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Yuliya22 [10]
2 years ago
9

Two examples of hydraulic brake​

Physics
1 answer:
rjkz [21]2 years ago
3 0

Answer:

1). Brake pedal or lever

2). A pushrod also called an actuating rod

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Technician A says that the intake and exhaust manifolds have to be removed before removing the engine from the vehicle. Technici
Amiraneli [1.4K]

Answer: Technician A is correct

Explanation:

The intake manifold is the compactment that all fuel and air supply to the cylinders. It's connected to the engine so it has to be disconnected while the exhaust manifold receives all the exhaust gases from the cylinders and releases the gas through a single or double exhaust gases outlet.

5 0
3 years ago
Antonina throws a coin straight up from a height of
vichka [17]

Answer:

s=vt-\frac{1}{2}gt^2

Explanation:

We could use the following suvat equation:

s=vt-\frac{1}{2}gt^2

where

s is the vertical displacement of the coin

v is its final velocity, when it hits the water

t is the time

g is the acceleration of gravity

Taking upward as positive direction, in this problem we have:

s = -1.2 m

g=-9.8 m/s^2

And the coin reaches the water when

t = 1.3 s

Substituting these data, we can find v:

v=\frac{s}{t}+\frac{1}{2}gt=-\frac{1.2}{1.3}+\frac{1}{2}(-9.8)(1.3)=-7.3 m/s

where the negative sign means the direction is downward.

5 0
3 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3
Pachacha [2.7K]

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC       Q₃ = 51 nC        Q₂ = 30.5 nC

x₁ = - 1.715m         x₃ = - 1.085m     x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0
3 years ago
The force needed to move an object 14 meters with 83 joules of work is what?
AleksandrR [38]
W=F*D
83J=F*14
83/14=F
5.92N
8 0
3 years ago
Two large metal plates of area 0.88 m2 face each other, 4.8 cm apart, with equal charge magnitudes but opposite signs. The field
Nuetrik [128]

Answer:

q=6.22*10^-10C

Explanation:

Two large metal plates of area 0.88 m2 face each other, 4.8 cm apart, with equal charge magnitudes but opposite signs. The field magnitude E between them (neglect fringing) is 80 N/C. Find |q|

E=α/∈, electric field within the plate

α=q/A

A=area of the plate

∈=is the permittivity

substituting , we have

The field magnitude E between them (neglect fringing)

E=q/A∈

q=EA∈

q=0.88*80*8.84*10^-12

q=6.22*10^-10C

3 0
3 years ago
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