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marta [7]
4 years ago
11

The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.6 square inches. If a force of 5.6 lb

is applied to this piston, what is the force on the brake pad of one wheel if the area of the pad is 1.8 square inches? (Hint: Since the force on each of the four wheels would be the same, you only need to calculate the force on one.)
Physics
1 answer:
balu736 [363]4 years ago
5 0

Answer:

16.8 lb is the force on the brake pad of one wheel.

Explanation:

Force applied on the piston = F_1=5.6 lb

Area of the piston = A_1=0.6 inches^2

Force applied on the brakes = F_2

Area of the brakes = A_2=1.8 inches^2

Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.

\frac{F_1}{A_2}=\frac{F_2}{A_2}

F_2=\frac{5.6 lb\times 1.8 inhes^2}{0.6 inches^2}=16.8 lb

16.8 lb is the force on the brake pad of one wheel.

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It is symbols to make it easier and to save space! 
7 0
3 years ago
Glass does not transmit ultraviolet radiation. Suggest what happens to ultraviolet radiation when it is incident on glass. (1 ma
o-na [289]

Answer:

Most of UV radiation is stopped by glass & this is why you will not get sunburns behind a glass. The glass simply filters out the UV radiation that is responsible for the sunburns & protect your skins from these energetic & somewhat harmful radiation

Explanation:

4 0
2 years ago
In a football game, a receiver is standing still, having just caught a pass. Before he can move, a tackler, running at a velocit
ivolga24 [154]

Answer:

m_{receiver}=115Kg*3.1/(1.6)-115Kg=107.8Kg    

Explanation:

The football players collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V=1.6m/s.

We need to use the conservation of momentum Law, the total momentum is the same before and after the collision, at the initial point the receiver does not have any speed

m_{tackler}*v_{tackler}=(m_{tackler}+m_{receiver})V     (1)

We solve in order to find the receiver mass:

m_{receiver}={m_{tackler}*v_{tackler}/V}-m_{tackler}    

m_{receiver}=115Kg*3.1/(1.6)-115Kg=107.8Kg    

5 0
3 years ago
You move a 25 N object 5 meters. How much work did you do ?
olga55 [171]
The answer is 125 Joules

The first thing to take note of is the work equation: W=F×D

Since we already have our force and our distance that will help make this problem easier.

So, W=25*5

W=125

Therefore, our answer is 125 Joules since work is measured in joules

Hope this helped!! :)


3 0
3 years ago
The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p
NISA [10]

Answer:

a)  The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be v_{s} = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

t = \frac{d}{v_{s} }

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be v_{r} = 2.5 m/s

S = v_{r} t

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}

d) Downstream velocity of the swimmer, v_{y} = v_{s} sin \theta\\

v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s

The vertical displacement is given by, y = v_{y} t

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s

The downstream horizontal distance of the swimmer, x = v_{x} t

x = 1.6 * 66.67

x = 106.67 m

7 0
3 years ago
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