A car accelerates from rest at a constant rate 5m/s^2. Which of the following statements is true

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

Brut [27]

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!

6 0

2 years ago

A scientist designed a foam container to help keep frozen foods from melting. Which best explains how the foam works?

FrozenT [24]

I think the answer should be D.<span>It reduces the amount of thermal energy that is transferred from outside to inside the container. </span>

6 0

3 years ago

Read 2 more answers

The symbol used for magnetic field is:

alina1380 [7]

The letter "B" is used for magnetic field

5 0

3 years ago

The Volts are the ________________

Vera_Pavlovna [14]

People who went out and sailed on the sea and then they went on missions

8 0

3 years ago

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

3 0

3 years ago