Altitude means the what?

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Fynjy0 [20]

D.
The reading between 7N and 8N would have to be 7.5N. Answers A and B are much to small and answer C is way to big.

8 0

3 years ago

A tank contains one mole of carbon dioxide gas at a pressure of 6.70 atm and a temperature of 23.0°C. The tank (which has a fixe

quester [9]

Answer: 888.45 K or 615.3 °c

Explanation:

According to Gay Lussacs law which states that at constant volume, pressure of an ideal gas is directly proportional to it's absolute temperature.

P/T = Constant

Therefore, P1/T1 = P2/T2

P1 = 6.7 atm

T1= 23°c = 273.15 + 23 = 296.15K

Since P2 is tripled, then,

P2 = 6.7 x 3= 20.1 atm

T2 = (20.1 x 296.15) ÷ 6.7

T2 = 888.45 K

Or in celcius 615.3°c

5 0

3 years ago

An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro

Mashutka [201]

1) The distance travelled by the electron is $1\cdot 10^{17} m$

2) The time taken is $4.0\cdot 10^{10}s$

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

$u=5\cdot 10^6 m/s$ is the initial velocity

v = 0 is the final velocity (it comes to a stop)

$a=-1.25\cdot 10^{-4} m/s^2$ is the acceleration

Solving for s, we find the distance travelled:

$s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m$

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

$u=5\cdot 10^6 m/s$

v = 0

$a=-1.25\cdot 10^{-4} m/s^2$

And solving for t, we find the time taken:

$t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s$

Learn more about accelerated motion:

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#LearnwithBrainly

7 0

2 years ago

A pickup truck is being driven down the highway carrying eight 100 gallon fish tank full of water. A hole is punctured in each t

irakobra [83]

Answer:

Explanation:

The velocity of the vehicle would increase because the the tanks (when filled with water) must have exerted a force which would reduce the velocity of the vehicle at a certain pressure on the gas pedal. Note that force equals mass multiplied by acceleration; as the mass decreases, so the force decreases. Thus, when the mass exerted by this tanks (on the vehicle) decrease as a result of the hole punctured in them, the force exerted by the tanks would also decrease causing an increase in velocity of the pick up truck when the same pressure is applied on the gas pedal throughout (before and after the puncture).

The conservation law that applied here is the law of conservation of energy which states that energy can neither be created nor destroyed but can be transformed from one form to another. This is because the energy the vehicle used in carrying the load (the tanks) was transformed to the energy that resulted in increasing it's velocity (no new energy was formed as the pressure on the gas pedal remained the same).

6 0

3 years ago

In a football game, the running back takes a handoff and begins running toward midfield at 3.21 yards/s . As he moves through hi

Inessa [10]

Given Information:

Initial speed = u = 3.21 yards/s

Acceleration = α = 1.71 yards/s²

Final speed = v = 7.54 yards/s

Required Information:

Distance = s = ?

Answer:

Distance = s = 13.61

Explanation:

We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.

We know from the equations of motion,

v² = u² + 2αs

Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.

Re-arranging the above equation for distance yields,

2αs = v² - u²

s = (v² - u²)/2α

s = (7.54² - 3.21²)/2×1.71

s = 46.55/3.42

s = 13.61 yards

Therefore, the runner traveled a distance of 13.61 yards before being tackled.

8 0

3 years ago