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myrzilka [38]
3 years ago
12

: To determine the focal length of a lens, the following except _ is needed

Physics
1 answer:
mixer [17]3 years ago
4 0
The answer should besiphon
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How does inertia relate to coin dropping????
vekshin1

Well Inertia means something wants to stay in place, and in reality that coin wants to stay in one place, If you placed it on an index card on a cup, and SLOWLY pulled it, it wouldn't be fast enough to overcome that force, if you pulled it quickly that coin would stay in place and drop into the cup.

3 0
3 years ago
How far will you travel if you run for 10. minutes at 2.0 m/s?
oee [108]
We know that 1 minute= 60 seconds (or 1 min= 60 s).

10 min* (60 s/ 1 min)* (2.0 m/ 1 s)= 1,200 m.
(Note that the units cancel out so you get the answer)

The final answer is 1,200 m.

Hope this helps~
6 0
3 years ago
IF THERE ARE ONLY 118 ELEMENTS, HOW DO YOU ACCOUNT FOR THE MANY MILLIONS OF THINGS THAT WE HAVE IN OUR UNIVERSE?
bogdanovich [222]

Answer:

theres only 118 elements that are discovered. now that they're the only ones out there

Explanation:

3 0
2 years ago
A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

4 0
3 years ago
8. An object accelerates 12.0 m/s2 when a force of 6.0 newtons is applied to it. What is the mass of the object? _______________
Sholpan [36]

Answer:

Mass of object is 0.5kg

Explanation:

Given the following data;

Force = 6N

Acceleration = 12m/s²

Mass =?

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

F = ma

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

Making mass (m) the subject, we have;

Mass (m) = \frac{F}{a}

Substituting into the equation;

Mass (m) = \frac{6}{12}

Mass, m = 0.5kg.

Therefore, the mass of the object is 0.5kg

5 0
3 years ago
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