Answer:
its bc of the way earth spins
Explanation:
Answer:
313.6 m downward
Explanation:
The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.
In fact, we have:
where
y(t) is the vertical position of the projectile at time t
h is the initial height of the projectile
is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally
t is the time
a = g = -9.8 m/s^2 is the acceleration due to gravity
We can rewrite the equation as
where the term on the left, 
, represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we find
So the bullet has travelled 313.6 m downward.
The magnitude of the magnetic field on the axis of the ring 5 cm from its center is 143 pT.
The radius of the nonconducting ring is R = 10 cm.
The ring is uniformly charged q = 10 μC.
The angular speed of the ring, ω = 20 rad/s
The ring is x = 5 cm from the center of the ring.
Now,
R = 10 cm = 0.1 m
q = 10.0 μC = 10 × 10⁻⁶ C
x = 5 cm = 0.05 m
The magnetic field on the axis of a current loop is given as:
B = [ μ₀ IR² ] / [4π(x² + R²)^{3/2} ]
Now, I = q / [2π/ω]
So, the magnitude of the magnetic field which is directed away from the center is:
B = [ μ₀ ωqR² ] / [4π(x² + R²)^{3/2} ]
B = [ μ₀ (200) (10 × 10⁻⁶) (0.1)² ] / [4π((0.05)² + (0.1)²)^{3/2} ]
B = 1.43 × 10⁻¹⁰ T
B = 143 pT
Learn more about the magnetic field here:
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Answer:
740, mm of Hg
Explanation:
The pressure of argon , in mm of Hg = difference in the level of mercury on either side of the manometer.
mercury column in the open end of the manometer is 22 mm below that in the side connected to the argon and 762 mmHg, end is open to atmospheric pressure.
therefore, The pressure of argon , in mm of Hg =762 -22 = 740, mm of Hg
