Answer:
vB' = 0.075[m/s]
Explanation:
We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.
Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.
where:
mA = 0.355 [kg]
vA = 0.095 [m/s] before the collision
mB = 0.710 [kg]
vB = 0.045 [m/s] before the collision
vA' = 0.035 [m/s] after the collision
vB' [m/s] after the collison.
The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.
Answer:
Magnetic field, B = 0.004 mT
Explanation:
It is given that,
Charge,
Mass of charge particle,
Speed,
Acceleration,
We need to find the minimum magnetic field that would produce such an acceleration. So,
For minimum magnetic field,
B = 0.004 T
or
B = 4 mT
So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.
Answer:
Explanation:
During the exchange of applied force, thermal energy is generated by the friction that exists between the ground and the tire.
Said force according to the statement is the reaction of half the force on the rear tire. In this way the normal force acted is,
The work done is given by the friction force and the distance traveled,
Where
The thermal energy released through the work done is,
Answer:
The velocity of the truck after the collision is 20.93 m/s
Explanation:
It is given that,
Mass of car, m₁ = 1200 kg
Initial velocity of the car,
Mass of truck, m₂ = 9000 kg
Initial velocity of the truck,
After the collision, velocity of the car,
Let is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.
So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.
Answer:
x = 0.0685 m
Explanation:
In this exercise we can use the relationship between work and energy conservation
W = ΔEm
Where the work is
W = F x
The energy can be found in two points
Initial. Just when the block with its spring spring touches the other spring
Em₀ = K = ½ m v²
Final. When the system is at rest
=
= ½ k₁ x² + ½ k₂ x²
We can find strength with Newton's second law
∑ F = F - fr
Axis y
N- W = 0
N = W
The friction force has the equation
fr = μ N
fr = μ W
The job
W = (F – μ W) x
We substitute in the equation
(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)
½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0
We substitute values and solve
½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0
x² 30 + 14.4 x - 1,125 = 0
x² + 0.48 x - 0.0375 = 0
We solve the second degree equation
x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2
x = [-0.48 ± 0.617] / 2
x₁ = 0.0685 m
x₂ = -0.549 m
The first result results from compression of the spring and the second torque elongation.
The result of the problem is x = 0.0685 m