Why is a convex lens useful for starting fires

1 answer:

7 0

because if you were stuck on an island and you had a bag of of stuff and you also had a convex lens in there it could be good if u can't start a fire by rubbing to sticks instead you can use the convex lens to start a fire

You might be interested in

the body Will remain at rest or move at constant velocity unless acted by external net force are unbalance force

Vikentia [17]

Answer:

law of motion states that a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. This is also known as the law of inertia. Inertia is the tendency of an object to remain at rest or remain in motion.

can u give me

your crown to me please brainliest me id you like my answer

4 0

2 years ago

During takeoff, an airplane climbs with a speed of 130 m/s at an angle of 47 degrees above the horizontal. The speed and directi

Reptile [31]

130cos(47deg) = 88.6598 m/s

5 0

3 years ago

The part of the earth system that includes earths solid, rocky, outer layer is the

Liono4ka [1.6K]

It would be the lithosphere

6 0

3 years ago

A ball is tossed from an upper-story window of a building. the ball is given an initial velocity of 8.00 m/s at an angle of 20.0

otez555 [7]

A) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
$x(t)=v_0 \cos \alpha t$
$y(t)=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
where
- the horizontal motion is a uniform motion, with constant speed $v_0 \cos \alpha$ , where $v_0 = 8.00 m/s$ and $\alpha=20.0^{\circ}$
- the vertical motion is an uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ , initial position h (the height of the building) and initial vertical velocity $v_0 \sin \alpha$ (with a negative sign, since it points downwards)

The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
$x(3.00 s)=v_0 \cos \alpha t=(8 m/s)(\cos 20^{\circ})(3.0 s)=22.6 m$

b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
$0=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
which becomes
$h=(8 m/s)(\sin 20^{\circ})(3.0 s)+ \frac{1}{2}(9.81 m/s^2)(3.0 s)^2=52.3 m$

c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
$y(t)=h-10$
If we substitute this into the equation of y(t), we have
$h-10 = h-v_0 \sin \alpha t- \frac{1}{2}gt^2$
$\frac{1}{2}gt^2+v_0 \sin \alpha t -10 =0$
$4.9 t^2 +2.74 t-10 =0$
whose solution is $t=1.18 s$ (the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.

4 0

4 years ago

If the ultraviolet photon has a wavelength of 249 nm and one of the photons emitted by the fluorescent material has a wavelength

melamori03 [73]

Answer:

601 nm

Explanation:

Energy of photon having wavelength of λ nm

= $\frac{1244}{\lambda}$ eV