Radioactive decay is given by:
N = No x e^(-λt)
We know that N/No has to be 0.05
λ = 0.15
0.05 = e^(-0.15t)
t = ln(0.05)/(-0.15)
t = 19.97 days
The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.
- The force per unit length between two parallel thin current-carrying
and
wires at distance ' r ' is given by
....(1) .
- If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.
A schematic of the information provided in the question can be seen in the image attached below.
From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

Using equation (1) , we get

I₃ = 2.4 A and the current is pointing in the downward direction
Learn more about the magnitude and direction of forces here:
brainly.com/question/14879801?referrer=searchResults
#SPJ4
Answer
acceleration due to gravity on Jupiter's moon,g' = 1.81 m/s²
weight of water melon on earth, W = 40 N
acceleration due to gravity on earth, g = 9.8 m/s²
a) Mass on the earth surface
M = 4.08 Kg
b) Mass on the surface of Lo
Mass of an object remain same.
Hence, mass of object at the surface of Lo = 4.08 Kg.
c) Weight at the surface of Lo
W' = m g'
W' =4.08 x 1.81
W' = 7.38 N
We have that the Number of stitches per sec and he mass of oscillation motion is mathematically given as
a) Nt=25stitches per sec
b) m=2.033e-5kg
<h3>
Number of
stitches per sec and he mass of oscillation motion</h3>
Question Parameters:
This <u>sewing </u>machine is capable of stitching 1,500 stiches in one minute.
If the <em>sewing </em>machine has a spring constant of 0.5 N/m,
Generally the equation for the Number of stitches per sec is mathematically given as
Nt=N/t
Therefore
Nt=1500/60
Nt=25stitches per sec
b)
Generally the equation for the Time t is mathematically given as

Therefore

m=2.033e-5kg
For more information on Mass visit
brainly.com/question/15959704
If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Given the data in the question;
- Length of the massless beam;

- Distance of support from the left end;

- First mass;

- Distance of beam from the left end( m₁ is attached to );

- Second mass;

- Distance of beam from the right of the support( m₂ is attached to );

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.
Hence, 
we divide both sides by 

Next, we make
, the subject of the formula
![x_1 = x - [ \frac{m_2x_2}{m_1} ]](https://tex.z-dn.net/?f=x_1%20%3D%20x%20-%20%5B%20%5Cfrac%7Bm_2x_2%7D%7Bm_1%7D%20%5D)
We substitute in our given values
![x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]](https://tex.z-dn.net/?f=x_1%20%3D%203.00m%20-%20%5B%20%5Cfrac%7B61.7kg%5C%20%2A%20%5C%200.273m%7D%7B31.3kg%7D%20%5D)


Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Learn more; brainly.com/question/3882839