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blondinia [14]
2 years ago
9

However, when is the best time to fill a balloon? Helium has a tendency to escape from any opening in the balloon and will event

ually deflate. This means that you should never wait until your event starts to inflate your balloons.
Physics
1 answer:
alekssr [168]2 years ago
3 0

Answer:The small, individual helium molecules can escape through the tiny holes in the latex far more easily than the conjoined oxygen or nitrogen molecules can. ... This is why your helium balloons deflate faster than the ones you fill with air.

Explanation:

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How many watt hours will 3-155amp hour 12 volt batteries wired in a parallel configuration produce?
Ede4ka [16]

Answer:

B 5580 W•hr

Explanation:

A Watt is a Volt times an Amp

3(12 V(155 A•hr)) = 5580 W•hr

4 0
3 years ago
Atomic math challenge will give brainly and thanks
nevsk [136]

Answer:

1. Hydrogen

Atomic # = 1

Atomic Mass = 1.00794  ( If you round it it's 1.008 )

# of protons = 1

# of neutrons = none

# of electrons = 1

8 0
3 years ago
Read 2 more answers
I need help, ASAP i’m failing and i have no clue what’s going on in my AP physics class at all.
garri49 [273]
What’s the question or problem ?
6 0
3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
2 years ago
A human being can be electrocuted if a current as small as 51.0 ma passes near the heart. an electrician working with sweaty han
boyakko [2]

The fatal current is 51 mA = 0.051 Ampere.

The resistance is 2,050Ω .

Voltage = (current) x (resistance)

            =  (0.051 Ampere) x (2,050 Ω)  =  104.6 volts .

==================

This is what the arithmetic says IF the information in the question
is correct.

I don't know how true this is, and I certainly don't plan to test it,
but I have read that a current as small as  15 mA  through the
heart can be fatal, not  51 mA .

If 15 mA can do it, and the sweaty electrician's resistance is
really 2,050 Ω, then the fatal voltage could be as little as  31 volts !

The voltage at the wall-outlets in your house is  120 volts in the USA !
THAT's why you don't want to stick paper clips or a screwdriver into
outlets, and why you want to cover unused outlets with plastic plugs
if there are babies crawling around.
6 0
3 years ago
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