Answer:
a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon
Explanation:
F= ma
v²=u² -2aS
(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)
a=1.36×10⁹m/s²
recall
F=ma
F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²
F= 2.55 × 10⁻¹⁹N
the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon
F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)
= 1.38 × 10⁸
Answer:
dJ = 1.7 m
Explanation:
The Equation of the Balancing the moments in the center of the seesaw is like this:
∑Mo = 0
Mo = F*d
Where:
∑Mo : Algebraic sum of moments in the center(o) of the balance
Mo : moment in the o point ( N*m)
F : Force ( N)
d : distancia of the force to the the o point ( N*m)
Data
mA = 60 kg : mass of the Anna
mJ = 70 kg : mass of theJon
dA = 2 m : Distance from Anna to the center of the seesaw
g: acceleration due to gravity
Calculation of the distance from Jon to the center of the seesaw (dJ)
∑Mo = 0 WA : Ana's weight , WJ : Jon's weight
W = m*g
(WA)(dA) - (WJ) (dJ) = 0
(mA*g)(dA) - (mJ*g)(dJ) = 0
We divide by g the equation:
(mA)(dA) - (mJ)(dJ)= 0
(mA)(dA) = (mJ)(dJ)
dJ = 1.7 m
Answer: 363 Ω.
Explanation:
In a series AC circuit excited by a sinusoidal voltage source, the magnitude of the impedance is found to be as follows:
Z = √((R^2 )+〖(XL-XC)〗^2) (1)
In order to find the values for the inductive and capacitive reactances, as they depend on the frequency, we need first to find the voltage source frequency.
We are told that it has been set to 5.6 times the resonance frequency.
At resonance, the inductive and capacitive reactances are equal each other in magnitude, so from this relationship, we can find out the resonance frequency fo as follows:
fo = 1/2π√LC = 286 Hz
So, we find f to be as follows:
f = 1,600 Hz
Replacing in the value of XL and Xc in (1), we can find the magnitude of the impedance Z at this frequency, as follows:
Z = 363 Ω
Answer:
0.125 cm
Explanation:
1/f = 1/d¡ + 1/d。
Find the focal point
(13.0^-1 + 20.8^-1) = 0.125 m
Focal point = 0.125 m
Answer:
As a mass greater than that of baseball, at the moment of the bowling wave the moment of the baseball ball is also greater
Explanation:
This problem is an application of momentum and momentum. When the astronaut pushed balls, he needed more force to move the ball of greater mass (bowling). The expression for soul is
p = m v
Besibol Blade
p1 = m1 v
Bowling ball
p2 = m2 v
As a mass greater than that of baseball, at the moment of the bowling wave the moment of the baseball ball is also greater
p2 >> p1