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4 years ago

10. A car starts from rest at a stop light and reaches 20 m/s in 3.5

Physics

2 answers:

adell [148]4 years ago

8 0

C
Acceleration= Final velocity-initial velocity
——————————————
Time

A=20 m/s
———- = 5.7 m/s2
3.5 s

liq [111]4 years ago

3 0

B i believe not really sure but that math i did seems about right

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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 right triangle ABC has sides /AB/ =9cm, /BC/=12cm find /AC/ and angles ACB and ABC if angle BAC= 90°

dezoksy [38]

fdi9bn 08989089089008i uuuuuuri9ij

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4 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

4 years ago

Describe how newtons third law is related to fluid pressure.

Hatshy [7]

Fish swimming forward in the water, the water gets pushed backward because the fish moving forward is forcing the water to move backward, the motion forward and backward are the same, they are opposite and equal.

3 0

4 years ago

Science is the process of pursuing knowledge?

tatyana61 [14]

yes science is the process of pursuing knowledge this is because without science we would not know about our brains and our brains help us peruse knowledge

4 0

3 years ago

Read 2 more answers