Answer:
the maximum length of the specimen before deformation is 0.4366 m
Explanation:
Given the data in the question;
Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²
cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m
tensile load F = 1810 N
maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m
Now to calculate the maximum length 
for the deformation, we use the following relation;
= [ Δl × E × π × D² ] / 4F
so we substitute our values into the formula
= [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )
= 3161.025289 / 7240
= 0.4366 m
Therefore, the maximum length of the specimen before deformation is 0.4366 m
Answer : The final velocity of the ball is, 12.03 m/s
Explanation :
By the 3rd equation of motion,
where,
s = distance covered by the object = 6.93 m
u = initial velocity = 2.99 m/s
v = final velocity = ?
a = acceleration = 
Now put all the given values in the above equation, we get the final velocity of the ball.
Thus, the final velocity of the ball is, 12.03 m/s
Answer:
is this a question for hoework
Answer:
When a horse pull a cart the action is on?
A horse is harnessed to a cart. If the horse tries to pull the cart, the horse must exert a force on the cart. By Newton's third law the cart must then exert an equal and opposite force on the horse. Newton's second law tells us that acceleration is equal to the net force divided by the mass of the system.
Explanation:
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = 
= 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ
