Someone please help me!

3. How do astronomers explain the lack of a meteorite in Tunguska? What evidence exists to support this

Vaselesa [24]

No traces of a meteorite were found, it many scientists concluded that the culprit was a comet. Comets, which are essentially muddy ice balls, could cause such a devastation and leave no trace.

But now, 105 years later, scientists have revealed that the Tunguska devastation was indeed caused by a meteorite. A group of Ukrainian, German, and American scientists have identified its microscopic remains. Why it took them so many years makes for a fascinating tale about the limits of science and how we are pushing them.

4 0

3 years ago

A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it

fomenos

Answer:68.15m/s

Explanation:

<u><em>Given: </em></u>

v₁=15m/s

a=6.5m/s²

v₁=?

x=340m

<u><em>Formula:</em></u>

v₁²=v₁²+2a (x)

<u>Set up:</u>

= $\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)$

<h2><u><em>Solution:</em></u></h2><h2><u><em>68.15m/s</em></u></h2>

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6 0

3 years ago

How would decreasing the volume of the reaction vessel affect each of the following equilibria?2NOBr(g)⇌2NO(g)+Br2(g)

mafiozo [28]

Answer:

The equilibrium position will shift towards the left hand side or reactants side

Explanation:

Decreasing the volume (increasing the pressure) of the system will shift the equilibrium position towards the lefthand side or reactants side. This is because, decreasing the volume (increasing the pressure) implies shifting the equilibrium position towards the side having the least number of moles.

There are two moles of reactants and a total of three moles of products(total). Hence decreasing the volume and increasing the pressure of the gas phase reaction will shift the equilibrium position towards the lefthand side.

5 0

3 years ago

An attempt to redirect internet traffic from a legitimate site to a different identical-looking site is categorized as which soc

Snezhnost [94]

Answer:PHARMING

Explanation:

Social engineering can be defined as an attempt to manage,change and regulate the future development and the behaviour of a society.

Common types of social engineering attack include; Watering hole, Pretexting, Phishing,Whaling attack, Baiting, PHARMING, and so on.

PHARMING is an attempt to redirect internet traffic from a legitimate site to a different identical-looking site. PHARMING is a spamming practice, a cyberattack and also, a type of phishing in which hackers use in stealing personal information from people on the internet.

PHARMING is done through the injection of malicious data or code into the victims' computer system. This injection is known as the DNS cache poisoning.

5 0

3 years ago

A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.

Zolol [24]

These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is $45^{\circ}$ .

Calling F the force of the wind, and $d=15~km=15000~m$ the distance covered by the boat, the work done by the wind is:
$W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J$

The total time of the motion is $t=1~h=3600~s$ and therefore the power of the wind is
$P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W$

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
$d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m$

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
$MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55$

Finally, the efficiency $\epsilon$ of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
$\epsilon = \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81$

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
$P= \frac{W}{t}$
But we can rewrite the work as
$W=Fdcos\theta$
where F is the force applied, d the displacement of rock and $\theta=60^{\circ]$ is the angle between the direction of the force and the displacement (3 m).
Therefore we can rewrite the power as
$P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta$
where $v=d/t=5~m/s$ is the velocity, Using the data of the exercise, we can then find the force, F:
$F= \frac{P}{v cos\theta} = \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N$

and now we can also calculate the work, which is
$W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J$

3 0

3 years ago