Answer:
a = 18.28 ft/s²
Explanation:
given,
time of force application, t= 10 s
Work = 10 Btu
mass of the object = 15 lb
acceleration, a = ? ft/s²
1 btu = 778.15 ft.lbf
10 btu = 7781.5 ft.lbf
m = 0.466 slug
now,
work done is equal to change in kinetic energy
now, acceleration of object
a = 18.28 ft/s²
constant acceleration of the object is equal to 18.28 ft/s²
To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.
Where
A = Cross-sectional Area
Y = Young's modulus
= Coefficient of linear expansion for steel
= Temperature Raise
Our values are given as,
Replacing we have,
Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N
To solve this problem we will apply the concepts related to the Impulse which can be defined as the product between mass and the total change in velocity. That is to say
Here,
m = mass
Change in velocity
As we can see there are two types of velocity at the moment the object makes the impact,
the first would be the initial velocity perpendicular to the wall and the final velocity perpendicular to the wall.
That is to say,
El angulo dado es de 45° y la velocidad de 25, por tanto
The change of sign indicates a change in the direction of the object.
Therefore the impulse would be as
The negative sign indicates that the pulse is in the opposite direction of the initial velocity.
Answer:
B_o = 1.013μT
Explanation:
To find B_o you take into account the formula for the emf:
where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.
By applying the derivative you obtain:
when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:
hence, B_o = 1.013μT