How do light travels

scientists have determined the composition of the lithosphere and asthenosphere but have struggled to discover much about the lo

dalvyx [7]

Explanation:

The lithosphere is composed of the crust and a portion of the upper mantle. the athenosphere is partially molten upper mantle material that behaves plasticly & can flow.

It is harder to work on the lower layers because its gets extremely hot & sometimes it can get so hot that it melts tools used to research these layers.

7 0

3 years ago

If the temperature outside increases from 30 degrees to 60 degrees, how has the kinetic energy of the air particles changed?

musickatia [10]

The kinetic energy increase because as temperature warms up particles move faster

7 0

3 years ago

Read 2 more answers

If 4.00 ✕ 10−3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.73 h, what is the curr

NeX [460]

Explanation:

Below is an attachment containing the solution.

3 0

4 years ago

Someone please help me it’s due today

stellarik [79]

Answer:

number 3 is the 3rd one and number 2 is the 4 one and number 1 is the 2nd one

4 0

3 years ago

17. A 1350 g projectile is launched with a force of 150 N. The length of the firing arm is 1.25 m.

creativ13 [48]

a) The exit velocity of the projectile is 16.7 m/s

b) The maximum height achieved by the projectile is 14.2 m

c) The total time of flight is 3.40 s

d) The distance covered by the projectile is 46.5 m

Explanation:

a)

We solve this first part of the problem by applying the work-energy theorem, which states that the work done on the projectile is equal to the gain in kinetic energy of the projectile. Mathematically:

$W=Fd = \frac{1}{2}mv^2-\frac{1}{2}mu^2=\Delta K$

where:

F = 150 N is the force applied

d = 1.25 m is the displacement of the projectile (the length of the firing arm)

m = 1350 g = 1.35 kg is the mass of the projectile

u = 0 is the initial velocity of the projectile

v is the exit velocity of the projectile

Solving for v, we find:

$v=\sqrt{\frac{2Fd}{m}}=\sqrt{\frac{2(150)(1.25)}{1.35}}=16.7 m/s$

b)

Assuming the projectile is fired vertically upward, then the initial kinetic energy of the projectile as soon as he leaves the cannon is fully converted into gravitational potential energy as it reaches the top of its trajectory. So we can write:

$K_i = U_f$

$\frac{1}{2}mv^2=mgh$

where:

$K_i$ is the initial kinetic energy

$U_f$ is the final potential energy

m = 1350 g = 1.35 kg is the mass of the projectile

v = 16.7 m/s is the velocity at which the projectile leaves the cannon

$g=9.8 m/s^2$ is the acceleration of gravity

h is the maximum height reached by the projectile

And solving for h, we find

$h=\frac{v^2}{2g}=\frac{(16.7)^2}{2(9.8)}=14.2 m$

c)

Assuming the projectile is launched vertically upward, then the total time of flight is twice the time it takes for reaching the maximum height. This time can be found by using the following suvat equation:

$v=u-gt$