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Pavel [41]
3 years ago
12

In a pneumatic lift, compressed air exerts a force on a piston with a radius of 10.00 cm. This pressure is transmitted to a seco

nd piston with a radius of 25.0 cm. This second piston lifts a crate. Ignoring the weight of the pistons, how large a force must the compressed air exert to lift a crate that is 66.5 N?
Physics
1 answer:
vitfil [10]3 years ago
7 0

Answer:

F₁ = 10.64 N

Explanation:

The pressure is transmitted equally from the first piston to the second piston:

P_1=P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{\pi r_1^2}=\frac{F_2}{\pi r_2^2}\\\\\frac{F_1}{r_1^2}=\frac{F_2}{r_2^2}\\\\F_1 = (\frac{F_2}{r_2^2})(r_1^2)

where,

F₁ = Force applied by compressed air on first piston = ?

F₂ = Weight lifted by second piston = 66.5 N

r₁ = radius of first piston = 10 cm

r₂ = radius of second piston = 25 cm

Therefore,

F_1 = \frac{66.5\ N}{(25\ cm)^2} (10\ cm)^2\\\\

<u>F₁ = 10.64 N</u>

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Marrrta [24]

The gravitation force is quartered when two objects' masses are halved without changing their distance.

Gravitational law states that the force of attraction and repulsion between two objects is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.

F=(KM1 M2)/r^2

K= Gravitation force constant

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r = distance between objects

When M1 and M2 are halved, it becomes M1/2 and M2/2

F=(K M1/2 x M2/2)/r^2

F=(K (M1 x M2)/4)/r^2

F=(KM1 x M2)/(4r^2 )

Recall

F=(KM1 x M2)/r^2

Therefore

F=F/4

Learn more about gravitational force here:

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7 0
2 years ago
A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni
mina [271]

Answer:

Induced emf, \epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}

Explanation:

The varying magnetic field with time t is given by according to equation as :

B=B_{max}e^{-t/\tau}

Where

B_{max}\ and\ t are constant

Let \epsilon is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}

\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}

So, the induced emf in the loop as a function of time is A\dfrac{B_{max}e^{-t/\tau}}{\tau}. Hence, this is the required solution.

7 0
3 years ago
A 25-kg child sits at the top of a 4-meter slide. After sliding down, the child is traveling at 5 m/s. How much PE does he start
Semmy [17]

Daniddmelo says it right there, don't know why he got reported.

The potential energy (PE) is mass x height x gravity. So it would be 25 kg x 4  m x 9.8 = 980 joules. The child starts out with 980 joules of potential energy. The kinetic energy (KE) is (1/2) x mass x velocity squared. KE = (1/2) x 25 kg x 5 m/s2 = 312.5 joules. So he ends with 312.5 joules of kinetic energy. The Energy lost to friction =  PE - KE. 980- 312.5 = 667.5 joules of energy lost to friction.

Please don't just copy and paste, and thank you Dan cause you practically did it I just... elaborated more? I dunno. 

4 0
3 years ago
the average speed of a runner in a 483. meter race is 3.0 meters per second. How long me runner to complete the race? Dont inclu
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Answer:

161

Explanation:

v=\frac{d}{t} slove for t

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Insert values of d and v

t=\frac{483}{3} \\

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3 0
3 years ago
What minerals attracts iron nails?
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3 0
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