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lana [24]
3 years ago
6

Calculate the applied force to the piston with a 12cm radius required to elevate a weight of 2.0X104N by the piston with a 36cm

radius in a hydraulic lift.
2.9 × 103 N

5.0 × 103 N

6.7 × 103 N

2.2 × 103 N

Physics
1 answer:
Alex3 years ago
6 0

Answer:

Option D is the correct answer.

Explanation:

Refer the figure given.

By Pascal's principle we have

            \frac{F_1}{A_1}=\frac{F_2}{A_2}

F2 = 2 x 10⁴ N

A_1=\frac{\pi\times (12\times 10^{-3})^2}{4}=1.13\times 10^{-4}m^2\\\\A_2=\frac{\pi\times (36\times 10^{-3})^2}{4}=1.02\times 10^{-3}m^2

Substituting

      \frac{F_1}{1.13\times 10^{-4}}=\frac{2\times 10^4}{1.02\times 10^{-3}}\\\\F_1=2.22\times 10^3N

Option D is the correct answer.

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ASHA 777 [7]

Explanation:

a) The Earth makes 1 rotation in 24 hours.  In seconds:

24 hr × (3600 s / hr) = 86400 s

b) 1 rotation is 2π radians.  So the angular velocity is:

2π rad / 86400 s = 7.27×10⁻⁵ rad/s

c) The earth's linear velocity is the angular velocity times the radius:

40075 km × 7.27×10⁻⁵ rad/s = 2.91 km/s

7 0
3 years ago
Which statement is true about a planet’s orbital motion?
lana66690 [7]

Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

                                    V = \sqrt{\frac{GM}{R + h} }

Where

                   V - velocity of the orbit at a height h from the surface

                    R - Radius of the second object

                    G - Gravitational constant

                    h - height from the surface

The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

                         1/2 mV^{2} = GMm/R

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3 0
3 years ago
If the focal length of a concave mirror is 18cm, find its radius of curvature.
Galina-37 [17]

Given :

The focal length of a concave mirror is 18 cm.

To Find :

The radius of curvature of the concave mirror.

Solution :

We know,

\text{Focal length}=\dfrac{\text{Radius of curvature}}{2}\\\\F=\dfrac{R}{2}\\\\R = 18\times 2\ cm\\\\R = 36 \ cm

Therefore, the radius of curvature of concave mirror is 36 cm.

Hence, this is the required solution.

8 0
3 years ago
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value v_2  =  4 \sqrt{10} \  m/s

Explanation:

From the question we are told that

   The  charge on the first sphere is  q_1  =  2\mu C  =  2*10^{-6} \  C

    The charge on the second sphere is  q_2 =  8 \mu C = 8*10^{-6} \  C

     The  mass of the second charge is m  =  1.50 \  g  =  1.50 *10^{-3} \ kg

      The  distance apart is  d =  0.4 \  m

      The  speed of the second  sphere is  v_1  =  20 \  ms^{-1}

Generally the total energy possessed by when q_2 and  q_1 are separated by 0.8 \  m is mathematically represented

     Q =  KE + U

Here KE   is  the kinetic energy which is mathematically represented as

     KE  =  \frac{1 }{2}  m (v_1)^2

substituting value

     KE  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (20 )^2

     KE  =  0.3 \  J

And  U is  the  potential  energy which is mathematically represented as

        U  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.8 }

      U  =  0.18 \  J

So

       Q =  0.3 +  0.18

       Q =  0.48 \  J

Generally the total energy possessed by when q_2 and  q_1 are separated by 0.4 \  m is mathematically represented

         Q_f =  KE_f + U_f

Here KE_f is  the kinetic energy which is mathematically represented as

     KE_f  =  \frac{1 }{2}  m (v_2^2

substituting value

     KE_f  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (v_2 )^2

     KE_f  =  7.50 *10^{ -4} (v_2 )^2

And  U_f is  the  potential  energy which is mathematically represented as

        U_f  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U_f  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.4 }

      U_f  =  0.36 \  J

From the law of energy conservation

     Q =  Q_f

So

    0.48 =  0.36 +(7.50 *10^{-4} v_2^2)

   v_2  =  4 \sqrt{10} \  m/s

     

   

6 0
3 years ago
II. IDENTIFY THE TYPE OF FORCE
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Answer:

Explanation:

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5.When a person is pushing a trolley then object experience a normal reaction from ground.

6.Gravitational force makes the planet to move in their orbits.

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