the answer is a unbalanced force is acting on the brick
Answer:
v₁ / v₂ = √μ
₂/μ
₁
Explanation:
The speed of a traveling wave on a string is given by
v = √ T /μ
Where v is the speed of the wave, T the tension and μ is the linear density of the string
Let's apply this equation to our case
v₁ = √ T / μ
₁
v2 = √ T /μ
₂
If the voltage of the two systems is the same
T = v₁² μ₁ = v₂² μ₂
We cleared
v₁² / v₂² = μ₂ / μ₁
v₁ / v₂ = √μ
₂/μ
₁
The first right-hand rule determines the directions of magnetic force, conventional current and the magnetic field. Given any two of theses, the third can be found.
The second Right-Hand Rule determines the direction of the magnetic field around a current-carrying wire and vice-versa<span> </span>
So, assuming that a magnetic field <span>exists and its direction is known and assuming that a charged particle moves in a specific direction through that field with velocity (v(, to determine the direction of force on the particle we should use the second right-hand rule.</span>
Answer:
a) x = 40 t
, y = 39 t
, z = 6 + 32 t - 16 t
², b) x = 80 feet
, y = 78 feet
, the ball came into the field
Explanation:
a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis
Since the cast is in the center of the field, let's place the coordinate system
x₀ = 0
y₀ = 0
z₀ = 6 feet
x-axis (towards end zone, GOAL zone)
x = xo + v₀ₓ t
x = 40 t
y-axis (field width)
y = y₀ +
t
y = 39 t
z axis (vertical)
z = z₀ + v_{oz} t - ½ g t²
z = 6 + 32 t - ½ 32 t²
z = 6 + 32 t - 16 t
²
b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive
z = 6
6 = 6 + 32 t - 16 t²
(t - 2)t = 0
t=0 s
t= 2 s
The ball position
x = 40 2
x = 80 feet
y = 39 2
y = 78 feet
the dimensions of the field from the coordinate system (center of the field) are
x_total = 150 feet
y _total = 80 feet
so we can see that the ball came into the field
Velocity of the sled is 3.2 m/s