Are you talking about a constructive force / constructive wave? Sorry I think that's what you are asking
Answer:
The normal force the seat exerted on the driver is 125 N.
Explanation:
Given;
mass of the car, m = 2000 kg
speed of the car, u = 100 km/h = 27.78 m/s
radius of curvature of the hill, r = 100 m
mass of the driver, = 60 kg
The centripetal force of the driver at top of the hill is given as;
where;
Fc is the centripetal force
is downward force due to weight of the driver
is upward or normal force on the drive
Therefore, the normal force the seat exerted on the driver is 125 N.
The atomic number of an atom says how many protons it has. This number cant change, since the atomic number is what gives elements their identities (in the periodic table, at least).
The mass number, on the other hand, says how many protons AND neutrons the atom has (so, the sum of P+ and N0). So, electrons have nothing to do with this number.
Atoms are neutrally charged, which means there has to be an equal number of positive and negative particles. The negative particles of an atom are its electrons, and since our atom has 4 protons, it must also have 4 electrons.
Answer:
e = 0.0898m
v = 2.07m/s
Explanation:
a) According to Hooke's law
F = ke
e is the extension
k is the spring constant
Since F = mg
mg = ke
e = mg/k
Substitute the given value
e = 1.1(9.8)/120
e = 10.78/120
e = 0.0898m
Hence it is stretched by 0.0898m from its unstrained length
2) Total Energy = PE+KE+Elastic potential
Total Energy = mgh +1/2mv²+1/2ke²
Substitute the given value
5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²
Solve for v
5.0 = 2.156+0.55v²+0.48338
5.0-2.156-0.48338= 0.55v²
2.36 =0.55v²
v² = 2.36/0.55
v² = 4.29
v ,= √4.29
v = 2.07m/s
Hence the required velocity is 9.28m/s
