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ehidna [41]
3 years ago
6

Technician A says that front bearing assembly locknuts for RWD vehicles are typically heavily torqued to maintain the bearing ad

justment. Technician B says that the axle nuts for a FWD wheel bearing assembly are often staked in place to maintain its specified torque. Who is correct?
A. Technician A
B. Technician B
C. Both Technician A and Technician B
D. Neither A nor B
Physics
1 answer:
Burka [1]3 years ago
7 0

Answer:

B. Technician B

Explanation:

The statement made by Technician A is not right because for an RWD vehicles, its front bearing assembly is not used for sustaining the bearing adjustment. The statement made by the Technician B is right because the main reason for staking the bearing assembly of the wheel of a FWD vehicle is to ensure that the torque is at a specific value. Thus, only Technician B is correct.

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It is a part of the sugary liquid in the plant that uses photosynthesis to produce.
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4 years ago
A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust rever
Amanda [17]

Answer:

257 kN.

Explanation:

So, we are given the following data or parameters or information in the following questions;

=> "A jet transport with a landing speed

= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"

= > The distance = 425 m along the runway with constant deceleration."

=> "The total mass of the aircraft is 140 Mg with mass center at G. "

We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"

Step one: determine the acceleration;

=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.

=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.

Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).

= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).

= 257 kN.

4 0
3 years ago
according to newton's law of universal gravitation, in which of the following situations does the gravitational attraction betwe
Alex17521 [72]

Explanation:

The force acting between two masses is given by :

F=G\dfrac{m_1m_2}{d^2}........(1)

Where

G is the universal gravitational constant

m_1\ and\ m_2 are masses

d is the distance between two masses

It is clear from equation (1) that the gravitational attraction between the bodies always increases if the masses of bodies increases and when the separation between masses decreases.

So, the correct answer is "the masses increase, and the distance between the centers of mass decreases". This is because the force of gravitation is directly proportional to the masses and inversely proportional to the separation.

4 0
3 years ago
A girl on roller skates accelerates at a rate of 2 m/sec/sec with a force of 100 N. What is her mass?​
Dvinal [7]

By the newtons Second Law:

F = ma

Solving for m:

m = F / a

m =  100 N / 2 m/s²

<h3>m = 50 kg → ANSWER</h3>
8 0
3 years ago
Can someone please help me
Zarrin [17]
They’re at a velocity of 2.0 m/s
6 0
3 years ago
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