The electric force between two charge objects is calculated through the Coulomb's law.
F = kq₁q₂/d²
The value of k is 9.0 x 10^9 Nm²/C² and the charge of proton is 1.602 x10^-19 C. Substituting the known values from the given,
2.30x10^-26 = (9.0 x 10^9 Nm²/C²)(1.602 x10^-19C)²/d²
The value of d is equal to 0.10 m.
Answer:
Angular acceleration will be 
Explanation:
We have given that mass m = 0.18 kg
Radius r = 0.32 m
Initial angular velocity 
And final angular velocity 
Time is given as t = 8 sec
From equation of motion
We know that 
So angular acceleration will be
Answer:
160N/m
Explanation:
According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,
F = ke where
F is the applied force
k is the spring constant
e is the extension
From the formula k = F/e
Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.
The spring constant k = ma/e where
m is the mass of the block = 0.4kg
a is the acceleration = 8.0m/s²
e is the extension of the spring = 2.0cm = 0.02m
K = 0.4×8/0.02
K = 3.2/0.02
K = 160N/m
The spring constant of the spring is therefore 160N/m
Answer:
13 km
Explanation:
The bird flies from the runner, to the finish line, and back to the runner. We can write two equations for the distance it travels:
d = 7.8 km + 7.8 km − 4.9 km/hr × t
d = 24.5 km/hr × t
Solve for t in the second equation and substitute into the first:
t = d / 24.5
d = 7.8 + 7.8 − 4.9 (d / 24.5)
d = 15.6 − 0.2 d
1.2 d = 15.6
d = 13
The bird flies a cumulative distance of 13 km.
Answer:
14.0 cm
Explanation:
Draw a free body diagram of the block. There are three forces: weight force mg pulling down, elastic force k∆L pulling down, and buoyancy ρVg pushing up.
Sum of forces in the y direction:
∑F = ma
ρVg − mg − k∆L = 0
(1000 kg/m³) (4.63 kg / 648 kg/m³) (9.8 m/s²) − (4.63 kg) (9.8 m/s²) − (176 N/m) ∆L = 0
∆L = 0.140 m
∆L = 14.0 cm
