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3 years ago

What is the mass of a bird if it has a potential energy of 147J when flying at a height of 30 m above the ground?

Physics

1 answer:

Artist 52 [7]3 years ago

5 0

Answer:

0.5kg

Explanation:

Given parameters:

Potential energy = 147J

Height = 30m

Unknown:

Mass of the bird = ?

Solution:

Potential energy is the energy due to the position of a body. Now, the expression for finding the potential energy is given as;

P.E = mgH

m is the mass

g is the acceleration due to gravity = 9.8m/s²

H is the height

147 = m x 9.8 x 30

m = 0.5kg

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A -5.0 μC charge experiences a 11 i^ N electric force in a certain electric field. [Recall that i^ is a unit vector in the x-dir

Pachacha [2.7K]

Answer:

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Explanation:

given,

charge = -5.0 μC

Electric force, F = 11 i^ N

force would a proton experience = ?

we know

$\vec{F} = q \vec{E}$

$11\hat{i} = -5 \times 10^{-6} \vec{E}$

$\vec{E} =-2.2 \times 10^{6}\hat{i}$

we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

$\vec{F} = q \vec{E}$

$\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}$

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Force experienced by the photon in the same field is equal to $\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

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2 years ago

A laser beam is incident at an angle of 30.0° from the vertical onto a solution of corn syrup in water. The beam is refracted to

dimaraw [331]

Answer with Explanation:

We are given that

Angle of incidence, $i=30^{\circ}$

Angle of refraction, $r=19.24^{\circ}$

a.Refractive index of air, $n_1=1$

We know that

$n_2sinr=n_1sini$

$n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517$

b.Wavelength of red light in vacuum, $\lambda=632.8nm=632.8\times 10^{-9} m$

$1nm=10^{-9} m$

Wavelength in the solution, $\lambda'=\frac{\lambda}{n_2}$

$\lambda'=\frac{632.8}{1.517}=417nm$

c.Frequency does not change .It remains same in vacuum and solution.

Frequency, $\nu=\frac{c}{\lamda}=\frac{3\times 10^8}{632.8\times 10^{-9}}$

Where $c=3\times 10^8 m/s$

Frequency, $\nu=4.74\times 10^{14}Hz$

d.Speed in the solution, $v=\frac{c}{n_2}$

$v=\frac{3\times 10^8}{1.517}=1.98\times 10^8m/s$

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3 years ago

A geosynchronous satellite moves in a circular orbit around the Earth and completes one circle in the same time T during which t

andreyandreev [35.5K]

Answer:

Explanation:

The time period of geosynchronous satellite must be equal to T .

The radius of its orbit will be ( R+ h )

orbital velocity V₀ = $\sqrt{\frac{GM}{( R+h)} }$

Time period T = 2π( R + h ) / V₀

= 2π( R + h ) x $\sqrt{\frac{( R+h)}{GM } }$

$\frac{T^\frac{2}{3}(GM)^\frac{1}{3} }{(2\pi )^\frac{2}{3} }$ = R +h

h = $\frac{T^\frac{2}{3}(GM)^\frac{1}{3} }{(2\pi )^\frac{2}{3} }$ - R.

7 0

3 years ago

Read 2 more answers

Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de

marshall27 [118]

Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos:

Altura o distancia recorrida: 40 m
Vo: 6 m/s
Aceleración de la gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes.

Entonces tenemos que:

$Vf ^{2} -Vo ^{2} =2 x g x h$

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo tenemos que:

$Vf x^{2} =(6 m/s) ^{2} +(2)x(9,81 m/s ^{2})x(40m) \\ Vf ^{2} =820,8 m^{2} /s ^{2} \\ \sqrt{Vf ^{2} } = \sqrt{820,8m^{2}/s ^{2} } \\ Vf=28,64 m/s$

Que tengas un buen día!

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3 years ago

a ball moves at a constant speed of 1/2 meter per second. How much time does it take the ball to move 1 meter?

lesya [120]

Answer:

2 seconds

Explanation:

if a ball travels 1/2 meter per second, and there's 2 halfs in a whole, 1/2 meter per second x 2 halfs in a whole meter is 2 seconds to travel a meter

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3 years ago