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Bond [772]
3 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]3 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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Hitman42 [59]

For vertical motion, use the following kinematics equation:

H(t) = X + Vt + 0.5At²

H(t) is the height of the ball at any point in time t for t ≥ 0s

X is the initial height

V is the initial vertical velocity

A is the constant vertical acceleration

Given values:

X = 1.4m

V = 0m/s (starting from free fall)

A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)

Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

H(t) = 1.4 - 4.905t²

We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:

1.4 - 4.905t² = 0

4.905t² = 1.4

t² = 0.2854

t = ±0.5342s

Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))

t = 0.53s

8 0
3 years ago
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Misha Larkins [42]
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2 years ago
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Compare between chromosphere, corona and photosphere
gladu [14]

Answer:

Explained

Explanation:

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5 0
3 years ago
an object near the surface of a planet falls 54m in 3s. what is the acceleration due to gravity on that planet
Harman [31]

Answer:

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Explanation:

7 0
3 years ago
A 2kg blob of putty moving at 4 m/s slams into a 6kg blob of putty at rest what is the speed of the to stick together blobs imme
andrew11 [14]
<h2>Answer</h2>

1m/s

<h2>Explanation</h2>

Given that:

<em>Mass of first blob = 2kg = m1</em>

<em>Velocity of blob = 4m/s = v1</em>

<em>Mass of second blob = 6kg = m2</em>

<em>Velocity of blob = 0m/s = v2</em>

<em />

To find:

<em>Final velocity = Vf</em>

<em />

<em>This question is of inelastic collision which is any collision between objects in which some energy is lost.</em>

<em />

<h3>Formula to be use:</h3><h2>(m1*v1) + (m2*V2) = Vf(m1 + m2)</h2>

(2*4) + (6*0) = Vf(2+6)

8 + 0 = Vf(8)

8 = Vf(8)

Vf = 1 m/s

So the speed of two blobs immediately after colliding = 1 m/s

3 0
3 years ago
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