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2

Find an expression for the displacement?

A08
Featured 5 months ago

Given expression for velocity of a particle by

v=(t−2)\ ms^-1

Comparing with kinematic expression

$v = u + a t$ ...............(1)

we get

$u = - 2 \setminus m {s}^{-} 1 \mathmr{and} a = 1 \setminus m {s}^{-} 2$

(1). Kinematic expression for displacement is

$s = {s}_{0} + u t + \frac{1}{2} a {t}^{2}$

Inserting given conditions we get the expression for displacement $r$ as

$r = {r}_{0} + \left(- 2\right) t + \frac{1}{2} \left(1\right) {t}^{2}$
$r = {r}_{0} - 2 t + \frac{1}{2} {t}^{2}$

To find ${r}_{0}$, use the given condition at $t = 6 \setminus s$

$r \left(6\right) = 11 = {r}_{0} - 2 \left(6\right) + \frac{1}{2} {\left(6\right)}^{2}$
$\implies 11 = {r}_{0} - 12 + 18$
$\implies {r}_{0} = 5 \setminus m$

$\therefore$ Expression for $r$ becomes

$r = 5 - 2 t + \frac{1}{2} {t}^{2}$ ........(2)

(2). From (2)

$r \left(8\right) = 5 - 2 \times 8 + \frac{1}{2} {\left(8\right)}^{2}$
$\implies r \left(8\right) = 5 - 16 + 32$
$\implies r \left(8\right) = 21 \setminus m$

(3). From (2)

$53 = 5 - 2 t + \frac{1}{2} {t}^{2}$
$\implies {t}^{2} - 4 t - 96 = 0$

Solving the quadratic by split the middle term we get

$t = - 8 \mathmr{and} 12$,

Ignoring $- v e$ root as time can not be negative and inserting in (1) we get

$v \left(12\right) = - 2 + 1 \times 12 = 10 \setminus m {s}^{-} 1$

3

What volume of water at 0°C can a freezer make into ice cubes in 1.0 hour,if the coefficient of performance of the cooling unit is 7.0 and the power input is 1.0 kilowatt?

Jane
Featured 5 months ago

$8.82 L$

Explanation:

Power input of the refrigerator is $1 k W$ means per second is supplies $1 \cdot {10}^{3} J$ of energy

So,in $1 h r$ it will supply ${10}^{3} \cdot 3600 = 36 \cdot {10}^{5} J = 8.56 \cdot {10}^{5}$ calories of heat energy.

Now,let the refrigerator will take the temperature of water to ${T}_{2}$ and the water is at ${T}_{1}$ temperature.

Given, $\beta = 7 , {T}_{1} = 0 + 273 = 273 K$

So,using the equation, $\beta = \frac{{T}_{2}}{{T}_{1} - {T}_{2}}$

We,get, ${T}_{2} = 238.875 K = - {34.125}^{\circ} C$

Now,heat energy required for conversion of $m$ $g r a m$ of water at ${0}^{\circ} C$ to same amount of ice at ${0}^{\circ} C$ i.e the latent heat required is $80 m$ (as,latent heat of water is $80$calories per gram)

So,to take this $m$ $g r a m$ of ice at ${0}^{\circ} C$ to $- {34.125}^{\circ} C$ heat energy required will be $m \cdot 0.5 \cdot \left(0 - \left(- 34.125\right)\right)$ calories (using the formula, heat required $H = m s d \theta$,where, $s$ is the specific heat,for ice it is $0.5$ C.G.S units and $d \theta$ is the change in temperature)

So,this entire amount of heat will be equals to $8.56 \cdot {10}^{5}$ calories

So, $80 m + m \cdot 0.5 \cdot 34.125 = 8.56 \cdot {10}^{5}$

we get,from the above equation, $m = 8819.06$ $g r a m$

Now,density of water is $1$ $\frac{g r a m}{c {m}^{3}}$

So,volume of this much water will be $\frac{8819.06}{1} = 8819.06 c {m}^{3}$ (as, $v = \frac{m}{d}$)

Now, $1 c {m}^{3} = 0.001 L$

so, $8819.06 c {m}^{3} = 8.81906 = 8.82 L$

2

A hot coffee at 80 c is left in a room at 20 c.it is found that after 20 minutes in the room, the temperature of coffee has decreased by 20 c. Determine the temperature of coffee after 30 minutes?

