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2 years ago

Shadow and eclipses result from

Physics

2 answers:

Basile [38]2 years ago

7 0

$\huge\mathsf{\red{\underline{\underline{Answer}}}}$

${\green{\dashrightarrow}}$ When an opaque obstacle is placed between a source of light and a screen, a shadow of the obstacle is formed on the screen. The kind of shadow depends on the size of the source of light. In other words, the earth casts its shadow on the moon. The solar eclipse occurs when the moon comes between the sun and the earth.

poizon [28]2 years ago

5 0

Lunar eclipses occur whenever the full moon passes into the shadow of the earth. Solar eclipses occur when the new moon slips in front of the sun and blocks it from view. The moon's shadow then falls upon the earth.

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A ray is incident on a film of thickness t and the index of refraction n=1.6 at an angle θ = 35 degrees. Find the angle of refra

lidiya [134]

Answer

Given,

refractive index of film, n = 1.6

refractive index of air, n' = 1

angle of incidence, i = 35°

angle of refraction, r = ?

Using Snell's law

n' sin i = n sin r

1 x sin 35° = 1.6 x sin r

r = 21°

Angle of refraction is equal to 21°.

Now,

distance at which refractive angle comes out

d = 2.5 mm

α be the angle with horizontal surface and incident ray.

α = 90°-21° = 69°

t be the thickness of the film.

So,

$tan \alpha = \dfrac{t}{d/2}$

$tan 61^0 = \dfrac{t}{2.5/2}$

t = 2.26 mm

Hence, the thickness of the film is equal to 2.26 mm.

4 0

2 years ago

Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims

777dan777 [17]

Answer:

The work done by the drag force is given by 29.96 J

Explanation:

Given :

Thrust force $F = 12.3$ N

Displacement $d = 10.2$ m

Mass of rocket $m = 0.663$ Kg

From work energy theorem,

$W = \Delta K$

$W_{t} - Wd - W_{g} = KE$

Where $W_{t} =$ thrust work $W_{g} =$ gravitational work

$KE = 12.3 \times 10.2 -Wd - 0.663 \times 9.8 \times 10.2$

$KE = 59.2 -Wd$

After cutoff kinetic energy is converted into potential energy,

$KE = Wd' + mg\Delta h$

Put value of KE

$59.2 -Wd = Wd' + 0.663 \times 9.8 \times 4.5$

Work done by drag force is given by,

$Wd'+Wd = 59.2 -29.23$

$= 29.96$ J

Therefore, the work done by the drag force is given by 29.96 J

5 0

2 years ago

amber walks towards the back of a train at 3mph. The train is moving foward at 65mph. Find the velocity of amber relative on the

andrew11 [14]

Answer: 62mph

Explanation:

8 0

3 years ago

a liquid with a specific heat of 1900J/Kg° C has 4750 joules of heat energy is added to it. if the temperature increases from 20

GREYUIT [131]

E=mcθ
where E is the energy added,m is the mass,c is the specfic heat capacity and θ is the change in temprature.
making m subject u get E/cθ=m
θ=(30-20)=10
plugging the values we get :
 $\frac{4750}{1900*10}$

solving it we get the answer that is 0.25kg or 250 grams

8 0

2 years ago

Any object that is given any initial velocity and which follows a path due to gravitational force acting on it and by the fricti

amm1812

Any object that is given any initial velocity and which follows a path due to gravitational force acting on it and by the frictional resistance of the atmosphere is called a projectile. This is because the object is projected and not influenced by anything except gravity.

5 0

2 years ago

Shadow and eclipses result from​

Shadow and eclipses result from