A typical amplitude for the ripples formed when a pebble is dropped in a
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Answer: 2.13 × 10⁻⁷ N
Explanation:
Gravitational force exists between any two bodies having mass.
Force of gravity is given by:
It is given that, mass of newborn baby is M = 2.50 kg
Mass of the doctor, m = 80.0 kg
Distance between the two, r = 0.250 m
Gravitational constant, G = 6.67 × 10⁻¹¹ N m²/kg²
⇒F = (6.67 × 10⁻¹¹ N m²/kg² × 2.50 kg × 80.0 kg )÷ (0.250 m)² = 2.13 × 10⁻⁷ N
Thus, the force of gravity between new born baby and doctor is 2.13 × 10⁻⁷ N.
Answer:
0.217 m/s
Explanation:
The protons in the beam passes undeflected when the electric force is equal to the magnetic force:
qE = qvB
where
q is the proton's charge
E is the magnitude of the electric field
v is the speed of the protons
B is the magnitude of the magnetic field
Re-arranging the equation,
And by substituting
E = 0.5 N/C
B = 2.3 T
We find