Answer:
q = 4.5 nC
Explanation:
given,
electric field of small charged object, E = 180000 N/C
distance between them, r = 1.5 cm = 0.015 m
using equation of electric field
k = 9 x 10⁹ N.m²/C²
q is the charge of the object
now,
q = 4.5 x 10⁻⁹ C
q = 4.5 nC
the charge on the object is equal to 4.5 nC
Answer:
d=0.137 m ⇒13.7 cm
Explanation:
Given data
m (Mass)=3.0 kg
α(incline) =34°
Spring Constant (force constant)=120 N/m
d (distance)=?
Solution
F=mg
F=(3.0)(9.8)
F=29.4 N
As we also know that
Force parallel to the incline=FSinα
F=29.4×Sin(34)
F=16.44 N
d(distance)=F/Spring Constant
d(distance)=16.44/120
d(distance)=0.137 m ⇒13.7 cm
Answer:
The airliner travels 1.65 km along the runway before coming to a halt.
Explanation:
Given
Resistive forces = (2.90 × 10⁵) N = 290000 N
Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg
Velocity of airliner = 75 m/s
Let the distance over moved by the airliner be equal to d
According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.
Work done by the resistive forces = (290000) × d = (290,000d) J
Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J
290000d = 478,125,000
d = (478,125,000/290,000)
d = 1648.7 m = 1.65 km
Hope this helps!!!
Answer:160.88 m
Explanation:
Given
Distance Between two person is 100
i.e. 
Let W be the width of River
From First Triangle
From Second Triangle
