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3 years ago

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What did newton's law of universal gravitation tell us about how gravity works?

1 answer:

vazorg [7]3 years ago

6 0

The force of gravity form the Sun will be stronger on a n object with more mass

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Why potential energy become equal to kinetic energy at height

Gennadij [26K]

Answer:

because potentil energy is redy to go but its bound up

And kinetic energy is in motion

Explanation:

7 0

3 years ago

Read 2 more answers

Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at

klasskru [66]

Answer:

b) twice the energy of each photon of the red light.

Explanation:

$\lambda$ = Wavelength

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

Energy of a photon is given by

$E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}$

Let $\lambda_1$ = 700 nm

$\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}$

For red light

$E_1=\dfrac{hc}{\lambda_1}$

For UV light

$E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}$

Dividing the equations

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1$

Hence, the answer is b) twice the energy of each photon of the red light.

7 0

3 years ago

Read 2 more answers

Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of

mylen [45]

Answer:

density = 5520 kg/m^3

Explanation:

given that

radius of earth = 6378 km

G = 6.67 x 10⁻¹¹ m³/kg.s²

g = 9.80 m/s²

we know,

$g = \dfrac{GM}{r^2}$

mass of earth

$M = \dfrac{gr^2}{G}$

$M = \dfrac{9.8 \times (6378 \times 10^3)^2}{6.67 \times 10^{-11}}$

M = 5.972 x 10²⁴ kg

density = $\dfrac{mass}{volume}$

V = volume of the earth = 4/3πr³

V = 4/3 x 3.14 x (6378 x 10³)³

V = 1.08 x 10²¹ m³

density = $\dfrac{5.972\times 10^{24}}{1.08\times 10^{21}}$

density = 5.52 x 10³ kg/m^3

density = 5520 kg/m^3

8 0

3 years ago

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the drag force?

timurjin [86]

Answer: The drag force goes up by a factor of 4

Explanation:

The <u>Drag Force</u> equation is:

$F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2}$ (1)

Where:

$F_{D}$ is the Drag Force

$C_{D}$ is the Drag coefficient, which depends on the material

$\rho$ is the density of the fluid where the bicycle is moving (<u>air in this case) </u>

$A_{D}$ is the transversal area of the body or object

$V$ the bicycle's velocity

Now, if we assume $C_{D}$ , $\rho$ and $A_{D}$ do not change, we can rewrite (1) as:

$F_{D}=C.V^{2}$ (2)

Where $C$ groups all these coefficients.

So, if we have a new velocity $V_{n}$ , which is the double of the former velocity:

$V_{n}=2V$ (3)

Equation (2) is written as:

$F_{D}=C.V_{n}^{2}=C.(2V)^{2}$

$F_{D}=4CV^{2}$ (4)

Comparing (2) and (4) we can conclude<u> the Drag force is four times greater when the speed is doubled.</u>

7 0

3 years ago

What are some practical applications for determining the motion of an object?

Mama L [17]

- rocket science
- automotive research
- space research

4 0

3 years ago