What would the person’s mass be on the earth? Part B pls

Which of the following is not used to measure wind?

lapo4ka [179]

Answer:

A) psychrometer

Explanation:

A psychrometer measures the relative humidity in the atmosphere not the wind.

6 0

4 years ago

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A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is

AveGali [126]

Explanation:

For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.

ΔU=Q−W

We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.

W=FΔx

W=3N×2m

W=6J

Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.

ΔU=Q−W

ΔU=10J−6J

ΔU=4J

Answer is 4J

i think this may help you very much

3 0

3 years ago

What is the smallest or most specific level of organization you and your neighbors are have in common

kow [346]

We have the same house

3 0

3 years ago

Find velocity vx(t) and coordinate x(t) for a particle of mass m which is subject to the force given by: Fx = F0 e −kt , where F

muminat

Answer:

$v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})$

$x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)$

Explanation:

Given that the force of the particle is,

$F_{x}=F_{0}e^{-kt}$

Now it can be further written as

$m\frac{dv}{dt}= F_{0}e^{-kt}\\\frac{dv}{dt}=\frac{ F_{0}}{m} e^{-kt}\\dv=\frac{ F_{0}}{m}e^{-kt}dt\\ v=\frac{ F_{0}}{-km}e^{-kt}+C$

Now the initial conditions are v=1 at t=0.

So,

$1=\frac{ F_{0}}{-km}e^{0}+C\\C=1+\frac{ F_{0}}{km}$

Now the velocity will become.

$v_{x} (t)=\frac{ F_{0}}{-km}e^{-kt}+1+\frac{ F_{0}}{km}\\v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})$

And,

$\frac{dx}{dt} =1+\frac{ F_{0}}{km}(1-e^{-kt})\\dx=(1+\frac{ F_{0}}{km}(1-e^{-kt}))dt\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+C\\$

And, another initial condition is x=0 at t=0

$0=0+\frac{ F_{0}0}{km}+\frac{ F_{0}t}{k^{2} m}e^{0}+C\\C=-\frac{ F_{0}t}{k^{2} m}$

Now,

$x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+-\frac{ F_{0}t}{k^{2} m}\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)$

5 0

3 years ago

Students were asked to place a mint in their mouths and determine how long it took for the mint to dissolve. The condition of th

Llana [10]

Students were asked to place a mint in their mouths and determine how long it took for the mint to dissolve. The condition of the mint varied in each student group. One group of students were asked to leave a whole mint in their mouth, not moving it around, and let it dissolve. Another group swirled a mint, while the other groups used mints broken into smaller pieces. See the chart for all of the manipulated variable. After reviewing that data table, what kind of result would you predict for the swirled, whole mint?

A) The time is likely between 10-30 seconds.

B) The time is likely between 40-80 seconds.

C) The time is likely between 90-160 seconds.

D) The time is likely between 100-200 seconds.

ANSWER: B) The time is likely between 40-80 seconds.

EXPLANATION:The time is likely between 40-80 seconds.

By swirling the mint, this is agitating and creating a higher frequency of collisions between the saliva particles and mint particles, increasing the rate of dissolution. Therefore, the time is likely to be less than the mint cut in half but probably more than the mint when it is in small pieces.

7 0

4 years ago

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