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
The formula you use is Density = (mass) / (volume) .
The calculation for this particular object is
Density = (68.895 g) / (23.5 mL - 14.7 mL)
Density = (68.895 g) / (8.8 mL)
<em>Density = 7.8 g/mL</em>
A cold front would bring rain or thunderstorms. Have a nice day!
<span>Answer:
Let m = mass of cannon
Then
10000 = ma
a = 10000/m
v^2 = u^2 + 2as
v^2 = 0 + 2as
84^2 = 2(2.21)(10000/m)
84^2 m = 4.42(10000)
m = 6.264172336
= 6.26 kg
Part 2
Range = u^2sin(2x38)/g
= 84^2sin(76)/9.8
= 698.6129229
= 698.6 m</span>
Answer:
Q= 722.5 *10⁻¹² C
Explanation:
Conceptual analysis
For a parallel plate capacitor, we can use the following formula :
E= (Q) /(ϵ₀*A) Formula (1)
Where:
E: electric field between the plates ( N/C)
Q: Charge of the plates (C)
ϵo : vacuum permittivity ( C²/ N.m²)
A : area oh the plates (m²)
Known data
A = 1.2 m²
E= 68 N/C
ϵo= 8.8542*10⁻¹² ( C²/ N.m²)
Problem development
We apply the formula (1) :
![68= \frac{Q}{(8.8542*10^{-12} )(1.2)}](https://tex.z-dn.net/?f=68%3D%20%5Cfrac%7BQ%7D%7B%288.8542%2A10%5E%7B-12%7D%20%29%281.2%29%7D)
Q= (68) (8.8542*10⁻¹²)(1.2)
Q= 722.5 *10⁻¹² C