Ask question

Ask question

All categories

3 years ago

13

A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil with a d

ensity of 800 kg/m3 and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20 m

1 answer:

const2013 [10]3 years ago

3 0

Answer:

The answer is below

Explanation:

Pascal's law states that pressure is exerted in all parts of a static fluid equally.

The area of the narrower arm with a radius of 0.05 m (5 cm) is given as:

Area of narrow arm = π(0.05)²

The area of the wider arm with a radius of 0.18 m (18 cm) is given as:

Area of narrow arm = π(0.18)²

The ratio if the wider arm area to the narrow arm area = π(0.18)² / π(0.05)² = 12.96

To move the 12000 N car, the amount of force needed = 12000/12.96 = 926 N

You might be interested in

An object that completes 20 vibrations in 10 seconds has a frequency of

nika2105 [10]

Answer:

<em> The object has frequency of 2 Hz and time period of 0.5 s.</em>

Explanation:

<em>Frequency</em> is defined as number of oscillation per second ie back and forth swings done in single second.

Here it is given that the object oscillates 20 times in 10 seconds.

So f = $\frac{20}{10}$ = 2Hz

The <em>time period</em> is defined as time taken by the object to complete one full oscillation.

T = $\frac{1}{f}$

T= $\frac{1}{2}$ =0.5 s

<em>Thus the object has frequency of 2 Hz and time period of 0.5 s.</em>

7 0

3 years ago

The acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

VladimirAG [237]

We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

From the question we are told

the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Generally the equation for the Force is mathematically given as

F=\frac{F}{dx}

Therefore

F=-kdx

k=600Nm^{-1}

now

K.E=0.5x ds^2

K.E=600*(-0.1^2)

K.E=3J

Therefore

the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

For more information on this visit

brainly.com/question/23379286

6 0

3 years ago

In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

alexandr402 [8]

Answer:

a) $v=0.999124c$

b) $E=7.566*10^{22}$

c) $E_a=760 times\ larger$

Explanation:

From the question we are told that

Distance to Betelgeuse $d_b=430ly$

Mass of Rocket $M_r=20000$

Total Time in years traveled $T_d=36years$

Total energy used by the United States in the year 2000 $E_{2000}=1.0*10^20$

Generally the equation of speed of rocket v mathematically given by

$v=\frac{2d}{\triangle t}$

$v=860ly/ \triangle t$

where

$\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\sqrt{(860)^2+(36)^2}$

$\triangle t=860.7532$

Therefore

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

$E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2$

$E=7.566*10^{22}$

c)Generally the equation of the energy $E_a$ required to accelerate the rocket mathematically given by

$E_a=\frac{E}{E_{2000}}$

$E_a=\frac{7.566*10^{22}}{1.0*10^{20}}$

$E_a=760 times\ larger$

8 0

3 years ago

At what speed does a 1,248 kg compact car have the same kinetic energy as a 18,777 kg truck going 26 km/h?

notka56 [123]

The speed of car is 100.8km/h

$KE = \frac{1}{2} m(v \: truck ) {}^{2}$

$= 0.5 \times 18777 \times \frac{26}{3.6} \times \frac{26}{3.6}$

$= 489708.79j$

$= \frac{1}{2} \times 1248 \times( v \: car) {}^{2}$

$= 624v {}^{2}$

$so \: v {}^{2} = \frac{489708.7}{624}$

$= 784$

$v = \sqrt{784}$

$v = 28m/s$

v car= 28×3.6

=100.8km/h

Hence, the speed of the car is 100.8km/h

learn more about speed from here:

brainly.com/question/28326855

#SPJ4

3 0

1 year ago

The lower the__ the frequency, the __the pitch <br><br> -lower <br> -higher

rjkz [21]

Answer:

umm the lower the frequency the higher the pitch

Explanation:

6 0

2 years ago