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3 years ago

15

If earth had no atmosphere, would a falling object ever reach terminal velocity?

1 answer:

stepan [7]3 years ago

5 0

No, because terminal velocity is when the acceleration of the Earth’s gravity is balanced by the air resistance of the atmosphere.

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Un ciclista recorre 10 km en 2 h.Calcula su velocidad media en km/h.¿cuantos metros recorre cada segundo

lys-0071 [83]

Answer:

La velocidad media es 5 $\frac{km}{h}$ , que equivale a 1.389 $\frac{m}{s}$

Explanation:

La velocidad es una magnitud física que expresa la relación entre el espacio recorrido por un objeto y el tiempo empleado para ello.

La velocidad media relaciona el cambio de la posición con el tiempo empleado en efectuar dicho cambio. Por lo que se calcula como la distancia recorrida por un objeto dividido por el tiempo transcurrido:

$velocidad=\frac{distancia}{tiempo}$

En este caso:

distancia= 10 km= 10,000 m (siendo 1 km= 1,000 m)
tiempo= 2 h= 7,200 s (siendo 1 h= 3,600 s)

Entonces, reemplazando en la definición de velocidad media:

$velocidad=\frac{10 km}{2 h}= \frac{10,000 m}{7,200 s}$

Resolviendo se obtiene:

$velocidad=5 \frac{km}{h}= 1.389\frac{m}{s}$

<u><em>La velocidad media es 5 </em></u> $\frac{km}{h}$ <u><em>, que equivale a 1.389 </em></u> $\frac{m}{s}$ <u><em></em></u>

4 0

2 years ago

When a pair of 10-n forces act on a box of candy, the net force on the box is?

Grace [21]

Any of the above depending on the direction of forces

8 0

2 years ago

Read 2 more answers

Osteoporosis is most likely to be affected by which cycle?

oksano4ka [1.4K]

<span>Its the phosphorus cycle</span>

8 0

3 years ago

The microwaves in a microwave oven are produced in a special tube called a magnetron. The electrons orbit the magnetic field at

earnstyle [38]

Answer:

The magnetic field strength is 0.086 T

The maximum kinetic energy is $1.62 \times 10^{-14}$ J

Explanation:

Given:

Frequency $f = 2.4 \times 10^{9}$ Hz

(A)

The magnetic field related to cyclotron is given by,

$B = \frac{2\pi mf}{e}$

Where $m =$ mass of electron $= 9.1 \times 10^{-31 }$ kg, $e = 1.6 \times 10^{-19}$ C

$B = \frac{6.28 \times 9.1 \times 10^{-31 }\times 2.4 \times 10^{9} }{1.6 \times 10^{-19} }$

$B = 0.086$ T

Therefore, the magnetic field strength is 0.086 T

(B)

Diameter of orbit $d= 2.5 \times 10^{-2}$ m

Radius of orbit $r = 1.25 \times 10^{-2}$ m

The maximum kinetic energy is given by,

KE = $\frac{1}{2} mv^{2}$

Where $v = r \omega = r 2\pi f$

KE = $\frac{1}{2} m(2\pi fr)^{2}$

KE = $\frac{1}{2} \times 9.1 \times 10^{-31 } \times (2 \times 3.14 \times 2.4 \times 10^{9} \times 1.25 \times 10^{-2} )^{2}$

KE = $1.62 \times 10^{-14}$ J

Therefore, the maximum kinetic energy is $1.62 \times 10^{-14}$ J

7 0

3 years ago

A 1.5kg mass attached to an ideal spring oscillates horizontally with an amplitude of 0.50m. The spring constant is 85 n/m. what

Stolb23 [73]

Answer:

frequency of the mass motion $f$ ≅ 1.2 Hz

Explanation:

Given that:

Mass m = 1.5 kg

Amplitude A = 0.50 m

Spring constant k = 85 n/m

The frequency can be calculated by using the formula:

$f = \frac{1}{2 \pi} \sqrt{\frac{k}{m} }$

$f = \frac{1}{2 \pi} \sqrt{\frac{85}{1.5} }$

$f = 1.198$ Hz

frequency of the mass motion $f$ ≅ 1.2 Hz

5 0

3 years ago