Nam D.
Featured 4 months ago

I get ${53}^{\circ} \text{C}$ after $30$ minutes have passed.

Explanation:

We use Newton's Law of Cooling, which states that

$\frac{\mathrm{dT}}{\mathrm{dt}} = K \left(T - {T}_{0}\right)$

Solving the differential equation, we get $T = {T}_{0} + A {e}^{K t}$.

where $t$ is the time in minutes, ${T}_{0}$ is the surrounding temperature, $T$ is the object temperature, and $K$ is the constant (varies).

Plugging in the values, we need to find $K$ and $A$ first.

We got:

${T}_{0} = 20$

$\therefore T = 20 + A {e}^{K t}$

When $t = 0 , T = 80$, and so

$80 = 20 + A {e}^{0}$

$80 = 20 + A$

We get:

$A = 60$

The equation becomes:

$\therefore T = 20 + 60 {e}^{K t}$

Now, we can find $K$.

Decreasing by ${20}^{\circ} \text{C}$, the new temperature is therefore $80 - 20 = {60}^{\circ} \text{C}$.

$60 = 20 + 60 {e}^{20 K}$

${e}^{20 K} = \frac{40}{60}$

$= \frac{2}{3}$

$20 K = \ln \left(\frac{2}{3}\right)$

$K = \frac{\ln \left(\frac{2}{3}\right)}{20}$

$\approx - 0.02$

So, the final equation is:

$T = 20 + 60 {e}^{- 0.02 t}$

After $30$ minutes, we plug in $t = 30$, and we get

$T = 20 + 60 {e}^{- 0.02 \cdot 30}$

$= 20 + 60 {e}^{-} 0.6$

$= 20 + 32.9$

$= 52.9$

$\approx {53}^{\circ} \text{C}$

So, the temperature after half an hour will be approximately $53$ degrees celsius.

2

A tank open on top has a depth h and is filled with water. A hole is made three-quarter from the ground and another hole is made half the depth. What is the difference in the range of water ejected through these holes?

Mr. Mike
Featured 3 months ago

The difference in range will be the height of the level of water in the tank times $\left(1 - \frac{\sqrt{3}}{2}\right)$.

Explanation:

You really want to use Torricelli's Law here.

$v = \sqrt{2 g \left(h - x\right)}$

where

$v$ = the horizontal velocity of the stream of water,

$g$ = the acceleration of gravity, and

$h$ = the vertical distance from the surface of the water to the ground, and

$x$ = is the vertical height of the hole from the ground (see diagram).

The range of the stream is the horizontal distance the stream lands away from the tank. For the halfway hole, I have labeled this ${R}_{1}$, and for the 3/4 hole, I have labeled this ${R}_{2}$.

How long will it take a water droplet to reach the ground once it has left the hole? Since it's vertical velocity component is controlled by gravity, the time it takes to land on the ground, $t$, will be

$t = \sqrt{\frac{2 x}{g}}$.

For the halfway hole, $x = \frac{h}{2}$, and for the 3/4 hole, $x = \frac{3 h}{4}$.

Now we need to figure out the horizontal distance that the drop will travel in that time. It will simply be its horizontal velocity times the time.

$R = v t$

We can find $v$ from Torricelli's Law and $t$ from above, so we will have

$R = \sqrt{2 g \left(h - x\right)} \sqrt{\frac{2 x}{g}} = 2 \sqrt{\left(h - x\right) x}$

The problem statement asks us to determine

R_1-R_2=2sqrt((h-h/2)h/2)-2sqrt((h-(3h)/4)(3h)/4

$= 2 h \left[\sqrt{\left(1 - \frac{1}{2}\right) \frac{1}{2}} - \sqrt{\left(1 - \frac{3}{4}\right) \frac{3}{4}}\right]$

$= 2 h \left(\frac{1}{2} - \frac{\sqrt{3}}{4}\right) = h \left(1 - \frac{\sqrt{3}}{2}\right)$

2

A great conical mound of height h is built by workers. If the workers simply heap up uniform material found at ground level, and the total weight of the finished mound is M, show that the work they do is (1/4)hM?

P dilip_k
Featured 4 months ago

Let the radius of the great conical mound of height $h$ built by workers be $r$.

Given that the weight of the finished mound is $M$. So the weight of finished mound per unit volume will be $m = \frac{M}{\frac{1}{3} \pi {r}^{2} h}$.

