Ask question

Ask question

All categories

3 years ago

15

Two technicians are discussing a problem where the brake pedal travels too far before the vehicle starts to slow. Technician A s

ays that the brakes may be out of adjustment. Technician B says that one circuit from the master cylinder may be leaking or defective. Which technician is correct?

1 answer:

iVinArrow [24]3 years ago

3 0

Answer:

Technician A

Explanation:

If Technician B was correct, and the master cylinder is defective - then no braking action would occur.

This is not true however, as some breaking action eventually occurs, meaning it must be out of adjustment.

You might be interested in

-4.3 light years what is the value of the 3??

Radda [10]

Answer:

A light-year is a unit of distance. It is the distance that light can travel in one year. Light moves at a velocity of about 300,000 kilometers (km) each second. So in one year, it can travel about 10 trillion km. More p recisely, one light-year is equal to 9,500,000,000,000 kilometers

3 0

2 years ago

Eric is creating a timeline of the formation of the solar system. Which sequence best describes the formation of the solar syste

vlada-n [284]

The answer is C. nebular are star nurseries. When the massive gas being collapsing in its own weight. Local areas of gas begin to coalesce under gravity. Due to enormous pressure, nuclear fusion begins and a protostar is formed. The protostar grows into the sun as more hydrogen fuses at the core. The planetesimal materials at the edges of the protostellar discs coalesce to form planets that orbit the star.

3 0

3 years ago

Read 2 more answers

If the distance d (in meters) traveled by an object in time t (in seconds) is given by the formula d=a+bt2, the si units of a an

ss7ja [257]

Distance , d = a+b $t^2$

The unit of d is in meter and t is in seconds.

So the unit of a a must be meter.

Now we have unit of b $t^2$ is meter.

So unit of b* $second^2$ = meter

Unit of b = meter/ $second^2$

So unit of a = m and unit of b = m/ $s^2$ .

8 0

2 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

3 years ago

Which two terms indicate the same area of the body? *

Kruka [31]

Answer:

For example, the toes are anterior to the heel, and the popliteus is posterior to the patella. Superior and inferior, which describe a position above (superior) or below (inferior) another part of the body. For example, the orbits are superior to the oris, and the pelvis is inferior to the abdomen.

Explanation:

4 0

2 years ago