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dybincka [34]
3 years ago
8

In which of the following situation is work being done?

Physics
1 answer:
Afina-wow [57]3 years ago
6 0

A wagon is pulled at an angle of 30 degrees to the horizontal.

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A coin and feather are dropped in a moon. what will fall earlier on ground.give reasons.if they are dropped in the earth,which o
Sergeeva-Olga [200]

Answer:

  • on the moon, they will fall at the time
  • on earth, the coin will fall faster to the ground

Explanation:

A coin and feather dropped in a moon experience the same acceleration due to gravity as small as 1.625 m/s², and because of the absence of air resistance both will fall at the same rate to the ground.

If the same coin and feather are dropped in the earth, they will experience the same acceleration due to gravity of 9.81 m/s² and because of the presence of air resistance, the heavier object (coin) will be pulled faster to the ground by gravity than the lighter object (feather).

4 0
3 years ago
Four objects are situated along the y axis as follows: a 1.99-kg object is at 2.99 m, a 2.96-kg object is at 2.57 m, a 2.43-kg o
Dominik [7]

Answer:

The center of mass for the object is  y_c = 1.063 \  m from the origin

Explanation:

From the question we are told that

   The mass of the first object is  m_1 =  1.99 \  kg

   The position of first object with respect to origin y_1 =  2.99 \ m

   The mass of the second object is  m_2 =  2.96 \  kg

   The position of second object with respect to origin y_2 =  2.57 \ m

   The mass of the third object is  m_3 =  2.43  \  kg

   The position of third object with respect to origin y_3 =  0 \ m

   The mass of the fourth object is  m_3 =  3.96  \  kg

   The position of fourth object with respect to origin y_3 =  -0.502  \ m

Generally the center of mass of the object along the x-axis is  zero  because all the mass lie on the y axis

Generally the location of the center mass of the object is mathematically represented as

    y_c = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3 + m_4 * y_4}{m_1 + m_2 + m_3 + m_4}

=>y_c = \frac{1.99 * 2.99 + 2.96 * 2.57 + 2.43 * 0 + 3.96 * (-0.502)}{1.99+ 2.96  + 2.43 + 3.96}

=>y_c = 1.063 \  m

3 0
3 years ago
During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t
leonid [27]

The instant it was dropped, the ball had zero speed.

After falling for 1 second, its speed was 9.8 m/s straight down (gravity).

Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.

Falling for 1 second at an average speed of 4.9 m/s, is covered <em>4.9 meters</em>.

ANYTHING you drop does that, if air resistance doesn't hold it back.

7 0
3 years ago
Read 2 more answers
f your risk-aversion coefficient is A = 4 and you believe that the entire 1926–2015 period is representative of future expected
siniylev [52]

Answer:

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

Explanation:

As the complete question is not given here ,the table of data is missing which is as attached herewith.

From the maximized equation of the utility function it is evident that

Weight=\frac{E_M-r_f}{A\sigma_M^2}

For the equity, here as

  • Weight is percentage of the equity which is to be calculated
  • {E_M-r_f} is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
  • A is the risk aversion factor which is given as 4.
  • \sigma_M is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59

By substituting values.

Weight=\frac{E_M-r_f}{A\sigma_M^2}\\Weight=\frac{8.30\%}{4(20.59\%)^2}\\Weight=0.4894 =48.94\%

So the weight of equity is 48.94%.

Now the weight of T bills is given as

Weight_{T-Bills}=1-Weight_{equity}\\Weight_{T-Bills}=1-0.4894\\Weight_{T-Bills}=0.5105=51.05\%\\

So  the weight of T-bills is 51.05%.

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

7 0
3 years ago
Water (2510 g ) is heated until it just begins to boil. if the water absorbs 5.01×105 j of heat in the process, what was the ini
natka813 [3]
E=energy=5.09x10^5J = 509KJ 
<span>M=mass=2250g=2.25Kg </span>
<span>C=specific heat capacity of water= 4.18KJ/Kg </span>
<span>ΔT= change in temp= ? </span>
<span>E=mcΔT </span>
<span>509=(2.25)x(4.18)xΔT </span>
<span>509=9.405ΔT </span>
<span>ΔT=509/9.405=54.1degrees </span>
<span>Initial temp = 100-54 = 46 degrees </span>
<span>Hope this helps :)</span>
5 0
3 years ago
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