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2 years ago

How much metal can be removed from a cracked drum to restore surface

Engineering

2 answers:

Alisiya [41]2 years ago

7 0

Answer:sc

Explanation:

adelina 88 [10]2 years ago

4 0

There is a turn to limit stamped on it but as others have said, if it is cracked, you need to replace.

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A poundal is the force required to accelerate a mass of 1 lbm at a rate of 1 ft/(s^2). Determine the acceleration of an object o

storchak [24]

Answer:

See it in the pic

Explanation:

See it in the pic

5 0

3 years ago

Technician A says if the input signal turn-on time is too fast for the input circuit, your program may operate as though the inp

ASHA 777 [7]

Answer:

b. Technician B only

Explanation:

A watchdog timer is a circuit that automatically monitors the MCU (Microcontroller Unit) for any anomaly, detects it and helps the MCU to recover from the malfunction it has detected.

If the input signal turn-on time is too fast for the input circuit, that is a malfunction and this activates the watchdog timer circuit to correct this malfunction immediately. So Technician B only is correct as the watchdog timer is activated immediately once there is a malfunction.

7 0

2 years ago

The base class Pet has attributes name and age. The derived class Dog inherits attributes from the base class Pet class and incl

Nonamiya [84]

Answer:

Explanation:

class Pet:

def __init__(self):

self.name = ''

self.age = 0

def print_info(self):

print('Pet Information:')

print(' Name:', self.name)

print(' Age:', self.age)

class Dog(Pet):

def __init__(self):

Pet.__init__(self)

self.breed = ''

def main():

my_pet = Pet()

my_dog = Dog()

pet_name = input()

pet_age = int(input())

dog_name = input()

dog_age = int(input())

dog_breed = input()

my_pet.name = pet_name

my_pet.age = pet_age

my_pet.print_info()

my_dog.name = dog_name

my_dog.age = dog_age

my_dog.breed = dog_breed

my_dog.print_info()

print(' Breed:', my_dog.breed)

main()

3 0

3 years ago

Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi

qaws [65]

Answer:

$\Delta V = 209.151\,L$ , $\Delta C = 217.517\,USD$

Explanation:

The drag force is equal to:

$F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A$

Where $C_{D}$ is the drag coefficient and $A$ is the frontal area, respectively. The work loss due to drag forces is:

$W = F_{D}\cdot \Delta s$

The reduction on amount of fuel is associated with the reduction in work loss:

$\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s$

Where $F_{D,1}$ and $F_{D,2}$ are the original and the reduced frontal areas, respectively.

$\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s$

The change is work loss in a year is:

$\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)$