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kow [346]
3 years ago
15

In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of 48 N to the input pi

ston, which has a radius r1. As a result, the output plunger, which has a radius r2, applies a force to the car. The ratio r2/r1 has a value of 9.0. Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.
Physics
1 answer:
erik [133]3 years ago
8 0

Answer:

Force that the output plunger applies to the car; F2 = 3888N

Explanation:

For a hydraulic device, the relationship between the force and the area using Pascal's principle is;

F1/A1 = F2/A2

Where;

F1 is force applied to the input piston

F2 is force that the output plunger applies to the car

A1 is Area of input piston

A2 is area of larger piston

We are given;

R2/R1 = 9

So,R2 = 9R1

F1 = 48N

Area of input piston;

A1 = π(R1)²

Area of output piston;

A2 = π(9R1)²

Since, (F1/A1) = (F2/A2)

Thus;

F1/(π(R1)²) = F2/(π(9R1)²)

If we simplify, π(R1)² will cancel out to give;

F1 = F2/9²

Thus;

F2 = 9² x F1

Plugging in 48N for F1, we have;

F2 = 9² x 48

F2 = 81 x 48

F2 = 3888N

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AlexFokin [52]

The question involves a ping-pong ball that is held submerged in a bucket by a string attached to the bottom of the bucket.

The answer is the tension of the string will increase. This is because making the water salty increases its density, and consequently, increases its buoyancy. This is why sea water is more buoyant than fresh water. Therefore the ping pong is pushed more upwards by the water when salt is added than initially. This gives the string more tension.


5 0
3 years ago
during a journey, a car travels at 40 km in 2.5 hours, next 62 km in 3 hours, then took a break for 30 minutes, again travelled
wlad13 [49]
45mph is the answer if you do the math right
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3 years ago
A 2.5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.7
Norma-Jean [14]

vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6


Therefore, v= 3.5 m/s.

7 0
3 years ago
Read 2 more answers
A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holdi
Elis [28]

Answer: 115.52\ N

Explanation:

Given

Length of plank is 1.6 m

Force F_1 is applied on the left side of plank

Force F_2 is applied 43 cm from the left end O.

Mass of the plank is m=13.7\ kg

for equilibrium

Net torque must be zero. Taking torque about left side of the plank

\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N

Net vertical force must be zero on the plank

\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N

8 0
2 years ago
Convert the following to relative uncertainties <br>a) 2.70 ± 0.05cm<br>b) 12.02 ± 0.08cm
DENIUS [597]

data which is expressed in form of following way

a = a_o + \Delta a

here in above expression

a_o = true value

\Delta a = uncertainty in the value

now the relative uncertainty is given as

\frac{\Delta a}{a_o}

now by above formula we can say

a) 2.70 ± 0.05cm

here

True value = 2.70

uncertainty = 0.05

Relative uncertainty = \frac{0.05}{2.70} = 0.0185

b) 12.02 ± 0.08cm

here

True value = 12.02

uncertainty = 0.08

Relative uncertainty = \frac{0.08}{12.02} = 0.00665

4 0
3 years ago
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