Answer:
Magnetic force, F = 0.24 N
Explanation:
It is given that,
Current flowing in the wire, I = 4 A
Length of the wire, L = 20 cm = 0.2 m
Magnetic field, B = 0.6 T
Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :


F = 0.24 N
So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.
Answer:

Explanation:
<u>Friction Force</u>
When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.
There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.
Please find the free body diagrams in the figure provided below.
The equilibrium condition for the mass 1 is

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa
![\displaystyle F_a=F_{r1}+F_{r2}.....[1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_a%3DF_%7Br1%7D%2BF_%7Br2%7D.....%5B1%5D)
The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

The friction forces are computed by


Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.
Replacing in [1]

Simplifying

Plugging in the values
![\displaystyle F_{a}=0.25(9.8)[400+2(100)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_%7Ba%7D%3D0.25%289.8%29%5B400%2B2%28100%29%5D)

Answer:
Q1: 3.2km
Q2: 4.8K
Explanation:
Q1:
So db is the distance of bird, and dr is the distance of runner
db = 2vr and the distance of bird is going to be 2 times greater than the runner.
formulas: db = 2vr & db = 2dr
- db = 2dr
- L + (L - x) = 2x
- 2L - x = 2x
- 2L = 3x
- x =
L
Insert it in x =
L
(2.4km) = 1.6km
Now we use formula db = 2dr
- db = 2L - x
- db = 2(2.4km) - 1.6km
- <u>db = 3.2km</u>
Q2:
Formulas: Vr = L /Δt & Vb = db/Δt
- Vr = L/ Δt ⇒ Δt =



(Km cancel each other)
- Vb = db/Δt ⇒ db = VbΔt
- 13.6km/hr

- <u>4.8km</u>
(hr cancel each other)
Hope it helps you :)
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
The acceleration is given as:
a = g sin(30°) where g is the gravitational acceleration
For g = 10 m/s^2, we get
a = 10 sin(30°) = 10 * 1/2 = 5 m/s^2