Answer:
The transverse component of acceleration is 26.32
where as radial the component of acceleration is 8.77 
Explanation:
As per the given data
u=π/4 rad
ω=u'=2 rad/s
α=u''=4 rad/s

So the transverse component of acceleration are given as

Here


So

The transverse component of acceleration is 26.32 
The radial component is given as

Here

So

The radial component of acceleration is 8.77 
Answer:
D= 1999.2 m
Explanation:
Given that
Average velocity ,v= 0.98 m/s
time ,t= 34 min
We know that
1 min = 60 s
That is why
t= 34 x 60 =2040 s
We know that
Displacement = Average velocity x time
D= v t
Now by putting the values in the above equation
D= 0.98 x 2040 m
D= 1999.2 m (eastward)
The direction of the displacement will be towards eastward.
That is why the displacement will be 1999.2 m or we can say that 1.9992 km.
Answer:
v_squid = - 2,286 m / s
Explanation:
This exercise can be solved using conservation of the moment, the system is made up of the squid plus the water inside, therefore the force to expel the water is an internal force and the moment is conserved.
Initial moment. Before expelling the water
p₀ = 0
the squid is at rest
Final moment. After expelling the water
= M V_squid + m v_water
p₀ = p_{f}
0 = M V_squid + m v_water
c_squid = -m v_water / M
The mass of the squid without water is
M = 9 -2 = 7 kg
let's calculate
v_squid = 2 8/7
v_squid = - 2,286 m / s
The negative sign indicates that the squid is moving in the opposite direction of the water
Answer:
Electric current produces a magnetic field. This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. ... Magnetic Field Generated by Current: (a) Compasses placed near a long straight current-carrying wire indicate that field lines form circular loops centered on the wire.
Explanation:
To solve this problem we will use the concepts related to hydrostatic pressure. Which determines the pressure of a body at a given depth of a liquid.
Mathematically this can be described as

Here
= Density
g = Gravity
h = Height (Depth)
If we replace the values given in the equation we will have to


Therefore the pressure at the bottom will be 9.8kPa