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soldi70 [24.7K]
2 years ago
5

Ty is three years old and as a result of a “stomach bug" has been vomiting for

Physics
1 answer:
nataly862011 [7]2 years ago
3 0

Answer:

Ty's blood is slightly alkaline

Explanation:

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A cube that has a volume of 1.00 m³ contains N distinguishable particles.
grandymaker [24]

Answer:

1 P = 0.5

2 P = 0.3

3 P = 0.01

Explanation:

The probability formula is

          P =V^N

Where P is the probability  V is the volume while N is the number of distinguishing particles

So for N = 1 and V = 0.500m^3

           P = (0.500m^3)^1\\

               = 0.5

For  N = 1 and V =  0.300m^3

           P = (0.300m^3)^1

               = 0.3

For   N = 1 and V = 0.0100m^3

     P =(0.0100m^3)^1

         = 0.01

         

6 0
3 years ago
What is the proper way to start a fire?
Katyanochek1 [597]

Answer:

Explanation:

STEP 1: Gather Your Tools. There's a bit more to building a great campfire than simply placing a few logs in a heap ... If your site has a fire ring, you'll probably have to push the ash and charcoal from ... (Remember, tinder is the really light, quick burning material.) 1. ... Then build a larger teepee of firewood over the kindling.

6 0
3 years ago
The unit for measuring electric power is the
bazaltina [42]
C.) The measuring unit of "Electrical Power" is "Watt"

Hope this helps!
4 0
3 years ago
If ball 4 has a mass of 2 kg and it is 5m high, what will be its gravitational potential energy? (g=10 N/kg) *
Nastasia [14]

Explanation:

Gravitational potential energy

= mgh

= (2kg)(10N/kg)(5m)

= 100J.

3 0
3 years ago
Read 2 more answers
A 460 W heating unit is designed to operate with an applied potential difference of 120 V (a) By what percentage will its heat o
dybincka [34]

Answer:

(a) = -0.16%

(b) = smaller

Explanation:

given

power = 460 W

potential difference = 120 V

(a) what percentage will   its heat output drop if the applied potential difference drops to 110 V ?

we know p = \frac{v^2}{R} .....................(i)

we need to find change in power

\Delta P = \frac{\Delta (V^2)}{R}  

\Delta P = \frac{2 V \Delta V}{R}..............(ii)

from equations we get

\frac{\Delta P}{P} =  \frac{2 \Delta V}{V}

\frac{\Delta P}{P} = 2 \frac{110 -120}{120}

\frac{\Delta P}{P} =  -2(\frac{10}{120})

\frac{\Delta P}{P} = - 0.16 %

(b)

if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power

and due to this increment of heating power resistance will decrease so actual drop in the power would  be smaller

7 0
3 years ago
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