Answer:
The velocity of the fish hitting the ground is , v = 45.795 m/s
Explanation:
Given data,
The mass of the fish, m = 5 kg
The height of the bird from the surface, h = 107 m
Using the III equation of motion,
v² = u² + 2gs
<em> v = √(u² + 2gs)</em>
Substituting the values,
v = √(0² + 2 x 9.8 x 107)
= 45.795 m/s
Hence, the velocity of the fish hitting the ground is, v = 45.795 m/s
Answer : 0.814 newton
Explanation:
force (magnetic) acting on the wire is given by
F= ? , I=2.2amp , B = 0.37 T
F = B i l sin (theta) = 0.37 x 2.2 x 2x 0.5 = 0.814N
<span>the arrangement of the outer planets is
</span>1. Mercury
<span>2. Venus </span>
<span>3. Earth </span>
<span>4. Mars </span>
<span>5. Jupiter </span>
<span>6. Saturn </span>
<span>7. Uranus </span>
8. Neptune
the inner most of the outer plannets is jupitor it is followed by saturn uranus and neptune
Answer:
E
= -4556.18 N/m
Explanation:
Given data
u = 3.6×10^6 m/sec
angle = 34°
distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in
Question)
from the projectile motion the horizontal distance traveled by electron is
x = u×cosA×t
⇒t = x/(u×cos A)
We also know that force in an electric field is given as
F = qE
q= charge , E= strength of electric field
By newton 2nd law of motion
ma = qE
⇒a = qE/m
Also, y = u×sinA×t - 0.5×a×t^2
⇒y = u×sinA×t - 0.5×(qE/m)×t^2
if y = 0 then
⇒t = 2mu×sinA/(qE) = x/(u×cosA)
Also, E = 2mu^2×sinA×cosA/(x×q)
Now plugging the values we get
E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})
E
= -4556.18 N/m