Now for the sake of our calculation let us consider the center of the circular base of the the mound $\left(O\right)$ as origin with diameter of the mound lying along X-axis and height along Y-axis.

It is obvious that the rate of decrease of radius of the conical mound with height will be given by $\frac{r}{h}$.

Hence at an arbitrary height $y$ its radius will be $\left(r - \frac{r y}{h}\right) = r \left(1 - \frac{y}{h}\right)$

So the volume of an imaginary circular disk of infinitesimal thickness $\mathrm{dy}$ at this height $y$ will be given by

$\mathrm{dv} = \pi {r}^{2} {\left(1 - \frac{y}{h}\right)}^{2} \mathrm{dy}$

So weight of this thin disk will be

$= m \cdot \mathrm{dv} = \frac{M}{\frac{1}{3} \pi {r}^{2} h} \cdot \pi {r}^{2} {\left(1 - \frac{y}{h}\right)}^{2} \mathrm{dy}$

$= \frac{3 M}{h} \cdot {\left(1 - \frac{y}{h}\right)}^{2} \mathrm{dy}$

So work done against gravitational pull for lifting this imaginary thin disk to a height of $y$ will be given by

$\mathrm{dw} = \text{weight"(mdv)xx"height} \left(y\right) = \frac{3 M}{h} \cdot {\left(1 - \frac{y}{h}\right)}^{2} \mathrm{dy} \cdot y$

The total work done $W$ to finish the mound will be obtained by integrating this $\mathrm{dw}$ as follows where $y$ varies from $0 \to h$

$W = \frac{3 M}{h} {\int}_{0}^{h} {\left(1 - \frac{y}{h}\right)}^{2} y \mathrm{dy}$

$\implies W = \frac{3 M}{h} {\left[{y}^{2} / 2 - \frac{2 {y}^{3}}{3 h} + {y}^{4} / \left(4 {h}^{2}\right)\right]}_{0}^{h}$

$\implies W = \frac{3 M}{h} \left[{h}^{2} / 2 - \frac{2 {h}^{3}}{3 h} + {h}^{4} / \left(4 {h}^{2}\right)\right]$

$\implies W = \frac{3 M}{h} \cdot \frac{1}{12} \left[6 {h}^{2} - 8 {h}^{2} + 3 {h}^{2}\right]$

$\implies W = \frac{3 M}{h} \cdot \frac{1}{12} \cdot {h}^{2}$

$\implies W = \frac{1}{4} h M$

Proved

Alternative method

Considering the center of mass of the conical mound which is placed at $\frac{1}{4} h$ we can calculate the work done in simpler way.

So work done to heap up uniform material found at ground level which is equivalent to lift the COM to a height of $\frac{h}{4}$ from ground will be

$W = M \times \frac{h}{4} = \frac{1}{4} h M$

2

A charge of 2 μC is located 48 cm above the origin. A second charge of -8 μC is located 48 cm below the origin. What is the magnitude and direction of the net electric field 20cm to the right of the origin?

Michael
Featured 4 months ago

$\textsf{E = 316.7 \times {10}^{3} \textcolor{w h i t e}{x} \text{N/C}}$ at an angle of $\textsf{{75.9}^{\circ}}$ below the horizontal or bearing $\textsf{{284.1}^{\circ}}$

Explanation:

The electric field at a particular point is the force experienced by a test charge of +1 Coulomb.

I will assign a charge of +1C at the location specified and find the net force that it experiences. The electric field is found by dividing that force by unit charge.

I will show 2 ways that you can do this.

(1)

Coulombs Law gives us:

$\textsf{F = k . \frac{{q}_{1} {q}_{2}}{r} ^ 2}$

$\textsf{k = 9.00 \times {10}^{9} \textcolor{w h i t e}{x} N . {m}^{2} \text{/} {C}^{2}}$

Here are the forces acting:

From Pythagoras we get:

$\textsf{{r}^{2} = {48}^{2} + {20}^{2}}$

$\textsf{r = 52 \textcolor{w h i t e}{x} c m = 0.52 \textcolor{w h i t e}{x} m}$

$\therefore$$\textsf{{F}_{1} = \frac{9 \times {10}^{9} \times 2 \times {10}^{- 6} \times 1}{0.52} ^ 2 = 66.57 \times {10}^{3} \textcolor{w h i t e}{x} N}$

$\textsf{{F}_{1} = 66.57 \textcolor{w h i t e}{x} k N}$

By inspection you can see that $\textsf{{F}_{2}}$ must be 4 x this:

$\textsf{{F}_{2} = 66.57 \times 4 = 266.27 \textcolor{w h i t e}{x} k N}$

Now we find the resultant of these two forces.

