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2 years ago

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How much of the universe is believed to consist of ordinary atoms that can be detected?

1 answer:

Margaret [11]2 years ago

4 0

The universe believed to be has no limit in space and the elements, molecules and different kind of gases are found in this area. So the percentage of that is believed that and ordinary atom can be detected is C. 3% i hope you understand my answer.

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An object has a kinetic energy ke, and potential energy pe. it also has a rest energy e0. what is the object's total energy e?

Dmitriy789 [7]

Look that one up in you text book PG:678 that is if you got the same book as my friend<span />

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3 years ago

An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t

Ede4ka [16]

Answer:

The magnitude of the flux of electric field through a square of surface area is zero.

Explanation:

$E=100 NC^{-1}\\\\A=2 m^2\\\\Electic\,\,flux\,\,flux\,\,is\,\,given\,\,as:\\\\\phi_E=E.A\,cos\,\theta$

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

$\phi_E=E.A\,cos\,(90^o)\\\\\phi_E=0$

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3 years ago

if you rub a balloon on your hair, it becomes charged, and as a result It can stick to your head without being held there. why d

7nadin3 [17]

Because the charges of static electricity and the eons coming from your hair pull together to make the balloon stick

7 0

3 years ago

A soccer player kicks a soccer ball with a force of 1.8 N. If the mass of the ball is .43 kg. How fast will the ball accelerate?

Vesnalui [34]

Answer:

The acceleration of the ball is 4.18 [m/s^2]

Explanation:

By Newton's second law we can find the acceleration of the ball

$F = m*a\\where:\\F = force applied [N] or [kg*m/s^2]\\m = mass of the ball [kg]\\a = acceleration [m/s^s]$

Now we have:

$a = F/m\\a = \frac{1.8 [kg*m/s^s]}{0.43[kg]} \\a = 4.18 [kg]$

4 0

3 years ago

Read 2 more answers

A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at

Vaselesa [24]

Answer: $24.70 ^{\circ}C$

Explanation:

Given

mass of lead piece $m_l=234 gm\approx 0.234 kg$

mass of water in calorimeter $m_w=611 gm\approx 0.611 kg$

Initial temperature of water $T_w=24^{\circ}C$

Initial temperature of lead piece $T_l=24^{\circ}C$

we know heat capacity of lead and water are $125.604 J/kg-k$ and $4.184 kJ/kg-k$ respectively

Let us take $T ^{\circ}C$ be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

$m_lc_l(T_l-T)=m_wc_w(T-T_w)$

$0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)$

$86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)$

$86-T=86.97T-2087.49$

$T=\frac{2173.491}{87.97}=24.70^{\circ}C$

3 0

2 years ago