The final answer is B hope its helps
Answer:
a=
Explanation:
The net force, 
of the box is expressed as a product of acceleration and mass hence
where m is mass and a is acceleration
Making a the subject, a= 
From the attached sketch,
∑ 
where 
is frictional force and 
is horizontal angle
Substituting ∑ as in the equation where we made a the subject
a= 
Since we’re given the value of F as 240N, as 41.5N, as 
and mass m as 30kg
a= 
Answer:
a ) 2.13 X 10⁶ m/s .
b ) 1.28 X 10⁶ m/s
Explanation:
When electrons are repelled by negative plates and attracted by positive plates , it will increase their kinetic energy.
Increase in their energy = 4.15 eV
= 4.5 X 1.6 X 10⁻¹⁹ J
= 6.64 x 10⁻¹⁹ J
Initial kinetic energy
= 1/2 mv²
= 1/2 x 9.1 x 10⁻³¹ x ( 1.76 x 10⁶)²
= 14.09 X 10⁻¹⁹ J
Total energy
= 6.64 x 10⁻¹⁹+14.09 X 10⁻¹⁹
= 20.73 x 10⁻¹⁹
If V be the increased velocity
1/2 m V² = 20.73 X 10⁻¹⁹
.5 X 9.1 X 10⁻³¹ V² = 20.73 X 10⁻¹⁹
V = 2.13 X 10⁶ m/s .
b ) When electrons are released from positive plate , their speed are reduced because of attraction between electrons and positively charge plates.
Initial kinetic energy
= 14.09 x 10⁻¹⁹ J (see above )
reduction in kinetic energy
= 6.64 x 10⁻¹⁹ J ( See above )
Total energy with electron -
= 14.09 x 10⁻¹⁹ - 6.64 x 10⁻¹⁹
= 7. 45 x 10⁻¹⁹ J .
If V be the energy of electrons reaching the negative plate,
1/2 m V² =7. 45 x 10⁻¹⁹
V = 1.28 X 10⁶ ms⁻¹
Answer:
σ = 1.09 mm
Explanation:
Step 1: Identify the given parameters
rod diameter = 20 mm
stiffness constant (k) = 55 MN/m = 55X10⁶N/m
applied force (f) = 60 KN = 60 X 10³N
young modulus (E) = 200 Gpa = 200 X 10⁹pa
Step 2: calculate length of the rod, L
K = \frac{A*E}{L}K=
L
A∗E
L = \frac{A*E}{K}L=
K
A∗E
A=\frac{\pi d^{2}}{4}A=
4
πd
2
d = 20-mm = 0.02 m
A=\frac{\pi (0.02)^{2}}{4}A=
4
π(0.02)
2
A = 0.0003 m²
L = \frac{A*E}{K}L=
K
A∗E
L = \frac{(0.0003142)*(200X10^9)}{55X10^6}L=
55X10
6
(0.0003142)∗(200X10
9
)
L = 1.14 m
Step 3: calculate the displacement of the rod, σ
\sigma = \frac{F*L}{A*E}σ=
A∗E
F∗L
\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}σ=
(0.0003142)∗(200X10
9
)
(60X10
3
)∗(1.14)
σ = 0.00109 m
σ = 1.09 mm
Therefore, the displacement at the end of A is 1.09 mm
For a better understanding of the question, please see attached picture that I've created.
We need to identify the other two angles by applying the Law of sine where we are given with the following values:
a = 3.2 unit
b = 2.4 unit
c = 4.6 unit
∠A = unknown
∠B = unknown
∠C = 110°
Solving for ∠B, we have:
sin C / c = sin B / b
sin 110 / 4.6 = sin B / 2.4
sin B = 2.4*sin 110 / 4.6
B = 29.36°
Solving for ∠A, we have:
sinC/c = sinA/ a
sin110 / 4.6 = sinA / 3.2
sinA = 3.2*sin110/ 4.6
A = 40.82°
Therefore, the missing angles are ∠A=40.82° and ∠B=29.36°.
