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lisabon 2012 [21]
4 years ago
10

In a car lift, compressed air exerts a force on a piston with a radius of 2.62 cm. This pressure is transmitted to a second pist

on with a radius of 10.8 cm. a) How large a force must the compressed air exert to lift a 1.44 × 104 N car
Physics
1 answer:
marin [14]4 years ago
8 0

Answer:

847.45 N

Explanation:

F₁=force of exerted by smaller piston

F₂=force of exerted by larger piston=1.44×10⁴ N

A₁=Area of smaller piston= 2.62 cm =0.0265 m

A₂=Area of larger piston= 10.8 cm =0.108 m

Pressure exerted by both the pistons will be equal

P_1=P_2\\\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}\\\Rightarrow F_1=\frac{F_2}{A_2} A_1\\\Rightarrow F_1=\frac{14400}{\pi\times  0.108^2}\pi\times  0.0262^2\\\Rightarrow F_1=847.45\ N

Hence, force exerted to lift a 14400 N car is 847.45 N

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4 years ago
Two cars start a race with initial velocity 7ms-1 and 4ms-1 respectively.Their acceleration are 0.4ms-2 and 0.5ms-2 respectively
Ronch [10]

The distance of the track is 600 m

Explanation:

The two cars move by uniformly accelerated motion, so we the distance they cover after a time t is described by the following suvat equation

d=ut+\frac{1}{2}at^2

where

u is the initial velocity

a is the acceleration

For car 1, we have

d_1 = u_1 t + \frac{1}{2}a_1 t^2

where

u_1 = 7 m/s is the initial velocity of car 1

a_1 = 0.4 m/s^2 is the acceleration of car 1

So the equation can be rewritten as

d_1 = 7t + 0.2t^2

For car 2, we have

d_2 = u_2 t + \frac{1}{2}a_2 t^2

where

u_2 = 4 m/s is the initial velocity of car 2

a_2 = 0.5 m/s^2 is the acceleration of car 2

So the equation can be rewritten as

d_2= 5t + 0.25t^2

The two cars finish the race at the same time: this means that they cover the same distance in the same time t, so we can write

d_1 = d_2\\7t + 0.2t^2=5t + 0.25t^2

And solving for t, we find

2t - 0.05t^2= 0\\t(2-0.05t)=0

Which gives two solutions:

t = 0 (initial instant)

t = 40 s

Therefore, the distance of the track is

d_1 = 7t +0.2t^2 = 7\cdot 40+0.2\cdot 40^2=600 m

Learn more about accelerated motion:

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4 years ago
At a processing facility outside of Detroit, a 7.0 kg box of goat cheese, initially at rest, is given a push up a smooth, inclin
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Answer:

d = 5.9 m

Explanation:

During the initial push, there are two forces acting on the box along  the ramp: the component of gravity force along the ramp (directed to the bottom of the ramp), and the pushing force (up the ramp).

As we have as a given, the time during which the pushing force is applied, we can use Newton's 2nd Law, expressed in its original form, as follows:

Fnet *Δt = Δp = m* (vf-v₀) (1)

Fnet, in this case, is the difference between Fapp, and the projection of Fg along the ramp, which is equal to Fg times the sinus of  the angle of the ramp with respect to the horizontal.

We choose as positive, the direction up the ramp, so we can write the follwing equation:

⇒ Fnet = Fapp - Fg*sin θ = 140 N - (7kg*9.8m/s²*sin 30º) = 140 N - 34.3 N

⇒ Fnet = 105.7 N

Replacing in (1):

105.7 N * 0.5 s = 7 kg* vf (v₀=0, as the box starts from rest)

Solving for vf:

vf = 105.7 N* 0.5 s / 7 kg = 7.6 m/s

When the push ends, the only force remaining along  the ramp, is the component of Fg that we have already obtained, that will cause the box to have a deceleration, which we can find out aplying Newton's 2nd Law, as follows:

m*g*sin 30º = m*a

As we have defined as positive direction the one up the ramp, a will be negative (as it is slowing down the box) , and can be calculated as follows:

a = -34.3 N / 7 kg = -4.9 m/s²

As this value is constant, we can use any kinematic equation in order to get  the distance traveled, farther the point where it disappeared the influence of the pushing force:

vf² - v₀² = 2*a*d

As we know that finally the box will come momentarily at rest (before falling under the influence of  gravity) , we have vf =0:

⇒ -v₀² = 2*a*d

For this part, v₀, is just the value for vf, that we got above:

v₀= 7.6 m/s

⇒ -(7.6)² =2*(-4.9 m/s²)*d

Solving finally for d (the answer we are looking for):

d = (7.6)² (m/s)² / 2*4.9 m/s² = 5.9 m

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Answer:

The circuit contains a filament bulb, connected with terminal to a terminal.

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