Using formula of charge
![Q=Q_{max}(1-e^{\dfrac{-t}{RC}})](https://tex.z-dn.net/?f=Q%3DQ_%7Bmax%7D%281-e%5E%7B%5Cdfrac%7B-t%7D%7BRC%7D%7D%29)
Put the value into the formula
![96\%Q_{max}=Q_{max}(1-e^{\dfrac{-t}{RC}})](https://tex.z-dn.net/?f=96%5C%25Q_%7Bmax%7D%3DQ_%7Bmax%7D%281-e%5E%7B%5Cdfrac%7B-t%7D%7BRC%7D%7D%29)
![0.96=(1-e^{\dfrac{-t}{\tau}})](https://tex.z-dn.net/?f=0.96%3D%281-e%5E%7B%5Cdfrac%7B-t%7D%7B%5Ctau%7D%7D%29)
![0.96=(1-e^{\dfrac{-t}{\tau}})](https://tex.z-dn.net/?f=0.96%3D%281-e%5E%7B%5Cdfrac%7B-t%7D%7B%5Ctau%7D%7D%29)
![0.04=e^{\dfrac{t}{\tau}}](https://tex.z-dn.net/?f=0.04%3De%5E%7B%5Cdfrac%7Bt%7D%7B%5Ctau%7D%7D)
![\dfrac{-t}{\tau}=ln(0.04)](https://tex.z-dn.net/?f=%5Cdfrac%7B-t%7D%7B%5Ctau%7D%3Dln%280.04%29)
![t=3.21\times9.8](https://tex.z-dn.net/?f=t%3D3.21%5Ctimes9.8)
![t=31.45\ sec](https://tex.z-dn.net/?f=t%3D31.45%5C%20sec)
Hence, The amount of time is 31.45 sec.