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lora16 [44]
3 years ago
13

Technician A says that the parking brake cable should be adjusted at each wheel. Technician B says that the parking brake cable

adjustment is usually done after adjusting the rear brakes. Which technician is correct?
a. Technician A only c. Both Technicians A and B
b. Technician B only d. Neither Technician A nor B
Physics
1 answer:
Gnesinka [82]3 years ago
5 0

Answer: b. Technician B only

Explanation: Usually after the rear brakes is adjusted, parking brake cable is then adjusted. The adjustment bolt is located on the parking brake lever.This adjustment bolt tightens the cable leading to the rear brakes. If one notice that brake lever needs to be pulled up or clicked many times before it engages, then theres need to adjust the parking brake cable.

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In which situation are the individual molecules moving the fastest ?
ch4aika [34]
In hot warter or substant bc hot makes molecues go,s faster then all the rest
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4 years ago
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Copy the diagram. add a voltmeter to show how you would measure the voltage of the cell
Kruka [31]

Answer: the answer is 23voltage

Explanation: because the voltage and time put together is 23

7 0
3 years ago
A 2.0-kilogram mass is located 3.0 meters above
leva [86]

The correct answer is: Option (3) 9.8 N/kg

Explanation:

According to Newton's Law of Gravitation:

F_g = \frac{GmM}{R^2} --- (1)

Where G = Gravitational constant = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

m = Mass of the body = 2 kg

M = Mass of the Earth = 5.972 × 10²⁴ kg

R = Distance of the object from the center of the Earth = Radius of the Earth + Object's distance from the surface of the Earth = (6371 * 10³) + 3.0 = 6371003 m

Plug in the values in (1):

(1)=> F_g = \frac{6.67408 * 10^{-11} * 2 * 5.972*10^{24}}{(6371003)^2} = 19.63

Now that we have force strength at the location, we can use:

Force = mass * gravitational-field-strength

Plug in the values:

19.63 = 2.0 * gravitational-field-strength

gravitational-field-strength = 19.63/2 = 9.82 N/kg

Hence the correct answer is Option (3) 9.8 N/kg

4 0
4 years ago
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Yellow light of wavelength 590 nm passes through a diffraction grating and makes an interference pattern on a screen 80 cm away.
riadik2000 [5.3K]

Answer:B

Explanation:

Given

Wavelength of light \lambda =590\ nm

Screen distance L=80\ cm

First fringe is at a distance y_1=1.9\ cm

No of lines per mm is given by N

N=\frac{1}{d}

where d=slit width

From N-slits Experiment

\sin \theta _m=\frac{m\lambda }{d}

d=\frac{m\lambda }{\sin \theta _m}-----1

Position of bright fringe is given by

y=\tan \theta _m\cdot L

\tan \theta _m=\frac{y}{L}

\theta _m=\tan^{-1}(\frac{y}{L})

Put the value of \theta _m  in eq. 1

d=\frac{m\lambda }{\sin (\tan^{-1}(\frac{y}{L}))}

Therefore N=d^{-1}

N=\frac{\sin (\tan^{-1}(\frac{y}{L}))}{m\lambda }

for m=1

N=\frac{\sin (\tan^{-1}(\frac{1.9\times 10^{-2}}{0.8}))}{1\times 590\times 10^{-9}}

N=40243\ line/m

N=40\ line/mm

   

4 0
4 years ago
If the dielectric constant is 14.1, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted
Lady bird [3.3K]

Answer:

\frac{Q}{Q_0}=1

Explanation:

Capacitance is defined as the charge divided in voltage.

C=\frac{Q}{V}(1)

Introducing a dielectric into a parallel plate capacitor decreases its electric field. Therefore, the voltage decreases, as follows:

V=\frac{V_0}{k}

Where k is the dielectric constant and V_0 the voltage of the capacitor without a dielectric

The capacitance with a dielectric between the capacitor plates is given by:

C=kC_0

Where k is the dielectric constant and C_0 the capacitance of the capacitor without a dielectric. So, we have:

Q=CV\\Q=kC_0\frac{V_0}{k}\\Q=C_0V_0\\Q_0=C_0V_0\\Q=Q_0\\\frac{Q}{Q_0}=1

Therefore, a capacitor with a dielectric stores the same charge as one without a dielectric.

7 0
3 years ago
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