The lantern rises up because the ______ of the _inside the lantern is _____ compared to the air around it

Physics

2 answers:

Vika [28.1K]2 years ago

7 0

Density, fire, less dense

ch4aika [34]2 years ago

5 0

Answer:

gass fire and the same density

Explanation:

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What is the energy in joules of a mole of photons associated with red light of wavelength 7.00 × 102 nm?

konstantin123 [22]

<span>The energy of a single photon is given by E = hc/lambda, where h is Planck's constant, c is the speed of light, and lambda is the wavelength. Plugging the values in gives E = 6.63E-34 x 3.00E8 / 700E-9 = 2.84E-19 Joules Now one mole of substance is equivalent to 6.02E23 particles, so one mole of these photons will be: 2.84E-19 x 6.02E23 = 1.71E5 Joules</span>

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3 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

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3 years ago

An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

$u -v= d$