We have a S ide A ngle S ide triangle ABC so we can apply The Cosine Rule:

$\textsf{{a}^{2} = {b}^{2} + {c}^{2} - 2 \text{bc} \cos A}$

From the diagram you can see that $\textsf{\tan \theta = \frac{48}{20} = 2.4}$

From which $\textsf{\theta = {67.38}^{\circ}}$

$\textsf{A = 2 \theta = 2 \times {67.38}^{\circ} = {134.76}^{\circ}}$

$\therefore$$\textsf{{F}_{r e s}^{2} = {66.57}^{2} + {266.27}^{2} - 2 \times 66.57 \times 266.27 \cos \left(134.76\right)}$

$\textsf{{F}_{r e s}^{2} = 4 , 431.5 + 70 , 899.7 - \left(- 24 , 962.5\right)}$

$\textsf{{F}_{r e s} = 316.69 \textcolor{w h i t e}{x} k N = 316.69 \times {10}^{3} \textcolor{w h i t e}{x} N}$

$\therefore$sf(color(red)(E=(316.69xx10^3)/(1)color(white)(x)"N/C")

To find angle C we can use The Sine Rule:

$\textsf{\frac{a}{\sin} A = \frac{c}{\sin} C}$

$\therefore$$\textsf{\frac{316.69}{\sin {134.76}^{\circ}} = \frac{266.7}{\sin} C}$

$\textsf{\sin C = \frac{266.27}{446.0} = 0.597}$

$\textsf{C = {36.65}^{\circ}}$

$\textsf{{36.65}^{\circ} + {134.76}^{\circ} + B = {180}^{\circ}}$

$\textsf{B = {8.58}^{\circ}}$

The angle the resultant makes with the horizontal is therefore sf(67.38^@+8.56^@=color(red)(75.94^@)

(2)

We can resolve $\textsf{{F}_{1}}$ and $\textsf{{F}_{2}}$ into their horizontal and vertical components which can be added and resolved using Pythagoras.

$\textsf{{F}_{x} = {F}_{1} \cos \theta - {F}_{2} \cos \theta}$

$\textsf{{F}_{1} = 66.57 \times 0.3846 - 266.27 \times 0.3846 \textcolor{w h i t e}{x} k N}$

$\textsf{{F}_{x} = 25.603 - 102.407 = 76.80 \textcolor{w h i t e}{x} k N}$

$\textsf{{F}_{y} = {F}_{1} \sin \theta + {F}_{2} \sin \theta}$

$\textsf{{F}_{y} = 66.57 \times 0.92307 + 266.27 \times 0.92307 \textcolor{w h i t e}{x} k N}$

$\textsf{{F}_{y} = 61.449 + 245.785 = 307.23 \textcolor{w h i t e}{x} k N}$

Now we can apply Pythagoras:

$\textsf{{F}_{r e s}^{2} = {76.80}^{2} + {307.23}^{2}}$

$\textsf{{F}_{r e s} = \sqrt{100 , 288.5} = 316.68 \textcolor{w h i t e}{x} k N}$

$\therefore$sf(color(red)(E=(316.68xx10^3)/(1)color(white)(x)"N/C")

$\textsf{\tan \alpha = {307.23}^{\circ} / {76.8}^{\circ} = 4}$

sf(color(red)(alpha=75.96^@)

As you can see there is close agreement between the 2 methods so thats all good.

2

What is the internal energy for an isothermal process?

Truong-Son N.
Featured 4 months ago

$\Delta U = {\int}_{{V}_{1}}^{{V}_{2}} {\left(\frac{\partial U}{\partial V}\right)}_{T} \mathrm{dV} = {\int}_{{V}_{1}}^{{V}_{2}} - P + T {\left(\frac{\partial P}{\partial T}\right)}_{V} \mathrm{dV}$

Now decide what gas law to use, or what $\alpha$ and $\kappa$ corresponds to your substance.

