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2 years ago

For a projectile with no air resistance, at the peak of its path, _______________.

Physics

1 answer:

3241004551 [841]2 years ago

4 0

Answer: "For a projectile with no air resistance, at the peak of its path, it's velocity is equal to zero"

Explanation:

Suppose that you throw an object up. The initial vertical velocity will be positive, and the acceleration (the gravitational acceleration) will point downwards, meaning that it opposes to the initial direction of the velocity, and that decreases the velocity as the time goes by.

There will be a point where the velocity ( that was positive until now, so until now the height of the object was increasing), is equal to zero, this means that at this moment the object stops moving, and after, because we still have negative acceleration, the velocity will start become negative, and the object will fall down.

Then that point where the velocity was zero is the peak of its path, then we can conclude that:

"For a projectile with no air resistance, at the peak of its path, it's velocity is equal to zero"

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Why don't astronauts use a fountain pen or a ball point pen with liquid ink in a spacecraft

mojhsa [17]

It's probably because there is no gravity which ballpoint pens need for inkflow to the tip.

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2 years ago

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$

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3 years ago

Question 6. B) and c)

maks197457 [2]

The formula for both is v(t) = v0 + a*t

b) v(8) = 0 + 6m/s^2 *8s = 48 m/s

now we know the beginning (2) and end speed (14), but not the time:

c) 14 = 2 + 1.5*t => t = (14-2)/1.5 = 8 seconds

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3 years ago

A fish can travel 30km/h in still water. How long will it take to arrive 12km downstream in a river flowing 6km/h?

kodGreya [7K]

30 + 6 = 36

36/12 = 3

So, it would take it 3 hours to go 12 kms downstream.

Note that we added 30 and 6 because it was going downstream. If it was going upstream, then we would have had to subtract.

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2 years ago

The pressure in Arabian sea is much bigger than near the surface?<br> please don't scam

sashaice [31]

Answer:

In general, the annual sea surface temperatures(SSTs) in the Bay of Bengal(BOB) are higher than the Arabian sea(AS). because, there are two main reasons for higher SST in the Bay of Bengal than the Arabian Sea. they are 1. stratification, 2.strong mixing

stratification is nothing but a phenomenon which stratifies(layers) the sea water when different density water(fresh water, rain water) add into the sea water. So the stratification in the bay of Bengal is comparatively high than the Arabian sea due to the high river discharge and precipitation in the BOB than the AS. the mixing process over the Arabian sea is higher than the Bay of Bengal due to the prevailing of strong winds over the AS (because of the presence of the mountains of east Africa) than Bay of Bengal (because of the winds over the BOB are sluggish in nature then the AS). But generally winds over the sea mixes easily the normal sea water than stratified/stabilized sea water column. That's why less mixing will takes place over the surface of BOB than the AS. So due to the presence of less mixing over the surface of the Bay of Bengal than the Arabian sea, the SST values over the Arabian sea are always lower than the Bay of Bengal. that's why the Arabian sea is colder than the Bay of Bengal.

Explanation:

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2 years ago

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