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3 years ago

Which is a gas at room temperature?

Physics

2 answers:

Alex3 years ago

8 0

The answer is B. Nitrogen

Sedaia [141]3 years ago

7 0

<span>Which is a gas at room temperature?
</span>B) nitrogen

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A sports car is advertised to be able to stop in a distance of 50.0 m from a speed of 80 km. What is its acceleration and how ma

Flauer [41]

Explanation:

Given that,

Initial speed of the sports car, u = 80 km/h = 22.22 m/s

Final speed of the runner, v = 0

Distance covered by the sports car, d = 80 km = 80000 m

Let a is the acceleration of the sports car. It can be calculated using third equation of motion as :

$v^2-u^2=2ad$

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{0-(22.22)^2}{2\times 80000}$

$a=-0.00308\ m/s^2$

Value of g, $g=9.8\ m/s^2$

$a=\dfrac{-0.00308}{9.8}\ m/s^2$

$a=(-0.000314)\ g\ m/s^2$

Hence, this is required solution.

8 0

3 years ago

How is the direction of a vector affected when it is multiplied by a scalar?

pantera1 [17]

The direction of a vector multiplied by a scalar is only affected if the scalar is negative, in which case the vector will now be in the opposite direction. If the scalar is positive, the vector will only change in magnitude

6 0

3 years ago

The pacific ocean has a surface area of about

Aleks04 [339]

Answer:

150,000,000 km 2

Explanation:

5 0

4 years ago

4. A little boy pushes a wagon with his dog in it. The mass of the dog and

Artyom0805 [142]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 70 × 0.85

We have the final answer as

Hope this helps you

6 0

3 years ago

Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p

boyakko [2]

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, $\sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}$

Density of metal, $\rho=10.5\ g/cm^3$

Atomic weight, A = 107.87 g/mol

Let n is the number of free electrons per cubic meter such that,

$n=1.3\ N$

$n=1.3(\dfrac{\rho N_A}{A})$

Where

$\rho$ is the density of silver atom

$N_A$ is the Avogadro number

A is the atomic weight of silver

$n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})$

$n=7.61\times 10^{22}\ cm^{-3}$

$n=7.61\times 10^{28}\ m^{-3}$

Hence, this is the required solution.

6 0

3 years ago