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Anna11 [10]
3 years ago
10

When used as a theatrical special effect to simulate fog, carbon dioxide gas tends to stay near the floor of the stage. What mig

ht account for this effect?
Physics
1 answer:
Bond [772]3 years ago
3 0
Carbon dioxide is denser than the oxygen surrounding it so it is pushed down.
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PLZ ANSWER ASAP!!!!!!!
jenyasd209 [6]

Explanation:

It is given that,

Mass of platinum bar, m = 750 g

Length of the bar, l = 5 cm

Breadth of the bar, b = 4 cm

Width of the bar, h = 1.5 cm

Volume of the bar, V=l\times b\times h

V=5\times 4\times 1.5=30\ cm^3

We need to find the density of the platinum bar. The density of any substance is given by :

d=\dfrac{m}{V}

d=\dfrac{750\ g}{30\ cm^3}

d=25\ g/cm^3

So, the density of the platinum bar 25\ g/cm^3. Hence, this is the required solution.

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Every organism may classified into seven level of classifications, such that each level contains organisms with similar characteristics. Kingdom is the largest and the broadest level of classification while species is the smallest and most specific level of classification. Therefore from the largest to the smallest the order is as follows:
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Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric e
kvv77 [185]

This question is incomplete, the complete question is;

A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.

Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.

<em>Hint </em>: ΔV = Ed <em> </em>

Answer:

the required potential difference, between the capacitor plates is 600 V

Explanation:

Given the data in the question;

B = 0.60 T

d = 2.0 mm = 0.002 m

v = 5.0 × 10⁵ m/s.

since particle pass straight through without deflection.

F_{net = 0

so, F_E = F_B

qE = qvB

divide both sides by q

E = vB

we substitute

E = (5.0 × 10⁵) × 0.6

E = 300000 N/C

given that; potential difference ΔV = Ed

we substitute

ΔV = 300000 × 0.002

ΔV = 600 V

Therefore, the required potential difference, between the capacitor plates is 600 V

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3 years ago
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Answer it might be this

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