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3 years ago

27. Improvements in space technology have greatly helped scientists better understand the stars, planets, and other

Physics

1 answer:

stellarik [79]3 years ago

8 0

Answer:

Explanation:

Technological advancements in astronomy have led to the development of thousands of products such as

new materials, medical devices, and communications satellites.

For example, the production of instruments like telescopes, microscopes, and other measuring devices now enables us to identify things that we couldn't examine with the naked eye.

Similarly, the development of communication satellites has led to the emergence and improvements in the mobile network industry.

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A 20 kg box has 1000 joules of potential energy on a shelf. How high is the shelf?

navik [9.2K]

Formula: PE = mgh
m = 20 kg, g = 9.8 m/s^2 h= ?
1000 = 20 * 9.8 * h
1000 = 196h
h = 5.10204082
The height is around 5m

5 0

2 years ago

Link between mass and force of gravity?

Leviafan [203]

Mass x Force of gravity = Weight

Weight is a force.

On Earth, the force due to gravity is ~9.81 N/Kg. So an object of 100 Kg will have a weight of ~981 N

8 0

3 years ago

Read 2 more answers

relationship beetween image distance,focal length ,power magnification, for both concave and convex lens.

scoray [572]

Answer:

Object distance means what is the distance between pole and object. Image distance means when image is formed then the distance between pole and image is called image distance. Focal length is the distance between pole and the principal focus of the mirror.

A lens is a clear object, usually made of glass or plastic, which is used to refract, or bend light. Lenses can concentrate light rays (bring them together) or spread them out. Common examples of lenses include camera lenses, telescope lenses, eyeglasses, and magnifying glasses. Lenses are often double lenses, meaning they have two curved sides. A convex lens is rounded outward, while a concave lens curves inward. (A great way to remember this is that a concave lens creates an indent like a cave!)

The image distance can be calculated with the knowledge of object distance and focal length with the help of lens formula. In optics, the relationship between the distance of an image (i), the distance of an object (o), and the focal length (f) of the lens are given by the formula known as Lens formula. Lens formula is applicable for convex as well as concave lenses. These lenses have negligible thickness. It is an equation that relates the focal length, image distance, and object distance for a spherical mirror. It is given as,

1/i + 1/o = 1/f

i= distance of the image from the lens

o= distance of the object from the lens

f= focal length of the lens

Explanation:

Hope it is helpful....

7 0

2 years ago

Read 2 more answers

A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle

NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area, $A=0.344 m^2$

Drag coefficient, $C_d=0.88$

Coefficient of rolling resistance, $\mu_r=0.007$

Friction force, $f=\mu_rmg=0.007\times 60\times 9.8=4.1 N$

Where $g=9.8 m/s^2$

Let speed of cyclist=v'

Drag force, $F_d=\frac{1}{2}\rho_{air}AC_dv^2$

Density of air, $\rho_{air}=1.225 kg/m^3$

$F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N$

Power,P= $(F_d+f)\times v'$

$340=(4.1+4.635) v'=8.735v'$

$v'=\frac{340}{8.735}=38.9m/s$

$v'=87.1 mph$

1 m=0.00062137 miles

1 hour=3600 s

7 0

3 years ago

La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.

MA_775_DIABLO [31]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

6 0

2 years ago