Answer:
A velocity of 
Explanation:
Since the magnitude of the vector is equal to the magnitude of velocity, velocity of the 2 cm vector represents a velocity of $10\times 2= 20 \: km/h$.
If l and m both are doubled then the period becomes √2*T
what is a simple pendulum?
It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.
A pendulum is a weight suspended from a pivot so that it can swing freely.
Here,
A certain frictionless simple pendulum having a length l and mass m
mass of pendulum = m
length of the pendulum = l
The period of simple pendulum is:

Where k is the constant.
Now the length and mass are doubled,
m' = 2m
l' = 2l



Hence,
If l and m both are doubled then the period becomes √2*T
Learn more about Simple Harmonic Motion here:
<u>brainly.com/question/17315536</u>
#SPJ4
Answer:
a

b

Explanation:
From the question we are told that
The speed of the spaceship is 
Here c is the speed of light with value 
The length is 
The distance of the star for earth is 
The speed is 
Generally the from the length contraction equation we have that
![l = l_o \sqrt{1 -[\frac{v}{c } ]}](https://tex.z-dn.net/?f=l%20%20%3D%20%20l_o%20%20%5Csqrt%7B1%20-%5B%5Cfrac%7Bv%7D%7Bc%20%7D%20%5D%7D)
Now the when at rest the length is 
So



Considering b
Applying above equation
![l =l_o \sqrt{1 - [\frac{v}{c } ]}](https://tex.z-dn.net/?f=l%20%20%3Dl_o%20%5Csqrt%7B1%20-%20%20%5B%5Cfrac%7Bv%7D%7Bc%20%7D%20%5D%7D)
Here 
So



Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:

ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Finally, you obtain for E:

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C