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3 years ago

Shadow and eclipses result from

Physics

2 answers:

Basile [38]3 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]3 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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Which of the following would be a good question that could be scientifically investigated?

Nataliya [291]

Do cherry popsicles freeze slower than orange popsicles

5 0

3 years ago

Read 2 more answers

A light bulb has a voltage of 36 and a current of 8 A. Calculate the resistance of the light bulb

timurjin [86]

R = U : I. U is in Voltage and I is in Ampère. That gives you R = 36 : 8 = 4,5 Ohm

4 0

3 years ago

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

$\tau=(25 N)(0.5 m)=12.5 Nm$

4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity ( $\omega$ ) is:

$L=I\omega$

In this problem, we have

$I=0.007875 kgm^2$

$\omega=6.28 rad/s$

So, the angular momentum is

$L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s$

6 0

3 years ago

During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 *

marta [7]

Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s

To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s

Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m

This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.

But anyways, good luck!

4 0

3 years ago

Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength

soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

$E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J$

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

$V = \sqrt{\frac{2K.E}{m} }\\\\V = \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5} \ m/s$

the shortest de Broglie wavelength for the electrons is given by;

$\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm$

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

7 0

4 years ago

Shadow and eclipses result from​

Shadow and eclipses result from