Answer:
Yes
Explanation:
The spring force is given as:
F = kd
F is the spring force
K is the spring constant
d is the magnitude of the stretch
Since k is a constant, therefore, doubling the stretch distance will double the force.
Both stretch distance and force applied can be said to be directly proportional to one another.
Answer:
A) ω = 6v/19L
B) K2/K1 = 3/19
Explanation:
Mr = Mass of rod
Mb = Mass of bullet = Mr/4
Ir = (1/3)(Mr)L²
Ib = MbRb²
Radius of rotation of bullet Rb = L/2
A) From conservation of angular momentum,
L1 = L2
(Mb)v(L/2) = (Ir+ Ib)ω2
Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.
(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2
(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2
Divide each term by Mr;
vL/8 = (L²/3 + L²/16)ω2
vL/8 = (19L²/48)ω2
Divide both sides by L to obtain;
v/8 = (19L/48)ω2
Thus;
ω2 = 48v/(19x8L) = 6v/19L
B) K1 = K1b + K1r
K1 = (1/2)(Mb)v² + Ir(w1²)
= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)
= (1/8)(Mr)v²
K2 = (1/2)(Isys)(ω2²)
I(sys) is (Ir+ Ib). This gives us;
Isys = (19L²Mr/48)
K2 =(1/2)(19L²Mr/48)(6v/19L)²
= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152
Thus, the ratio, K2/K1 =
[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19
The average speed is 8 miles
Answer: F
Out of the page.
Explanation:
For an electron with a charge of -e, the magnitude of the force on it is F = BeV
Where
F = force on the electron
e = charge ( electrons )
V = velocity
B = magnetic field
F is the force acting on all the electrons in a wire which gives rise to the F = BIL
Where
I = current
L = length of the wire
The force F is always at the right angle to the particle's velocity and its direction can be found using the left hand rule.
When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.
Answer:
Explanation:
m = ρV = 1.03( 1000 kg/m³)(π(2² m²)(3.0 m)) = 12360π kg
m ≈ 38,830 kg
