Answer:
The velocity of other mass is 3.60 m/s.
Explanation:
Given that,
Mass of first block = 8 kg
mass of second block = 4.3 kg
Speed = 6.7 m/s
We need to calculate the speed of first mass
Using conservation of momentum

where, m₁ =mass of first block
m₂ =mass of second block
m₁ =mass of first block
v₂ =speed of second block
Put the value into the formula



Negative sign represent the opposite direction of initial value.
Hence, The velocity of other mass is 3.60 m/s.
Answer:
The magnitude of the field is 8.384×10^-4 T.
Explanation:
Now, i start solving this question:
First, convert the potential difference(V) 2 kv to 2000 v.
As, we have the final formula is qvB = mv^2/r. It came from the centripetal force and the magnetic force and we know that these two forces are equal. When dealing with centripetal motion use the radius and not the diameter so
r = 0.36/2 = 0.18 m.
As, we are dealing with an electron so we know its mass is 9.11*10^-31 kg and its charge (q) is 1.6*10^-18 C.
We can solve for its electric potential energy by using ΔU = qV and we know potential energy initial is equal to kinetic energy final so ΔU = ΔKE and kinetic energy is equal to 1/2mv^2 J.
qV = 1/2mv^2
(1.6*10^-19C)(2000V) = (1/2)(9.11*10^-31kg) v^2
v = 2.65×10^7 m/s.
These all above steps we have done only for velocity(v) because in the final formula we have 'v' in it. So, now we substitute the all values in that formula and will find out the magnitude of the field:
qvB = mv^2/r
qB = mv/r
B = mv/qr
B = (9.11*10^-31 kg)(2.65×10^7 m/s) / (1.6*10^-19 C)(0.18 m)
Hence, B = 8.384*10^-4 T.
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

Substitute numerical values:

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
brainly.com/question/2004529
#SPJ4
Hello,
The exosphere is the lowest<span> in density of the </span>layers of the atmosphere.
The higher up you get into the atmosphere, the lower the density becomes. Thus, the troposphere has the highest density of the five layers.
Faith xoxo
Answer: A: wear several thin layers and a waterproof top layer
Explanation: