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2 years ago

Acceleration is defined as the rate of change of position true or false

Physics

2 answers:

Svetradugi [14.3K]2 years ago

7 0

Answer:

True

Explanation:

Likurg_2 [28]2 years ago

6 0

The answer is : True

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You have a mixture of substances with diifferent sized molecules. How could you seperate them?

Trava [24]

Distillation, Magnetism, Filtration, Crystallization, Extraction,

3 0

3 years ago

Read 2 more answers

A -5.0 μC charge experiences a 11 i^ N electric force in a certain electric field. [Recall that i^ is a unit vector in the x-dir

Pachacha [2.7K]

Answer:

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Explanation:

given,

charge = -5.0 μC

Electric force, F = 11 i^ N

force would a proton experience = ?

we know

$\vec{F} = q \vec{E}$

$11\hat{i} = -5 \times 10^{-6} \vec{E}$

$\vec{E} =-2.2 \times 10^{6}\hat{i}$

we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

$\vec{F} = q \vec{E}$

$\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}$

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Force experienced by the photon in the same field is equal to $\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

3 0

3 years ago

A 65.0 kg diver is 4.90 m above the water, falling at speed of 6.40 m/s. Calculate her kinetic energy as she hits the water. (Ne

mojhsa [17]

Answer:

4452.5 J.

Explanation:

The diver have both kinetic and potential energy.

Ek = 1/2mv² ................. Equation 1

Where Ek = Kinetic Energy of the diver, m = mass of the diver, v = velocity of the diver.

Given: m = 65 kg, v = 6.4 m/s.

Substitute into equation 1

Ek = 1/2(65)(6.4²)

Ek = 1331.2 J.

Also,

Ep = mgh ............................ Equation 2

Where Ep = Potential energy of the diver when its above the water, h = height of the diver above the water, g = acceleration due to gravity.

Given: m = 65 kg, h = 4.9 m, g = 9.8 m/s²

Substitute into equation 2.

Ep = 65(4.9)(9.8)

Ep = 3121.3 J.

Note: When she hits the water, the potential energy is converted to kinetic energy.

E = Ek+Ep

Where E = Kinetic energy of the diver when she hits the water.

E = 1331.2+3121.3

E = 4452.5 J.

3 0

3 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

3 years ago

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