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3 years ago

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If the temperature at the surface of earth (at sea level) is 40°c, what is the temperature at 2000 m if the normal lapse rate is

6.4°c/1000 m?

1 answer:

Viefleur [7K]3 years ago

8 0

6.4*2=12.8 thus, your answer would be 12.8 degrees Celsius.

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An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o

Alekssandra [29.7K]

Answer:

$0.53m/s^2$

Explanation:

We are given that

Mass of wire=m=3.6 g= $3.6\times 10^{-3} kg$

1 kg=1000 g

Length of wire=l=1.6 m

Mass of object=m'=3 kg

Time,t=60.1 ms= $60.1\times 10^{-3} s$

$1 ms=10^{-3} s$

Speed,v= $\frac{distance}{time}=\frac{1.6}{60.1\times 10^{-3}}=26.62 m/s$

$g=\frac{v^2m}{m'l}$

Using the formula

$g=\frac{(26.62)^2\times 3.6\times 10^{-3}}{3\times 1.6}=0.53m/s^2$

8 0

4 years ago

PLEASE HELP!!!!!!!! More science questions coming soon!!

Naily [24]

Answer:

1.) A - 1,2,1,2

2.) Both forces can act without objects touching.

3.) C

Both can be modeled as waves having amplitude, frequency and wave lengths.

4.) A, B and C

Explanation:

1.) SO2 + ___ H2 ----------> _____ S + _______H2O

Looking at above equation, we have one molecule of surfur at the right hand side (RHS), and one molecule of surfur at the left hand side (LHS).

Two atoms of hydrogen and both RHS and LHS

But oxygen is not balanced. We have two atoms of oxygen at the RHS while having one at the LHS.

So let's make oxygen 2 atoms at LHS by adding 2 to water molecules and hydrogen molecules at the RHS.

SO2 + ___ 2H2 ----------> _____ S + _______2H2O

The correct answer is A - 1,2,1,2

2.) Both magnetic force and gravitational force obey inverse square law with distance. They are not directly proportional but inversely proportional to the square distance. B is the correct answer because Both forces can act without objects touching.

3.) Light waves are transverse waves which can travel through a vacuum without a medium while sound waves are longitudinal waves which cannot travel through vacuum without a medium. But both can be modeled as waves having amplitude, frequency and wave lengths.

4.) If an object is slowing down, due to conservation of energy,

The potential energy could be increasing, like a ball thrown into the air.

The kinetic energy could be lost to friction or air resistance.

The ball could be returning to its natural resting state.

7 0

3 years ago

If a neutral material like fur is rubbed on another neutral object like a rubber ball what is the resulting charges of the objec

alexdok [17]

Answer:

b

Explanation:

they both have a neutral charge so they couldn't be positive or negative since that wouldn't come from anywhere

8 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

2 years ago

Read 2 more answers

Air is warmer and less dense than surrounding air at the equator because the equator receives more?

Hatshy [7]

The answer is C: Solar energy

4 0

3 years ago

Read 2 more answers