Well, from the total differential at constant temperature,

$\mathrm{dU} = {\cancel{{\left(\frac{\partial U}{\partial T}\right)}_{V} \mathrm{dT}}}^{0} + {\left(\frac{\partial U}{\partial V}\right)}_{T} \mathrm{dV}$,

so by definition of integrals and derivatives,

$\Delta U = {\int}_{{V}_{1}}^{{V}_{2}} {\left(\frac{\partial U}{\partial V}\right)}_{T} \mathrm{dV}$ $\text{ } \boldsymbol{\left(1\right)}$

The natural variables are $T$ and $V$, which are given in the Helmholtz free energy Maxwell relation.

$\mathrm{dA} = - S \mathrm{dT} - P \mathrm{dV}$$\text{ } \boldsymbol{\left(2\right)}$

This is also related by the isothermal Helmholtz relation (similar to the isothermal Gibbs' relation):

$\mathrm{dA} = \mathrm{dU} - T \mathrm{dS}$$\text{ } \boldsymbol{\left(3\right)}$

Differentiating $\left(3\right)$ at constant temperature,

${\left(\frac{\partial A}{\partial V}\right)}_{T} = {\left(\frac{\partial U}{\partial V}\right)}_{T} - T {\left(\frac{\partial S}{\partial V}\right)}_{T}$

From $\left(2\right)$,

${\left(\frac{\partial A}{\partial V}\right)}_{T} = - P$

and also from $\left(2\right)$,

${\left(\frac{\partial S}{\partial V}\right)}_{T} = {\left(\frac{\partial P}{\partial T}\right)}_{V}$

since the Helmholtz free energy is a state function and its cross-derivatives must be equal. Thus from $\left(3\right)$ we get

$- P = {\left(\frac{\partial U}{\partial V}\right)}_{T} - T {\left(\frac{\partial P}{\partial T}\right)}_{V}$

or we thus go back to $\left(1\right)$ to get:

$\overline{\underline{|}} \stackrel{\text{ ")(" "DeltaU = int_(V_1)^(V_2) ((delU)/(delV))_TdV = int_(V_1)^(V_2)-P + T((delP)/(delT))_VdV" }}{|}$

And what remains is to distinguish between the last term for gases, liquids and solids...

GASES

Use whatever gas law you want to find it. If for whatever reason your gas is ideal, then

${\left(\frac{\partial P}{\partial T}\right)}_{V} = \frac{n R}{V}$

and that just means

${\left(\frac{\partial U}{\partial V}\right)}_{T} = - P + \frac{n R T}{V}$

$= - P + P = 0$

which says that ideal gases have changes in internal energy as a function of only temperature. One would get

$\textcolor{b l u e}{\Delta U = {\int}_{{V}_{1}}^{{V}_{2}} 0 \mathrm{dV} = 0}$.

Not very interesting.

Of course, if your gas is not ideal, this isn't necessarily true.

LIQUIDS AND SOLIDS

These data are tabulated as coefficients of volumetric thermal expansion $\alpha$ and isothermal compressibility $\kappa$,

$\alpha = \frac{1}{V} {\left(\frac{\partial V}{\partial T}\right)}_{P}$

$\kappa = - \frac{1}{V} {\left(\frac{\partial V}{\partial P}\right)}_{T}$

$\frac{\alpha}{\kappa} = \left[. . .\right] = {\left(\frac{\partial P}{\partial T}\right)}_{V}$

at VARIOUS temperatures for VARIOUS condensed phases. Some examples at ${20}^{\circ} \text{C}$:

• ${\alpha}_{{H}_{2} O} = 2.07 \times {10}^{- 4} {\text{K}}^{- 1}$
• ${\alpha}_{A u} = 4.2 \times {10}^{- 5} {\text{K}}^{- 1}$ (because that's REAL useful, right?)
• ${\alpha}_{E t O H} = 7.50 \times {10}^{- 4} {\text{K}}^{- 1}$
• ${\alpha}_{P b} = 8.7 \times {10}^{- 5} {\text{K}}^{- 1}$

• ${\kappa}_{{H}_{2} O} = 4.60 \times {10}^{- 5} {\text{bar}}^{- 1}$

• ${\kappa}_{A u} = 5.77 \times {10}^{- 7} {\text{bar}}^{- 1}$
• ${\kappa}_{E t O H} = 1.10 \times {10}^{- 4} {\text{bar}}^{- 1}$
• ${\kappa}_{P b} = 2.33 \times {10}^{- 6} {\text{bar}}^{- 1}$

