Ask question

Ask question

All categories

Ray Of Light [21]

3 years ago

10

sports photographers often use large aperture, long focal length lenses. what limitations do these lenses impose on the photogra

phs?

1 answer:

Brut [27]3 years ago

6 0

Answer:

The depth of focus achievable with those lenses is small.

Explanation:

A larger aperture makes it much harder to focus on more than one object. When using a telephoto lens (the ones the question is referring to), the depth of focus is very small. For example, using a telephoto lens to take a photo of a runner might get the runner in focus, but certainly not the track, or the audience behind them. If you look at photos, especially older photos, of Olympians in almost any sport you can see this.

Hope this helps!

You might be interested in

A cannonball explodes in mid-air, fragmenting into several pieces. How does the total

Sloan [31]

Hi there!

I. They are the same.

Due to the Conservation of Momentum, an explosion as such means that the TOTAL momentum of the system (all of the pieces) is conserved.

8 0

2 years ago

A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem

Viktor [21]

Answer:

A

The energy dissipated in the resistor ${U_k} = \frac{U}{k}$

B

The energy dissipated in the resistor ${U_k} = kU$

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{q ^2}{C}$

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation $q = CV$ is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

$U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2} kCV ^2$

In this equation, $Uk$ is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2$

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Substitute $U = \frac{1}{2}\frac{{{q^2}}}{C}$ in the equation of ${U_k}$

$U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}$

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

$U_{k} = \frac{1}{2}kCV^2$

Here, V is voltage in the circuit.

Substitute $U =\frac{1}{2} CV^2$ in the equation of ${U_k}$

So,

$U_{k} = \frac{1}{2} kCV^2\\$

$= k(\frac{1}{2} CV^2)$

$U_{k} = kU$

4 0

3 years ago

Is there any induced current in the pickup coil when the polarity of the bar magnet is flipped?

stiks02 [169]

Answer:

Faraday's law , he direction of the magnetic field changes by 180º, in the polarity inversion processes, induces a voltage.

Explanation:

For this exercise let's use Faraday's law

E = - dФ / dt

Ф = B.A = B A cos θ

where B is the magnetic field, A is the area and θ is the angle between the field line and the normal to the area.

We can see that an electromotive force (voltage) is indexed when there is a variation of the field B, a variation of the area and change of the angle or when there is a combinational of them.

In this case, the magnitude of the field is constant, as the wire is rigid metal, the area is constant, but the direction of the magnetic field changes by 180º, in the polarity inversion processes, for which reason each change induces a voltage.

If a voltage is created in the ring, which has a resistance, a current is also generated in it.

Therefore the answer is If a current is created in the hoop

8 0

3 years ago

How to do it? Urgent

mixer [17]

a)

F= ma

a=v/t

F=5*(35/5)

F=35N

b)

a=F/m

a=(35-2)/5

a=33/5

a=6.6N

8 0

3 years ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

jeyben [28]

Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

Let $g$ be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. $V_0$ will remain constant throughout the projectile motion.

So, if the time of flight is $t$ , then

$D=V_0t\; \cdots (i)$

Now, from the equation of motion

$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

$\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)$

Similarly, for the launched height $2H$ , the new time of flight, $t'$ , is

$t'=\sqrt {\frac {4H}{g}}$

From equation (iii), we have

$\Rightarrow t'=\sqrt 2 t\;\cdots (iv)$

Now, the spheres may be launched at speed $V_0$ or $2V_0$ .

Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

$D_1=V_0t'$

$D_1=V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_1=\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_1=1.41 D$ (approximately)

This is in the $3$ points range as given in the figure.

Similarly, $D_2=2V_0t'$

$D_2=2V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_2=2\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

7 0

3 years ago