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Vanyuwa [196]
2 years ago
12

1. The first thing to do in the event of brake failure is to a. use the parking brake to stop your carb. pump the brake pedal ra

pidlyc. shift to a lower geard. steer against the curb 2. Engine failure can be caused by a. a broken timing gearb. a lack of fuelc. extreme heatd. all of the above 3. If your engine stalls a. your power brakes won't work at allb. your power steering won't work at allc. your power brakes and power steering won't work very welld. you should pump your power brakes 4. If your engine is flooded, you will probably a. smell gasolineb. see steam coming out from under the hoodc. have your engine stall after going through a large puddle of waterd. all of the above 5. The most common kind of steering failure is a. total system failureb. power-assist failurec. both a
Physics
1 answer:
mars1129 [50]2 years ago
8 0

Answer:

The correct answers are -

1. b. pump the brake pedal rapidly

2. d. all of the above

3. c. your power brakes and power steering won't work very well

4. a. gasoline

5. b. power-assist failure

Explanation:

Brakes help in stopping the vehicle by causing resistance to wheels. IF brake failure it is advised to pump the breaks rapidly in order to cause resistance. Engine failure can be caused by various reasons such as a broken timing gear, no fuel, or getting extremely heated.

In the case of stalling of engine most common effects as power steering not working properly or power brakes not working well. The flooded engine means the engine has too much gasoline that it wont starts.

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an athlete sprints from 150 m south of the finish line to 65 m south of the finish line in 5.0s what is his average velocity
MatroZZZ [7]
  • Total displacement=150-65m=85m
  • Time=5s

\\ \rm\longrightarrow Avg\:Velocity=\dfrac{Total\:Displacement}{Total\:Time}

\\ \rm\longrightarrow Avg\:Velocity=\dfrac{85}{5}

\\ \rm\longrightarrow Avg\:Velocity= 17m/s

6 0
2 years ago
Read 2 more answers
A car travels from Boston to Hartford in 4 hours. The two cities are 240 kilometers apart. What was the average speed of the car
Y_Kistochka [10]

Answer:

Average speed is 60 km/hour

Explanation:

When we need to calculate average speed, we use this equation:

V = \frac{x_{f} - x_{o}}{t_{f} - t_{o}}

Where:   x_{o} = 0 km   position at the beginning

              x_{f} = 240 km   at the end

              t_{o} = 0 hours

              t_{f} = 4 hours

Then:     V = \frac{240 km - 0 km}{4 hours - 0 hours}

              V = \frac{240 km}{4 hours}

Finally    V = 60 km/hour

6 0
3 years ago
Which best describes a major characteristic of both volcanoes and earthquakes? they are centered at the poles?
Arturiano [62]
<span>A major characteristic of both volcanoes and earthquakes is that they are located in the same geographic area. Most earthquakes are along the edges of tectonic plates. This is where most volcanoes are too. Most earthquakes directly beneath a volcano are caused by the movement of magma.</span>
6 0
3 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t
KatRina [158]

Answer:

T₂ = 123.9 N,  θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

       T₁ₓ - T_{2x} = 0

        T_{2x}= T₁ₓ

Y axis y

      T_{1y} + T_{2y} - 200N = 0

      T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

         sin 60 = T_{1y} / T₁

         cos 60 = t₁ₓ / T₁

         T_{1y} = T₁ sin 60

         T1x = T₁ cos 60

         T_{1y}y = 100 sin 60 = 86.6 N

         T₁ₓ = 100 cos 60 = 50 N

for voltage 2 it is done in the same way

         T_{2y} = T₂ sin θ

         T₂ₓ = T₂ cos θ

we substitute

         

           T₂ sin θ= 200 - 86.6 = 113.4

           T₂ cos θ = 50              (1)

to solve the system we divide the two equations

           tan θ = 113.4 / 50

           θ = tan⁻¹ 2,268

           θ = 66.2º

we caption in equation 1

           T₂ cos 66.2 = 50

           T₂ = 50 / cos 66.2

           T₂ = 123.9 N

8 0
2 years ago
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