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dezoksy [38]
3 years ago
6

A mass of 250 N is on a piston of 2.0 m^2. What force is needed to lift this piston if the area of the second piston is 0.5 m^2?

Physics
1 answer:
prohojiy [21]3 years ago
8 0

Answer:

<h3>62.5N</h3>

Explanation:

The pressure at one end of the piston is equal to the pressure on the second piston.

Pressure = Force/Area

F1/A1 = F2/A2

Given

F1 = 250N

A1 = 2.0m²

A2 = 0.5m²

F2 = ?

Substituting the given values in the formula;

250/2 = F2/0.5

cross multiply

250*0.5 = 2F2

125 = 2F2

F2 = 125/2

F2 = 62.5N

Hence the  force needed to lift this piston if the area of the second piston is 0.5 m^2 is 62.5N

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NeTakaya

Answer:

Force on front axle = 6392.85 N

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Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

F_1 + F_2 = W

F_1 + F_2 = (1530\times 9.81)

F_1 + F_2 = 15009.3 N

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

F_1(1.15) = F_2(2.70 - 1.15)

F_1 = 1.35 F_2

now from above equation

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now we have

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now the other force is given as

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4 0
3 years ago
We have two solenoids: solenoid 2 has twice the diameter, half the length, and twice as many turns as solenoid 1. The current in
leva [86]

Answer:

the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.

Explanation:

Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:

B=\mu_0\, \frac{N}{L} I

Then, if we assign the subindex "1" to the quantities that define the magnetic field (B_1) inside solenoid 1, we have:

B_1=\mu_0\, \frac{N_1}{L_1} I_1

notice that there is no dependence on the diameter of the solenoid for this formula.

Now, if we write a similar formula for solenoid 2, given that it has :

1) half the length of solenoid 1 . Then L_2=L_1/2

2) twice as many turns as solenoid 1. Then N_2=2\,N_1

3) three times the current of solenoid 1. Then I_2=3\,I_1

we obtain:

B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1

5 0
3 years ago
What is the relationship between the normal force and weight
GalinKa [24]

Answer: they have the same magnitude.

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normal force = mg

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4 0
3 years ago
Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez
katrin [286]

Answer:

a= 4.4×10 m/s^2

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c= speed of light  = 3×10^8 m/s

P = 1000/ 3×10^8

P = 3.33×10^(-6) Pa

Force F = P×A

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F = 3.33×10^{-6}×6.65×10(-29)

= 2.22×10^{-6}

acceleration a  = F/m = 2.22×10^{-6}/ 5.10×10^{-27}

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4 0
3 years ago
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Answer:

false

Explanation:

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3 0
3 years ago
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