Which of the following best describes the gas trapped in a sealed balloon?

1 answer:

4 0

"C) The movement of the gas particles keeps the balloon inflated" best describes the gas trapped in a sealed balloon, although the movement varies with the heat of the gas.

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The centers of two 15.0 kg spheres are separated by 3.00 m. The magnitude of the gravitational force between the two spheres is

kompoz [17]

we have to use newtons law of gravitation which is
F=GMm/r^2
G=6.67 x 10^-11N kg^2/m^2
M=(15kg)
m=15 kg
r=(3.0m)^2
putting values we have
=(6.67 x 10^-11N kg^2/m^2)(15kg)(15kg)/(3.0m)^2 
=1.67 x 10^-9N

7 0

3 years ago

Why is the density of iron higher than the density of oxygen?

pentagon [3]

Answer:

Density of iron is higher than the density of oxygen because iron is solid and the particles in it are very compact but oxygen is not dense as it is gas and particles in it are loosely packed.

7 0

2 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

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