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3 years ago

The ideal gas model is valid if which of the following conditions is true?

Physics

1 answer:

VashaNatasha [74]3 years ago

3 0

in ideal gas we have few things that we need to follow as following

1. Force of interaction between gas molecules are negligible.

2. There is no effect of gravity on them

3. All collisions are perfectly elastic collision.

4. there will be no energy loss

5. All newton's law are valid for them.

6. all molecules moves with same speed in random direction.

So here in order to follow all above conditions we have to maintain low pressure and high temperature in the gas due to which the density of gas becomes low.

So correct answer will be

<em>The gas density is low and the temperature is high.</em>

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1. A vector is given by its components, Ax = 2.5 and A, = 7.5. What angle

Allushta [10]

Answer:

A) 72° or 71.56°

Explanation:

We have two components on the X-axis and y-axis respectively. So we can use the tangent of the angle to be able to find the angle with respect to the horizontal component.

Ax = 2.5

Ay = 7.5

tan(α) = 7.5/2.5

$\alpha = tan^{-1} (3)\\$

α = 71.56°

4 0

3 years ago

A car accelerates from 20.0 m/s to 28.0 m/s over a distance of 50.0 m. What is the car’s acceleration?

Blababa [14]

Answer:

Explanation:

$V^2=V^2_o+2a(x_f-x_i)$

$28^2=20^2+2a(50)\\784=400+100a\\384=100a\\a=3.84m/s^2$

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3 years ago

A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58 above the horizontal. The end of the ram

sweet-ann [11.9K]

Answer:

a) Maximum height reached above ground = 2.8 m

b) When he reaches maximum height he is 2 m far from end of the ramp.

Explanation:

a) We have equation of motion v²=u²+2as

Considering vertical motion of skateboarder.

When he reaches maximum height,

u = 6.6sin58 = 5.6 m/s

a = -9.81 m/s²

v = 0 m/s

Substituting

0²=5.6² + 2 x -9.81 x s

s = 1.60 m

Height above ground = 1.2 + 1.6 = 2.8 m

b) We have equation of motion v= u+at

Considering vertical motion of skateboarder.

When he reaches maximum height,

u = 6.6sin58 = 5.6 m/s

a = -9.81 m/s²

v = 0 m/s

Substituting

0= 5.6 - 9.81 x t

t = 0.57s

Now considering horizontal motion of skateboarder.

We have equation of motion s =ut + 0.5 at²

u = 6.6cos58 = 3.50 m/s

a = 0 m/s²

t = 0.57

Substituting

s =3.5 x 0.57 + 0.5 x 0 x 0.57²

s = 2 m

When he reaches maximum height he is 2 m far from end of the ramp.

8 0

4 years ago

The magnitude of force A is 467 N. The magnitude of force B is 169 N. If the net force acting on the object is 0 N, what is the

Maurinko [17]

Answer:

Explanation:...

5 0

3 years ago

In a ballistics test, a 28-g bullet pierces a sand bag that is 30 cm thick. If the initial bullet velocity was 55 m/s and it eme

skelet666 [1.2K]

Answer:

Frictional force will be equal to 126.04 N

Explanation:

We have given mass of the bullet m = 28 gram = 0.028 kg

Initial velocity u = 55 m /sec

Width of sand = 30 cm = 0.3 m

So initial kinetic energy $E=\frac{1}{2}mu^2=\frac{1}{2}\times 0.028\times 55^2=42.35j$

Final velocity of the bullet v = 18 m /sec

So final kinetic energy $E=\frac{1}{2}mv^2=\frac{1}{2}\times 0.028\times 18^2=4.536j$

So change in kinetic energy = 42.35 - 4.536 = 37.814 j

From work energy theorem this change in kinetic energy will be equal to work one by frictional force

So $f\times 0.3=37.814$

$f=126.04 N$

4 0

4 years ago