Answer:
The tension is 
Explanation:
From the question we are told that
The total mass is 
The radius is 
The density of air is 
Generally the upward force acting on the balloon is mathematically represented as

=> 
=> 
Here V is the volume of the spherical helium filled balloon which is mathematically represented as

=> 
=> 
So


That would be a frequency of 1.2666... beats per second. This can be phrased as your heart beats at 1.27 Hz.
Answer:A
Explanation:
In R-L circuit current is given by
![i=i_0\left [ 1-e^{\frac{-t}{L/R}}\right ]](https://tex.z-dn.net/?f=i%3Di_0%5Cleft%20%5B%201-e%5E%7B%5Cfrac%7B-t%7D%7BL%2FR%7D%7D%5Cright%20%5D)
where i=current at any time t

R=resistance
L=Inductance
at t=0
approaches to 1
therefore ![i=i_0\left [ 1-1\right ]](https://tex.z-dn.net/?f=i%3Di_0%5Cleft%20%5B%201-1%5Cright%20%5D)
i=0
when t approaches to
,
approaches to zero
thus 
thus we can say that initially circuit act as broken wire with zero current
and it increases exponentially with time and act as ordinary connecting wire
Answer:
39.81 N
Explanation:
I attached an image of the free body diagrams I drew of crate #1 and #2.
Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.
∑Fₓ = maₓ
∑Fᵧ = maᵧ
Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:
- ∑Fᵧ = maᵧ
- T₁ - m₂g = m₂aᵧ
Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.
Let's solve for T in the equation...
- T₁ = m₂aᵧ + m₂g
- T₁ = m₂(a + g)
We'll come back to this equation later. Now let's go to the free body diagram for crate #1.
We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.
The normal force is equal to the x-component of the force of gravity.
- (F_n · μ_k) - m₁g sinΘ = m₁aₓ
- (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
- [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
- [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
- [2.539595871] - [-58.0962595] = 6aₓ
- 60.63585537 = 6aₓ
- aₓ = 10.1059759 m/s²
Now let's go back to this equation:
We have 3 known variables and we can solve for the tension force.
- T = 2(10.1059759 + 9.8)
- T = 2(19.9059759)
- T = 39.8119518 N
The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.
A horizontal line means the object is. not changing its position it is not moving, it is at rest.