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2 years ago

Which picture represents 2NO2?

Physics

2 answers:

wariber [46]2 years ago

3 0

2NO2 means that there is 2 oxygen atoms and one nitrogen with two sets of that. So its the third one

KIM [24]2 years ago

3 0

Answer:
The last option

Explanation:
Let's take this step by step.
First, we are looking for two moles of NO₂
This means that we can exclude the first two options since each of them shows only one mole.

Now, the compound we are looking for is NO₂. This means that it has two oxygen moles and one nitrogen mole.
You have stated that nitrogen is the blue one. Therefore, the compound we are looking for will be the one having one blue ball (representing nitrogen) and two red balls (representing oxygen).

Now taking a look at the choices, we will find that the last one fits our criteria.
1- It has two moles
2- each mole as two red balls and one blue ball

Therefore, the last option is the correct one

Hope this helps :)

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What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal

murzikaleks [220]

Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"

Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of $v_f = 16.5 m/s$ , so $\Delta v=16.5 m/s$ , in a time interval of $\Delta t = 181 ms=0.181 s$ . The acceleration of the ball in the horizontal direction (x-axis) is therefore

$a_x = \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2$

And the distance covered by the ball during this time interval, before it is released, is:

$S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m$

(b) For this part we need to consider also the weight of the ball, which is $W=mg=2.28 N$

From this, we find its mass: $m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg$

Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have

$F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N$

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

$F_y=W=mg=2.28 N$ (upward)

So, the magnitude of the force is

$F= \sqrt{F_x^2+F_y^2}= \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N$

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

$\tan \alpha = \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11$

From which we find $\alpha=6.2^{\circ}$

7 0

2 years ago

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When the wind kicks up dust and sand, the dust grains are charged. The small grains tend to get a negative charge, and the large

diamong [38]

Answer:

Explanation:

Small grains are negatively charged by the wind while big grains is positively charged and remains at the ground . This process creates an electric field due to the presence of oppositely charged particles.

When ever electric field exists it is directed from a positive charge to a negative charge so the here electric field is towards an upwards direction.

4 0

2 years ago

550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c

Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work

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2 years ago

the grid in a triode is kept negatively charged to prevent… a. the variations in voltage from getting too large. b. electrons be

Nezavi [6.7K]

D:the electrons from being attracted to the grid instead of the anode

5 0

2 years ago

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Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The mass o

insens350 [35]

Answer:

295.42 N

Explanation:

From Newton's law of universal gravitation.

F = Gmm'/r².................. Equation 1

Where F = Gravitational force, G = Universal constant, m = mass of the human, m' = mass of mass, r = radius of mass.

Given: m = 80 kg, m' = 6.4×10²³ kg, r = 3.4×10⁶ m.

Constant: G = 6.67×10⁻¹¹ Nm²/Kg²

Substitute into equation 1

F = 6.67×10⁻¹¹(80)(6.4×10²³ )/( 3.4×10⁶)²

F = 3415.04×10¹²/(11.56×10¹²)

F = 3415.04/11.56

F = 295.42 N

Hence the gravitational force = 295.42 N

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2 years ago