Answer:
b. 4.9 × 1011
Step-by-step explanation:
Using scientific notation is similar to expressing in standard form. Given that (7 × 105)2
We open the parenthesis. This may first be expressed as
7² × 10⁽⁵⁾²
Then expand,
= 49 × 10¹⁰
To put in scientific notation, 49 = 4.9 × 10
Hence the expression becomes
= 4.9 × 10 × 10¹⁰
Using the laws of indices
= 4.9 × 10¹¹ in scientific notation
Answer:
7/3 ft
Step-by-step explanation:
Given data
Lenght of fence= 7 miles
Amount completed= 2/3 of the total length
let us find the amount completed already = 2/3*7= 14/3 miles
Hence the length of fence remaining for completion is
=7- 14/3
=21-14/3
=7/3
Hence the amount in length undon is 7/3 ft
Answer:
20 seconds
Step-by-step explanation:
I assume you mean 
for your function.
As the given function is in the form of 
(some textbooks may write this formula differently), the period is equal to 
. Hence, the period of the given function is 
, or 20 seconds.
Answer:
r=294.9m
Step-by-step explanation:
The forces on the particle are
Now , we sum all these forces to get the net force
we can use the fact F=m*a and integrate the acceleration
![a(t)=\frac{1}{m}F(t)\\\\v(t)=\int a(t)dt=\frac{1}{m}\int{F_{T}}dt\\\\v(t)=\frac{1}{m}[(57t-t^{2}+\frac{5}{3}t^{3})\hat{i}+(12t-2t^{2})\hat{j}+(-t^{2}-t)\hat{k}]\\\\r(t)=\int v(t)dt=\frac{1}{m}[(\frac{57}{2}t^{2}-\frac{1}{3}t^{3}}+\frac{5}{4}t^{4})\hat{i}+(6t^{2}-\frac{2}{3}t^{3})\hat{j}+(-\frac{1}{3}t^{3}-\frac{1}{2}t^{2})]](https://tex.z-dn.net/?f=a%28t%29%3D%5Cfrac%7B1%7D%7Bm%7DF%28t%29%5C%5C%5C%5Cv%28t%29%3D%5Cint%20a%28t%29dt%3D%5Cfrac%7B1%7D%7Bm%7D%5Cint%7BF_%7BT%7D%7Ddt%5C%5C%5C%5Cv%28t%29%3D%5Cfrac%7B1%7D%7Bm%7D%5B%2857t-t%5E%7B2%7D%2B%5Cfrac%7B5%7D%7B3%7Dt%5E%7B3%7D%29%5Chat%7Bi%7D%2B%2812t-2t%5E%7B2%7D%29%5Chat%7Bj%7D%2B%28-t%5E%7B2%7D-t%29%5Chat%7Bk%7D%5D%5C%5C%5C%5Cr%28t%29%3D%5Cint%20v%28t%29dt%3D%5Cfrac%7B1%7D%7Bm%7D%5B%28%5Cfrac%7B57%7D%7B2%7Dt%5E%7B2%7D-%5Cfrac%7B1%7D%7B3%7Dt%5E%7B3%7D%7D%2B%5Cfrac%7B5%7D%7B4%7Dt%5E%7B4%7D%29%5Chat%7Bi%7D%2B%286t%5E%7B2%7D-%5Cfrac%7B2%7D%7B3%7Dt%5E%7B3%7D%29%5Chat%7Bj%7D%2B%28-%5Cfrac%7B1%7D%7B3%7Dt%5E%7B3%7D-%5Cfrac%7B1%7D%7B2%7Dt%5E%7B2%7D%29%5D)
and we evaluate in r(2) an we take the norm to obtain the distance
![r(2)=\frac{1}{m}[\frac{394}{3}\hat{i}+\frac{56}{3}\hat{j}-\frac{14}{3}\hat{k}]\\|r(2)|=\frac{1}{m}\sqrt{[(\frac{394}{3})^{2}+(\frac{56}{3})^{2}+(\frac{14}{3})^{2}]}\\|r(2)|=\frac{132.73}{0.45}=294.9m](https://tex.z-dn.net/?f=r%282%29%3D%5Cfrac%7B1%7D%7Bm%7D%5B%5Cfrac%7B394%7D%7B3%7D%5Chat%7Bi%7D%2B%5Cfrac%7B56%7D%7B3%7D%5Chat%7Bj%7D-%5Cfrac%7B14%7D%7B3%7D%5Chat%7Bk%7D%5D%5C%5C%7Cr%282%29%7C%3D%5Cfrac%7B1%7D%7Bm%7D%5Csqrt%7B%5B%28%5Cfrac%7B394%7D%7B3%7D%29%5E%7B2%7D%2B%28%5Cfrac%7B56%7D%7B3%7D%29%5E%7B2%7D%2B%28%5Cfrac%7B14%7D%7B3%7D%29%5E%7B2%7D%5D%7D%5C%5C%7Cr%282%29%7C%3D%5Cfrac%7B132.73%7D%7B0.45%7D%3D294.9m)
I hope this is useful for you
regards
