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11Alexandr11 [23.1K]
3 years ago
9

Technician A says that you can usually depressurize a brake accumulator by turning the ignition switch ""off"" to disable the el

ectric pump and depress the brake pedal 30–50 times. Technician B says that you can use a scan tool to depressurize a brake accumulator. Who is correct?
Physics
1 answer:
Rzqust [24]3 years ago
3 0

Answer:

Both Technician A and Technician B

Explanation:

In order to gain a better understanding of the solution above let define some terms

   Break Accumulator

We can define a break accumulator as storage that that helps generate the required pressure in order for the breaking system to respond faster this accumulator is charged by turning the steering wheel slowly at once from  lock to lock now this build the pressure in the accumulator and one way to depressurize is it is by  turning the ignition switch ""off""

 Now a scan tool is a device that can interface with a car it can also be used to diagnose a car an get the diagnostic information to help in the cars diagnoses and also be used to reprogram a car

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The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.
Salsk061 [2.6K]

Answer:

5.94\ \text{m/s}

1.7

0.577

Explanation:

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

\theta = Angle of slope = 30^{\circ}

v = Velocity of child at the bottom of the slide

\mu_k = Coefficient of kinetic friction

\mu_s = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}

The speed of the child at the bottom of the slide is 5.94\ \text{m/s}

Length of the slide is given by

l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}

v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}

The force energy balance of the system is given by

mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73

The coefficient of kinetic friction is 1.7.

For static friction

\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577

So, the minimum possible value for the coefficient of static friction is 0.577.

8 0
3 years ago
A certain superconducting magnet in the form of a solenoid of length 0.72 m can generate a magnetic field of 3.5 T in its core w
ira [324]

Answer:

55407

Explanation:

we have given that magnetic field B=3.5 T

current through the coil=90 A

Length of solenoid =0.72 m

we know the formula of magnetic field

B=\frac{\mu _0NI}{l}

so N=\frac{Bl}{\mu _0I}=\frac{3.5\times 0.72}{4\pi \times 10^{-6}\times 90}=55407.277=55407\ turns

so the number of turn in solenoid will be 55407

3 0
3 years ago
A student is swimming south applying a force of 256 N. The water exerts a westward force of 104 N. If the student has a mass of
grigory [225]

Answer:

a=2.9\ m/sec^2

Explanation:

<u>Net Forces and Acceleration</u>

The second Newton's Law relates the net force F_r acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

\vec F_r=m\vec a

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.

The magnitude of the resulting force F_r is computed as the hypotenuse of a right triangle

|F_r|=\sqrt{256^2+104^2}

|F_r|=276.32\ Nw

The acceleration can be obtained from the formula

F_r=ma

Note we are using only magnitudes here

\displaystyle a=\frac{F_r}{m}

\displaystyle a=\frac{276.32Nw}{95.3Kg}

\boxed{a=2.9\ m/sec^2}

7 0
3 years ago
If you sight Polaris with an astrolabe and determine its altitude is 20° above the horizon, then what does this tell you about y
oksano4ka [1.4K]
If you sight Polaris at 20 degrees above your Northern Horizon then you know that your latitude is 20 degrees north of the equator.
3 0
3 years ago
Read 2 more answers
A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its l
mylen [45]

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be \who_w = 1000 kg/m^3

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

F_s + F_b = W

Where F_s = kxN is the spring force, F_b = W_w = m_wg = \rho_w V_s g is the buoyancy force, which equals to the weight W_w of the water displaced by the submerged portion of the cylinder, which is the product of water density \rho_w, submerged volume V_s and gravitational constant g. W = mg is the weight of the metal cylinder.

kx + \rho_w V_s g = mg

The submerged volume would be the product of cross-section area and the submerged length x

V_s = Ax = \pi(d/2)^2x

Plug that into our force equation and we have

kx + \rho_w \pi(d/2)^2x g = mg

x(k + \rho_w g \pi d^2/4) = mg

x = \frac{m}{(k/g) + (\rho_w\pi d^2/4)} = \frac{1.5}{(33/9.8) + (100*\pi * 0.048^2/4)} = 0.423 m

6 0
3 years ago
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