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3 years ago

12

What is the momentum of a 0.5 kg ball which is thrown at 20 m/s?

1 answer:

tatuchka [14]3 years ago

7 0

A =-16.7 Ns B= -167.N

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If 14:15=42:x find the value of x

Yuliya22 [10]

Answer:

x=45

Explanation:

by taking 42 and dividing it by 14 you get 1/3, because 14 is 1/3 of 42 you can then see that the ratio is multiplied by 3. si then you can just multiply 15 by 3 to get 45

3 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

A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal

Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, $a=g=-9.8\ m/s^2$ (Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

$S=ut+\frac{1}{2}at^2$

Plug in the given values and solve for 'u'. This gives,

$0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s$

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0

3 years ago

Assume that the equator of the Earth is 24,000 miles long (that’s its circumference). Now, pretend that you are standing somewhe

slega [8]

Answer:

1000 mph

Explanation:

P = Perimeter = 24000 mi

r = Radius of the equator

t = Time taken to complete one rotation = 24 h

Perimeter of a circle is given by

$P=2\pi r\\\Rightarrow r=\dfrac{P}{2\pi}\\\Rightarrow r=\dfrac{24000}{2\pi}\ mi$

Angular speed is given by

$\omega=\dfrac{2\pi}{t}\\\Rightarrow \omega=\dfrac{2\pi}{24}$

Velocity if given by

$v=r\omega\\\Rightarrow v=\dfrac{24000}{2\pi}\times \dfrac{2\pi}{24}\\\Rightarrow v=1000\ mph$

The person would be going at a speed of 1000 mph

3 0

3 years ago

What is the Potential Energy of ball 1??

Dvinal [7]

GPE= height x mass x gravitational field strength
5 x 10 x 9,8=490J

7 0

3 years ago