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stiv31 [10]
3 years ago
13

Are the small sized gears only force, speed and distance, only speed

Physics
1 answer:
sesenic [268]3 years ago
3 0
If the gears are of different sizes, they can be used to increase the power of a turning force. The smaller wheel turns more quickly but with less force, while the bigger one turns more slowly with more force. Cars and bicycles use gears to achieve amazing speeds our bodies could never match without help.

You might be interested in
A train traveling at 25 m/s is blowing its whistle at 440 Hz as it crosses a level crossing. You are waiting at the crossing and
ohaa [14]

Answer:

b) 472HZ, 408HZ

Explanation:

To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:

f_o=f\frac{v_s+v_o}{v_s-v}\\\\f_o=f'\frac{v_s-v_o}{v_s+v}\\\\

fo: frequency of the source = 440Hz

vs: speed of sound = 343m/s

vo: speed of the observer = 0m/s (at rest)

v: sped of the train

f: frequency perceived when the train leaves us.

f': frequency when the train is getTing closer.

Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:

f=f_o\frac{v_s-v}{v_s+v_o}=(440Hz)\frac{340m/s-25m/s}{340m/s}=408Hz\\\\f'=f_o\frac{v_s+v}{v_s-v_o}=(440Hz)\frac{340m/s+25m/s}{340m/s}=472Hz

hence, the frequencies for before and after tha train has past are

b) 472HZ, 408HZ

6 0
3 years ago
(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro
Kipish [7]

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:  

v = \frac{m + M}{m} \cdot \sqrt{2gh}

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.   </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:    

\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}           (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.  </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}  

\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}

\frac{v_{2}}{v_{1}} = 1.41  

Therefore, the second projectile was 1.41 times faster than the first.  

I hope it helps you!

8 0
3 years ago
A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp
Jobisdone [24]

Answer:

The tension is  T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by hinge   Fx= \frac{11}{4\sqrt{3} } Mg

Explanation:

   From the question we are told that

          The mass of the beam  is   m_b =M

          The length of the beam is  l = L

           The hanging mass is  m_h = 3M

            The length of the hannging mass is l_h = \frac{3}{4} l

            The angle the cable makes with the wall is \theta = 60^o

The free body diagram of this setup is shown on the first uploaded image

The force F_x \ \ and \ \ F_y are the forces experienced by the beam due to the hinges

      Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

     So

           \sum F =0

Now about the x-axis the moment is

              F_x -T cos \theta  = 0

     =>     F_x = Tcos \theta

Substituting values

            F_x =T cos (60)

                 F_x= \frac{T}{2} ---(1)

Now about the y-axis the moment is  

           F_y  + Tsin \theta  = M *g + 3M *g ----(2)

Now the torque on the system is zero because their is no rotation  

   So  the torque above point 0 is

          M* g * \frac{L}{2}  + 3M * g \frac{3L}{2} - T sin(60) * L = 0

            \frac{Mg}{2} + \frac{9 Mg}{4} -  T * \frac{\sqrt{3} }{2}    = 0

               \frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}

               T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }

                   T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by the hinge is

             F_x= \frac{T}{2} ---(1)

Now substituting for T

              F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}

                  Fx= \frac{11}{4\sqrt{3} } Mg

4 0
3 years ago
A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu
Setler79 [48]
<h2>Given :</h2>

  • total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

1.6 \times 10 {}^{ - 19} \:  C

For producing 1 Cuolomb charge we need :

  • \mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }

  • \dfrac{10 {}^{19} }{1.6}

  • \dfrac{10\times 10 {}^{19} }{16}

  • \dfrac{100 \times 10 {}^{18} }{16}

  • \mathrm{6.24 \times 10 {}^{18}  \:  \: electrons}

Now, for producing 0.009 C of charge, the number of electrons required is :

  • 0.009 \times 6.24 \times  {10}^{18}

  • 0.05616 \times 10 {}^{18}

  • \mathrm{5.616 \times 10 {}^{16}  \:  \: electons}

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

\mathrm{5.616 \times 10 {}^{16} \:  \: electrons}

Number of electrons passing through it in 1 Second is :

  • \dfrac{5.616 \times  {10}^{16} }{3.6}

  • \mathrm{1.56 \times 10 {}^{16}  \:  \: electrons}

Now, in 10 seconds the number of electrons passing through it is :

  • 10 \times  \mathrm{1.56 \times 10 {}^{16}  \:  \: }

  • \mathrm{1.56 \times 10 {}^{17}  \:  \: electrons}

_____________________________

\mathrm{ \#TeeNForeveR}

6 0
2 years ago
Argelia has a stack of schoolbooks sitting in the backseat of her car. When Argelia makes a sharp right turn, the books slide to
Kaylis [27]

Answer:

The books slide to the left due to inertia, and then they come to a rest due to the force applied by the car door.

Explanation:

The entire motion of the books can be explained by using the first two laws of Newton:

- 1st Newton's law (also called Law of Inertia): an object at rest stays at rest and an object in motion stays in motion with constant velocity if no unbalanced forces act on it

- 2nd Newton's law: when an object is acted upon unbalanced forces, an acceleration is induced in the motion of the object, according to the equation

F=ma

where F is the net force on the object, m is its mass and a is its acceleration.

Coming back to our problem:

- At the beginning, Argelia makes a sharp turn right. Due to the law of inertia, the books (which are not fixed to the car) continue their motion straight as it was before the curve: so, since the car is moving right, they appear to go the left of the car.

- When the books hit the car door, they stop moving due to the 2nd Newton's law: in fact, the car door applies an unbalanced force against the books, and as a result the books have a negative acceleration (=deceleration), so they slow down and eventually they stop.

4 0
3 years ago
Read 2 more answers
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