What has affected almost evey aspect of modern sports

Help ASAP!<br> Everything on screenshot

BARSIC [14]

Answer:

For the first one, its B) cities B and C

I'm not so sure, but I hope this helps.

7 0

3 years ago

Allison is looking through a telescope at an object in space. The object looks like a very small planet, and it does not have a

Verizon [17]

Answer:

Allison is probably looking at the asteroid.

Explanation:

Asteroids are the giant gas balls of inner solar system that are present in the space.
Asteroids orbits around the sun in very irregular ans strange path unlike the other planets.
This is also the reason why it's called as the minor planets.
It could have been comet but if it had been a comets he must have seen the tail but asteroids do not have any tail like structure.

8 0

3 years ago

The diagram below shows a wave with its wavelength indicated in red.

Shtirlitz [24]

Answer:

answer is D.it will decrease

8 0

3 years ago

Read 2 more answers

On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much lik

Anon25 [30]

a) $4.62\cdot 10^{14} J$

b) 0.110 megatons

c) 8.46 bombs

Explanation:

a)

The energy lost by the meteorite is equal to the difference between its final kinetic energy and its initial kinetic energy:

$\Delta K=K_f-K_i$

Which can be rewritten as:

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

where:

$m=3.2\cdot 10^6 kg$ is the mass of the meteorite

$v=0$ is the final speed of the meteorite

$u=17 km/s = 17,000 m/s$ is the initial speed of the meteorite

Substituting the values into the equation, we found the loss in energy of the meteorite:

$\Delta K=0-\frac{1}{2}(3.2\cdot 10^6)(17000)^2=-4.62\cdot 10^{14} J$

So, the energy lost by the meteorite is $4.62\cdot 10^{14} J$

b)

The energy equivalent to 1 megaton of TNT is

$E_{TNT}=4.2\cdot 10^{15} J$

Here the energy lost by the meteorite is

$E=4.62\cdot 10^{14} J$

Therefore, in order to write the energy lost by the meteorite as a multiple of the energy of 1 megaton of TNT, we have to divide the energy lost by the meteorite by the energy equivalent to 1 TNT; we find:

$\frac{E}{E_{TNT}}=\frac{4.62\cdot 10^{14}}{4.2\cdot 10^{15}}=0.110$