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Bond [772]
2 years ago
14

A hydraulic lift is designed to lift cars up to 2000 kg in mass.If the lift under the car is 1.0 m by 1.2m and the area of the i

nput piston is 10 cm by 10 cm, how much force must be exerted on the input piston to lift a 2000-kg car? A) 136 N B 24x 10^4 N C) 160 N D) 196 N E) 17N
Physics
1 answer:
Lerok [7]2 years ago
6 0

Answer:

C) 160 N

Explanation:

m = mass of the car = 2000 kg

Weight of the car is given as

W = mg \\W = (2000) (9.8)\\W = 19600 N

F_{i} = Force exerted on the input piston

A_{i} = Area of input piston = 10 cm x 10 cm = 0.1 m x 0.1 m = 0.01 m²

A_{lift} = Area of lift = 1 m x 1.2 m = 1.2 m²

Using Pascal's law

\frac{F_{i}}{A_{i}} = \frac{W}{A_{lift}}\\\frac{F_{i}}{0.01} = \frac{19600}{1.2}\\F_{i} = 160 N

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Every winter I fly home to Chicago. It takes 3 hours. What is my average speed?
Tanya [424]

It depends on where you live when you're not visiting Chicago. We need to know the distance of the trip.

Your average speed on the trip is . . .

(total distance in miles) / (3 hours)

miles per hour

5 0
2 years ago
A race car makes one lap around a track of radius 50 m in 9.0 s. What is the average velocity? *
Oksi-84 [34.3K]

Given that,

Radius of track, r = 50 m

time , t = 9 s

velocity, v = ?

Distance covered by car in one lap around a track is equal to the circumference of the track.

C = 2 π r = 2 * 3.14 * 50

C = 314.159 m

Distance covered by car, s = 314.159 m

Velocity = distance/ time

V = 314.159 / 9

V = 34.9 m/s

The average velocity of car is 34.9 m/s.

7 0
1 year ago
An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in th
pogonyaev

Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, c=345J/^oC

Initial temperature, T_i=88^{\circ}C

Final temperature, T_f=45^{\circ}C

We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :

Q=mc\Delta T

Let the mass of the object is 10 g or 0.01 kg

So,

Q=0.01\times 345\times (88-45)

Q = 148.35 Joules

So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.                                                

5 0
2 years ago
A dog of mass 10 kg sits on a skateboard of mass 2 kg that is initially traveling south at 2 m/s. The dog jumps off with a veloc
Tasya [4]

Answer:

17 m/s south

Explanation:

m_1 = Mass of dog = 10 kg

m_2 = Mass of skateboard = 2 kg

v = Combined velocity = 2 m/s

u_1 = Velocity of dog = 1 m/s

u_2 = Velocity of skateboard

In this system the linear momentum is conserved

(m_1+m_2)v+m_1u_1+m_2u_2=0\\\Rightarrow u_2=-\dfrac{(m_1+m_2)v+m_1u_1}{m_2}\\\Rightarrow u_2=-\dfrac{(10+2)2+10\times 1}{2}\\\Rightarrow u_2=-17\ m/s

The velocity of the skateboard will be 17 m/s south as the north is taken as positive

3 0
2 years ago
Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el
Rina8888 [55]
<span>this  may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>
8 0
2 years ago
Read 2 more answers
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