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Phoenix [80]
3 years ago
12

A vehicle with a manual transaxle can be cranked and started without depressing the clutch pedal. Technician A says the clutch i

s not releasing when the pedal is depressed. Technician B says the most likely cause is a defective clutch safety switch. Who is correct?
Physics
1 answer:
SashulF [63]3 years ago
6 0

Answer: Technician B

Explanation: In manual cars,the clutch safety is designed to stop the vehicle from moving when you start the gnition. It prevents power from flowing into the circuit . This is found in the pedal mechanism of cars so depressing the clutch pedal will likely cause a defective in the clutch safety. You will begin to perceive the clutch burning and white fumes coming out from the pedal.

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a girl that has a mass of 55kg is standing on a floor, how much supporting force does the floor have?
KengaRu [80]
A girl standing on a floor would have two opposite forces acting on it. These forces are the weight and the normal force. Since no other forces are acting and that the girl is at rest, then the weight must equate to the normal force. Therefore, the supporting force would be:

F = mg = 55kg (9.81 m/s^2) = 539.55 N
4 0
3 years ago
Un trineo de 20 kg descansa en la cima de una pendiente de 80 m de longitud y 30° de inclinación. Si µ = 0.2, ¿cuál es la veloci
Mariulka [41]

Answer:

v= 26.70 m/seg

Explanation:  Ver anexo ( diagrama de cuerpo libre)

De acuerdo a la segunda ley de Newton

∑ F  =  m*a

∑ Fx  =  m* a(x)             ∑ Fy  =  m* a(y)

También sabemos que el coeficiente de roce dinámico es:

  μ  = 0.2 = F(r)/N            siendo N la fuerza normal.

Si descomponemos la fuerza P = mg  =  20Kg* 9.8m/seg²

P =  196 [N]    en sus componentes sobre los ejes x y y tenemos

Py  =  P* cos30  =  196* √3/2  =  98*√3

Px  = P* sen30   =  196*1/2  =  98

La sumatoria sobre el eje y es :

∑ F(y)  =  m*a         Py  - N  = 0          98*√3  = N       ( no hay movimiento en la dirección y)

∑ F(x)  = m*a    P(x)  -  Fr  =  m*a

Fr  =   μ *N  =  0.2* 98*√3

Fr  =  19.6*√3  [N]

98 -  19.6*√3  =  m*a

98  -  33.52  = m*a

a =  (98  -  33.52 ) / 20

a = 3.22 m/seg²

Para calcular la velocidad del trineo al pié del plano, sabemos que al pié del plano el trineo ha recorrido 80 m, y que de cinemática

v²  =  v₀²  +  2*a*d             ( se pueden chequear unidades para ver la consistencia de la ecuación  v  y  v₀    vienen dados en m/seg  entonces  v²  y  v₀²  vienen en m²/seg²,  el producto de a (m/seg²) por la distancia d (m) resulta en m²/seg²  entonces es consistente la relación

v²   =  0   +  2*3.22*80       ( la velocidad inicial es cero)

v²  = 515.2  m²/seg²

v  =  √515.2  m/seg

v= 26.70 m/seg

6 0
3 years ago
Please help: I don't know how to do these problems
antiseptic1488 [7]
d =2.55.68m and t = 11.36s
In my opinion
3 0
3 years ago
Why to astronauts appear weightless while they are filmed performing activities inside the orbiting space shuttle?
vagabundo [1.1K]
<h2>Answer: The astronauts are falling at the same rate as the space shuttle as it orbits around earth</h2>

The astronauts seem to float because they are in free fall just like the spacecraft.

However, although they are constantly falling on the Earth, they do not fall because the ship orbits at a sufficient speed (in the same direction of rotation of the Earth) so that the centrifugal force is balanced with the Earth's gravitational pull.

In other words:

The spaccraft and the astronauts are in free fall but the Earth's surface will never be reached as long as they does not decrease the speed.

Then, as they accelerate toward Earth (regardless of their mass), it curves beneath them and never comes close.

That's why astronauts, having the same acceleration as the spacecraft, feel weightless and see themselves floating.

8 0
3 years ago
Read 2 more answers
A cubical Gaussian surface surrounds two positive charges, each has a charge q 1 1 = + 3.90 × 10 − 12 3.90×10−12 C, and three ne
Masteriza [31]

Answer:

The electric flux is zero because charge is zero.

Explanation:

Given that,

Positive charge q_{1}=3.90\times10^{-12}\ C

Negative charge q_{2}=-2.60\times10^{-12}\ C

We need to calculate the total charged

Using formula of charge

Q_{enc}=2q_{1}+3q_{2}

Put the value into the formula

Q_{enc}=2\times3.90\times10^{-12}+3\times(-2.60\times10^{-12})

Q_{enc}=0

We need to calculate the electric flux

Using formula of electric flux

\phi=\dfrac{Q_{enc}}{\epsilon_{0}}

Put the value into the formula

\phi=\dfrac{0}{8.85\times10^{-12}}

Hence, The electric flux is zero because charge is zero.

7 0
3 years ago
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