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podryga [215]
3 years ago
15

Find the 24th term of the arithmetic sequence 11, 14, 17, … . Express the 24 terms of the series of this sequence using sigma no

tation. Find the sum of the first 24 terms of the series.
Physics
1 answer:
Anna71 [15]3 years ago
5 0

Answer:

The 24th term is 80 and the sum of 24 terms is 1092.

Explanation:

Given that,

The arithmetic series is

11,14,17,........24

First term a = 11

Difference d = 14-11=3

We need to calculate the 24th term of the arithmetic sequence

Using formula of number of terms

t_{n}=a+(n-1)d

Put the value into the formula

t_{24}=11+(24-1)\times3

t_{24}=80

t_{24}=u_{24}=80

We need to calculate the sum of the first 24 terms of the series

Using formula of sum,

S_{n}=\dfrac{n}{2}(a+u_{24})

Put the value into the formula

S_{n}=\dfrac{24}{2}\times(11+80)

S_{n}=1092

Hence, The 24th term is 80 and the sum of 24 terms is 1092.

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Read 2 more answers
A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

with this values of a you can use the following formula:

a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0
2 years ago
A vehicle that goes from 5m/s to 45m/s in 8s. what is its acceleration?
GaryK [48]

Answer: 5m/s^2

Explanation:

V= 45m/s

U = 5m/s

t = 8s

a =?

V = u + at

45 = 5 + 8a

8a = 45 — 5

8a = 40

a = 40 / 8

a = 5m/s^2

3 0
3 years ago
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