Answer:
Explanation:
First of all, we analyze the system of blocks before starting to move.
Hence, the incline angle θ for which both blocks begin to slide is 10.20°.
Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.

Where:



Therefore, the required stretch or compression in the connecting spring is 0.10 ft.
I hope it helps you!
Answer:
A) 10 m/s
Explanation:
We know that according to conservation of momentum,
m1v1 + m2v2 = m1u1 + m2u2 ..............(equation 1)
where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.
From the given data
If truck and car are two bodies
truck : m1 = 2000 Kg v1 = 5 m/s u1 = 0
car : m2 = 1000 kg v2 = 0 u2 = ?
final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.
substituting the values in equation 1, we get
(2000 x 5) + 0 = 0 + (1000 x u2)
u2 =
x 5
= 10 m/s
Hence after collision, car moves at a velocity of 10 m/s
Answer:
240 cm³
Explanation:
Weight = Buoyancy
mg = ρVg
m = ρV
(0.9 g/cm³ × 2400 cm³) = (1 g/cm³) V
V = 2160 cm³
The submerged volume is 2160 cm³, so the volume above the surface is 240 cm³.
Answer:
e^-4t e^-5t
Explanation:
solving s²+9s+20 quadratically we have (s+4)(s+5)
x(s) can be written as x(s) =(1/s+4)(1/s+5)
if we take the laplace inversve
L-¹ (s)=L-1(1/s+4) L-1(1/s+5)
we have e^-4t e^-5t