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3 years ago

8

the value of acceleration due to gravity option A is same on equator and poles option B is list on poles option C is list on equ

ator option d increase from pole to equator which is the answer

1 answer:

stiks02 [169]3 years ago

8 0

Answer:

D. Increases from pole to equator

Explanation:

I majored in Science

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A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for

xxTIMURxx [149]

Answer:

5N

Explanation:

We have a simple problem of momentum here.

ΔMomentum= mΔv= FΔt

Solve for F

mΔv/Δt=F

Plug in givens

1*(2-1.5)/0.1=F

F=5N

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3 years ago

Tridil is infusing at 15 ml/hr on an infusion pump. The drug is mixed 50 mg in 500 ml D5W. How many MCG/minute is the patient re

olga nikolaevna [1]

Answer:

patient receiving drug 25 MCG/minute

Explanation:

given data

infusing = 15 ml/hr

drug = 50 mg

D5W = 500 ml

to find out

How many MCG/minute

solution

we know infusing rate is 15 ml/hr = 0.25 ml/min

so 0.25 ml drug content = 50 /500 × 0.25

0.25 ml drug content = 0.025 mg

so here

rate of drug will be 0.025 mg

rate of drug = 0.025 mg = 25 × $10^{-6}$ gm/min

rate of drug = 25 MCG/minute

so patient receiving drug 25 MCG/minute

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3 years ago

If the relative humidity is 25 nd the saturation mixing ratio is 24g/kg, what is the mixing ratio?

levacccp [35]

The mixing ratio is 6.

To find the answer, we have to know about the mixing ratio.

<h3>What is mixing ratio?</h3>

The mixing ratio must be calculated in a complex manner.
A saturated vapor pressure (es) for values of air temperature and an actual vapor pressure (e) for values of dewpoint temperature must be determined in order to determine the mixing ratio.
The air temperature and/or dewpoint temperature must first be converted to degrees Celsius (°C) before the vapor pressures can be calculated.
The equation below can be used to determine the relative humidity (rh), as well as the actual mixing ratio and saturated mixing ratio,

$Rh=\frac{w}{w_s} *100$

where; w is the mixing ratio and w(s) is the saturation mixing ratio.

In our question, it is given that,

$Rh=25\\w_s=24$

Thus, the mixing ratio will be,

$w=\frac{Rh*w_s}{100} =\frac{25*24}{100}=6$

Thus, we can conclude that, the mixing ratio is 6.

Learn more about mixing ratio here:

brainly.com/question/8791831

#SPJ4

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2 years ago

Why is the ocean warmer than the air at night

Sliva [168]

You see, during the day the ocean collects heat from the sun. So the air above the ocean get warm at night, but the rest of the air on the land gets cooler because water has the ability to collect energy from the Sun.

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3 years ago

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

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3 years ago