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Ierofanga [76]
3 years ago
14

'Suppose your hand moves upward by 0.50m while you are throwing the ball. The ball leaves your hand with an upward velocity of 2

0.0 m/s. Find the magnitude of the force (assumed constant) that your hand exerts on the ball. Ignore air resistance.' Now my question is not 'What is the magnitude' but rather: why did I get (roughly) the same answer using F=ma when you were supposed to use the total mechanical energy (W+K1+U1=K2+U2). So I'm more confused about how the 2 formulas are 'related', what the force actually represents in both, when to use what and if there is an actual difference.

Physics
1 answer:
REY [17]3 years ago
6 0

The formula F = ma is used when the statement of second law of motion is applied to the problem. If the statement is applied to the problem, The fomula will be used. If not, use the formula W +  K_{1} +  U_{2} =  K_{2} +  U_{2}

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The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in
mihalych1998 [28]

Answer:

a) Ra = 0.517 Ω

Rb = 0.032 Ω

Rc = 0.129 Ω

b) Ia = 5.8A

Ib = 93.75A

Ic = 23.2 A

Explanation:

a) The resistance is equal to:

Resistance for case a:

R_{a} =\frac{pL_{a} }{A_{a} } =\frac{p*4*L_{0} }{2L_{0}*L_{0}  } =\frac{2p}{L_{0} }

Where

p = 1.5x10⁻²Ωm

L0 = 5.8 cm = 0.058 m

R_{a} =\frac{2*1.5x10^{-2} }{0.058} =0.517ohm

Resistance for case b:

R_{b} =\frac{pL_{b} }{A_{b} } =\frac{pL_{0}}{2L_{0}4L_{0} } =\frac{p}{8L_{0}} =\frac{1.5x10^{-2} }{8*0.058} =0.032ohm

Resistance for case c:

R_{c} =\frac{pL_{c}}{A_{c} } =\frac{p2L_{0}}{L_{0}4L_{0}} =\frac{p}{2L_{0}} =\frac{1.5x10^{-2} }{2*0.058} =0.129ohm

b) The current is equal to:

Current for case a:

I_{a} =\frac{V}{R_{a} } =\frac{3}{0.517} =5.8A

Current for case b:

I_{b} =\frac{V}{R_{b} } =\frac{3}{0.032} =93.75A

Current for case c:

I_{c} =\frac{V}{R_{c} } =\frac{3}{0.129} =23.2A

4 0
3 years ago
If a refrigerator is heat pump that follows the first law of thermodynamics, how much heat was removed from food inside of the r
aliya0001 [1]
The <span>first law of thermodynamics</span><span> is a version of the law of </span>conservation of energy<span>, adapted for </span>thermodynamic systems<span>. The law of conservation of energy states that the total </span>energy<span> of an </span>isolated system<span> is constant; energy can be transformed from one form to another, but cannot be created or destroyed. so assuming no heat losses then heat removed is also 333 J</span>
5 0
3 years ago
What can subatomic particles be broken down into?
AfilCa [17]
Subatomic particles cannot be broken down any further
5 0
3 years ago
(n = 1.33). When visible light (400nm – 700 nm) shines normally on the film an observed from above, the bright reflected light l
Yuki888 [10]

Complete Question

A thin film in air is made by putting a thin layer of acetone (n = 1.25) on a layer of water

(n = 1.33). When visible light (400 nm – 700 nm) shines normally on the film an observed from above, the bright reflected light looks kind of “purplish” with red light of wavelength 650 nm mixed some blue giving it that appearance and “greenish” light of wavelength 520 nm is completely destroyed. Determine the thickness of the acetone film. (you only need to worry about the red and green wavelengths, not the blue)

Answer:

The thickness of acetone is  d = 104 nm

Explanation:

A diagram showing this process is shown on the first uploaded image

From the question we are told that

     The refractive index of acetone is n_a =1.25

     The refractive index of water is  n_w = 1.33

      The wavelength of the reflected light is  \lambda_r = 650nm =  650  *10^{-9}m

      The wavelength of the destroyed light is  \lambda_g = 520nm =  520 *10^{-9}m

       

Looking at the given data we can see that the

             n_a < n_w

This means that the light which the acetone-water layer would reflect will have a phase shift of \pi

  Again this make us to understand that the light reflected at the acetone layer will also have a phase shift of \pi

Since they would be having the same phase shift the two light would interfere

  For interference the condition for minima is mathematically represented as

             2 n_a d = (m + \frac{1}{2} ) \lambda_g

Where d is the thickness of acetone

                d = \frac{\lambda_g}{4 n_a}

Substituting values

              d = \frac{520 *10^{-9}}{4 * 1.25}

               d = 104 *10^{-9}m

               d = 104 nm

       

     

     

5 0
3 years ago
How does light reflect from the surface of carbon, copper wire, and the silicon lump? Does it appear metallic or nonmetallic (is
HACTEHA [7]

Answer:

When a light wave hits a metallic surface, the electrons on the surface are pushed and ... in a process called "total internal reflection", making that surface look shiny too.

Explanation:

metallic shiny nonmetallic dull:)

4 0
3 years ago
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