In that case,

${\left(\frac{\partial U}{\partial V}\right)}_{T} = - P + \frac{T \alpha}{\kappa}$

Thus,

$\textcolor{b l u e}{\Delta U} = {\int}_{{V}_{1}}^{{V}_{2}} - P + \frac{T \alpha}{\kappa} \mathrm{dV}$

$= \textcolor{b l u e}{\frac{\alpha T \Delta V}{\kappa} - {\int}_{{V}_{1}}^{{V}_{2}} P \left(V\right) \mathrm{dV}}$

Next, the pressure will vary a lot if the volume varies a little for a solid or liquid...

Hence, to find $\Delta U$, we first need to measure the temperature of the system, and reference $\alpha$ and $\kappa$.

Then, we'll need to measure how the volume changes (to high precision!) with the change in pressure to empirically find the pressure as a function of the volume for the given solid or liquid.

1

Can someone pls show me the VR (Voltage-Resistance) Graphs for an NTC thermistor, filament lamp and conductors?

Sean
Featured 2 months ago

Explanation:

.

Let's take a look at an NTC thermistor circuit:

The input of $12$ volts goes through a voltage regulator that regulates and supplies ${V}_{s} = 5$ volts as supply voltage to the circuit. The temperature sensor measures the change in the output voltage as a result of the change in the resistance of the element in the sensor and converts it into the desired unit .

We know that the circuit has a fixed resistor $\left(R\right)$ of $10 k \Omega$ and the resistor in the thermistor element $\left({R}_{t}\right)$ also has a value of $10 k \Omega$ at the nominal ambient temperature of ${21}^{\circ} C$.which varies with temperature.

The output voltage ${V}_{o}$ varies as ${R}_{t}$ varies. The relationship between them is:

${R}_{t} = R \left({V}_{s} / {V}_{o} - 1\right)$

It is difficult to find the Voltage-Resistance graph of an NTC thermistor in books but we can develop it here. We will designate the horizontal axis as the ${V}_{o}$-axis and the vertical one as the ${R}_{t}$-axis.

Furthermore, in order to obtain a reasonably clear scale, we will consider the two resistances in the equation in units of $10 K \Omega$ instead of $\Omega$. This is like dividing both sides of the equation by $10000$. As such, our equation becomes::

${R}_{t} = R \left(\frac{5}{V} _ o - 1\right)$ or

${R}_{t} / R = \frac{5}{V} _ o - 1$ which we will graph.

The graph will be:

But since we are dealing with positive voltages only, we will concern ourselves with the right hand side of the graph where both ${V}_{o}$ and ${R}_{t}$ are positive. We will also limit the domain to $0 \le {V}_{o} \le 5$. Therefore, we will have:

As the graph shows, when ${V}_{o} = 0$, ${R}_{t} = \infty$ and when ${V}_{o} = 5$, ${R}_{t} = 0$. In between these values, the graph shows how the output voltage varies as the resistance of the thermistor element changes.

As far as filament lamps go, here are some graphs:

As far as metallic conductors go, their voltage-current, current-resistance and voltage-resistance relationships are linear and their graphs are straight lines that go through the origin. They are called Ohmic conductors:

Hope this helps.

1

A basketball player stands 15m away from the basket to make a 3-point shot. The basket is 3m (10 ft) above the ground. Assume the ball leaves the player's hands at a height of 2m, and with velocity vector Vo=(Vox, Voy)?

Gió
Featured 2 months ago

I got this; however I am not sure I interpreted the question correctly...

Explanation:

Consider the diagram:

where:
$H = 3 m$
$h = 2 m$
$d = 15 m$

Let us divide our problem into a component along the $x$ axis and $y$ axis; along the $y$ axis we have the acceleration of gravity $g$ operating on the ball while horizontally no acceleration will interfere (assuming negligible air restance).

