Answer:
35.35 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 20 m/s
Angle of projection (θ) = 30°
Acceleration due to gravity (g) = 9.8 m/s²
Range (R) =.?
The range (i.e how far away) of the ball can be obtained as follow:
R = u² Sine 2θ /g
R = 20² Sine (2×30) / 9.8
R = 400 Sine 60 / 9.8
R = (400 × 0866) / 9.8
R = 346.4 / 9.8
R = 35.35 m
Therefore, the range (i.e how far away) of the ball is 35.35 m
Answer:
a.96cm
b.77cm
Explanation:
Highest intensity,
=19cm
Lowest intensity,
=29cm
48cm is the distance for half-wave, wavelength =48cm*2=96cm
b. From a, above we know that half-wavelength is 48cm
Also, the distance of known maximum is 29cm
Next maximum is calculated as:

The next maximum will occur at 77cm
<h2>The emf produced is 7.2 V</h2>
Explanation:
When coil is placed in the magnetic field , the flux attached with it can be found by the relation . Flux Ф = the dot product of magnetic field and area of coil .
Thus Ф = B A cosθ
here B is magnetic field strength and A is the area of coil .
The angle θ is the angle between coil and field direction .
When coil rotates , the angle varies . By which the flux varies . The emf is produced in coil due to variation of flux . The relation for this is
The emf produced ξ = -
= B A sinθ 
Now in the given problem
5 = 0.38 x A x
I
Now if the magnetic field is 0.55 T and all the other terms are same , the emf produced
ξ = 0.55 x A x
Ii
dividing II by I , we have
=
= 1.45
or ξ = 7.2 V