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inysia [295]
3 years ago
13

Swinging a tennis racket against a ball is an example of a third class lever. Please select the best answer from the choices pro

vided. T F
Physics
2 answers:
kari74 [83]3 years ago
7 0
that statement is true

a Third class lever applied when the effort place between the load and the fulcrum.

For example, in a forearm serve
Fulcrum : The elbow
Effort : The effort that putted by the biceps muscle
Load : The arm

Kobotan [32]3 years ago
4 0

Answer:

The given statement is true.

Explanation:

One of the examples of a simple machine is Levers. It reduces the human effort to do work and make work easier. There are three classes of the lever.

The given case is "swinging a tennis racket against a ball". It is an example of the third class lever. In this type of class, the effort is placed in between the load and the fulcrum.

Hence, the swinging a tennis racket against a ball is an example of a third class lever. This statement is true.

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Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
solmaris [256]

Answer:

a

\lambda = 3.68 *10^{-36} \  m

b

\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

Here  h is the Planck constant with the value

      h = 6.62607015 * 10^{-34} J \cdot s

So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

=> \lambda = 3.68 *10^{-36} \  m

Generally the energy of the proton is mathematically represented as

         E_p =  \frac{1}{2}  *   m_p  *  v^2_p

Here m_p  is the mass of proton with value  m_p  =  1.67 *10^{-27} \  kg

=>     8.0*10^{-13} =  \frac{1}{2}  *   1.67 *10^{-27}  *  v^2

=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

5 0
3 years ago
An object has a mass of 6kg. calculate it's gpe​
m_a_m_a [10]

Explanation:

When m=<em>mass</em>

G=<em>a</em><em>c</em><em>c</em><em>e</em><em>l</em><em>e</em><em>r</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>d</em><em>u</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>gravity</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>H</em><em>=</em><em>h</em><em>e</em><em>i</em><em>g</em><em>h</em><em>t</em>

<em>U</em><em>s</em><em>i</em><em>n</em><em>g</em><em> </em><em>f</em><em>o</em><em>r</em><em>m</em><em>u</em><em>l</em><em>a</em>

<em>M</em><em>g</em><em>h</em>

<em>(</em><em>M</em><em>=</em><em>6</em><em>, </em><em>g</em><em>=</em><em>10</em><em>,</em><em>h</em><em>=</em><em>?</em><em>) </em>

6×10×h

=60joules

7 0
3 years ago
Both slicing a tomato and a chemical change such as burning toast cannot be reversed. however why is slicing a tomato sill consi
kondaur [170]
Slicing a tomato is still considered a physical change because you are only altering the shape of it by slicing it. It does not change color or make bubble when slicing the tomato. Therefore, it is still a physical change
hope this helps:)
4 0
3 years ago
A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.1
Stolb23 [73]

Answer:

it could be 69

Explanation:

because it is a solution

4 0
3 years ago
A circular coil 14.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this locati
sp2606 [1]

Answer:

a)T = 2.9*10^{-5} N-m

b) north edge will rise up

Explanation:

torque on the coil is given as

T = NIABsin\theta

where N is number of loop =  9 loops

i is current = 7.80 A

-B -earth magnetic field = 5.00*10^{-5} T

A- area of circular coil

A = \frac{\pi d^{2}}{4}

A =\frac{\pi .14^{2}}{4}

A =0.015 m2

PUTITNG ALL VALUE TO GET TORQUE

T = 9*7.8*0.015*5*10^{-5} sin{90-56}

T = 2.9*10^{-5} N-m

b) north edge will rise up

3 0
3 years ago
Read 2 more answers
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