Let us start considering the initial vertical component of ${v}_{o}$ needed to reach the height of the basket along $y$ (part a, distance $s$):

we can consider the expression from Kinematics:

${v}_{f}^{2} = {v}_{i}^{2} + 2 a s$

with:

${v}_{f} = {v}_{B} = 0$ ( barely reach the height of the basket )
${v}_{i} = {v}_{0 y}$
$s = {y}_{f} - {y}_{i} = 3 m - 2 m = 1 m$
$a = - 9.8 \frac{m}{s} ^ 2$

we get:

$0 = {v}_{0 y}^{2} - 2 \cdot 9.8 \cdot 1$

so that: ${v}_{0 y} = \sqrt{19.6} = 4.4 \frac{m}{s}$

Let us now use the relationship from Kinematics:

${v}_{f} = {v}_{i} + a t$

$0 = {v}_{0 y} - 9.8 t$

$t = {v}_{0 y} / 9.8 = \frac{4.4}{9.8} = 0.45 s$ to go from A to B.

$- - - - - - - - - - - -$

Let us now consider part b; to get ${v}_{0 x}$ we know that ${v}_{0 y} = 4.4 \frac{m}{s}$ and the ball has to travel $15 m$ horizontally to get to the basket.

Along $x$ there is no acceleration so we have:

${v}_{0 x} = \frac{d}{t}$

where $t$ is the time of flight that must be equal to the vertical time of flight from A to B!
So we end up with:

${v}_{0 x} = \frac{d}{{v}_{0 y} / 9.8} = \frac{15}{0.45} = 33.3 \frac{m}{s}$

$- - - - - - - - - - - -$

On the Moon $g$ wold be smaller (${g}_{\text{Moon}} = 1.6 \frac{m}{s} ^ 2$) so that we get (part a):

$0 = {v}_{0 y}^{2} - 2 \cdot 1.6 \cdot 1$

so that: ${v}_{0 y \text{Moon}} = \sqrt{3.2} = 1.8 \frac{m}{s}$

1

Electricity problem..... please can you solve this...?

ali ergin
Featured 2 months ago

$\text{the solution has shown at explanation section but please check the math operations.}$

Explanation:

$a :$

• The electric charge of -3.0 $\mu$C pulls the electric charge of 5.0 $\mu$C(due to Coulomb's law)) .
• This force has shown as red vector as symbolically.
• The force vector has shown at negative direction.

$\text{r:6.0 cm=6.10^-2 m (let convert cm to m)}$

${q}_{1} = - 3.0 \cdot {10}^{-} 6 C \text{ (let convert "mu"C to Coulomb)}$

${q}_{2} = + 5.0 \cdot {10}^{-} 6 C$

$k = {9.10}^{9} N \cdot {m}^{2} \cdot {C}^{-} 2$

• The magnitude of the force can be calculated using Coulomb's law.

$F = {9.10}^{9} \frac{- 3.0 \cdot {10}^{-} 6 \cdot 5.0 \cdot {10}^{-} 6}{{\left(6 \cdot {10}^{-} 2\right)}^{2}}$

$F = \cancel{9} {.10}^{9} \frac{- 3.0 \cdot {10}^{-} 6 \cdot 5.0 \cdot {10}^{-} 6}{\cancel{36} \cdot {10}^{-} 4}$

$F = - \frac{150}{4} = - 37.5 N$

$b :$

• the charge -4 $\mu$C pushes by the charge -3 $\mu$C toward positive direction(due to Coulomb's law).

• This force has shown as blue vector as symbolically.

$\textcolor{b l u e}{{F}_{1}} = {9.10}^{9} \frac{\left(- {3.10}^{-} 6\right) \cdot \left(- 4 \cdot {10}^{-} 6\right)}{{\left(1.5 \cdot {10}^{-} 2\right)}^{2}}$

$\textcolor{b l u e}{{F}_{1}} = 480 N$

• the charge -4 $\mu$C pulls by the charge -5 $\mu$C toward positive direction(due to Coulomb's law).

• This force has shown as green vector as symbolically.

$\textcolor{g r e e n}{{F}_{2}} = {9.10}^{9} \frac{\left(- {4.10}^{-} 6\right) \cdot \left(5 \cdot {10}^{-} 6\right)}{{\left(4.5 \cdot {10}^{-} 2\right)}^{2}}$

$\textcolor{g r e e n}{{F}_{2}} = - 8.89 N \left(\text{The negative sign means the force is attractive.}\right)$

• Since vectors ${F}_{1}$ and${F}_{2}$ has the same direction, we must find the vectorial sum of these.

• F=${F}_{1}$+${F}_{2}$

• F=480+8.89=488.89 N