Answer:
gravitational waves are ripples in spece-time caused primarily when objects are accelerated and the energy for the acceleration is transpoted as gravitational radiation.
they are difficult to detect because they require very sensitive technology or you will have to wait unitl black holes collide.
Current in the wire = 2 A
Explanation:
the magnetic field is given by
B= \frac{\mu i}{2\pi r}
μo= 4π x 10⁻⁷ Tm/A
i= current
r=0.02 m
B = magnetic field= 2 x 10⁻⁵ T
2 x 10⁻⁵= (4π x 10⁻⁷)(i) / (2π*0.02)
i=2 A
If you move a magnet through a loop of wire, induction will happen. The more loops you make, the stronger the effect becomes.
Answer:
drag force 
Velocity (V) = 40.169 m/s
Explanation:
The drag force
is given by the formula:

where:
= drag coefficient depending on the Reynolds number
Reynolds number Re = 
Let's Assume that the air is in room temperature at 25 °C ; Then
density of the air
= 1.1845 kg/m³
viscosity of fluid or air
= 1.844 × 10⁻⁵ kg/ms
diameter of the baseball D = 7.4 cm
Velocity V = 44.3 m/s
Replacing them into the equation of Reynolds number ; we have :

A = Projected Area
From the diagram attached below which is gotten from NASA for baseball;
the drag coefficient which depends on Reynolds number is read as:
= 0.3
Projected Area A = 
A = 
A = 0.0043 m²
Finally, drag force is then calculated as ;

b)

since acceleration a = 
Then;

Also;
velocity (V) = 
Then;



Then;


Integrating the above equation ; we have :

when time (t) = 0 ; then 
44.3 = - 10 × 0 + C
C = 44.3

Time (t) =

∴ Velocity ; 

Velocity (V) = 40.169 m/s