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3 years ago

Question 1

Physics

1 answer:

m_a_m_a [10]3 years ago

3 0

Answer:

Gravity

Explanation:

Pretty self-explanatory

We are kept on earth because of gravity and it keeps planets orbiting around the sun.

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A satellite is in a circular Earth orbit of radius r. The area A enclosed by the orbit depends on r2 because A = πr2. Determine

Salsk061 [2.6K]

Answer:

a. $T=r^{3/2}$

b. $K=\frac{1}{r}$

c. $v=\frac{1}{\sqrt{r}}$

d. $v=\sqrt{r}$

Explanation:

To make analysis about the satellite circular earth the depends or r and A

$T^2=\frac{4\pi}{GM}*r^3$

$K=\frac{GM*m}{2*r}$

$T^2=r^3$

$T=r^{3/2}$

$K=\frac{GM*m}{2}*\frac{1}{r}$

$K=\frac{1}{r}$

$K=\frac{1}{2}*m*v^2$

$v^2=\frac{2*K}{m}=\frac{2*Gm*m}{2*m*r}$

$v=\frac{1}{\sqrt{r}}$

$v=\sqrt{2*GMr}$

$v=\sqrt{r}$

3 0

3 years ago

In part one of this experiment, a 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the s

statuscvo [17]

To solve this problem it is necessary to apply the concepts related to Hooke's Law as well as Newton's second law.

By definition we know that Newton's second law is defined as

$F = ma$

m = mass

a = Acceleration

By Hooke's law force is described as

$F = k\Delta x$

Here,

k = Gravitational constant

x = Displacement

To develop this problem it is necessary to consider the two cases that give us concerning the elongation of the body.

The force to keep in balance must be preserved, so the force by the weight stipulated in Newton's second law and the force by Hooke's elongation are equal, so

$k\Delta x = mg$

So for state 1 we have that with 0.2kg there is an elongation of 9.5cm

$k (9.5-l)=0.2*g$

$k (9.5-l)=0.2*9.8$

For state 2 we have that with 1Kg there is an elongation of 12cm

$k (12-l)= 1*g$

$k (12-l)= 1*9.8$

We have two equations with two unknowns therefore solving for both,

$k = 3.136N/cm$

$l = 8.877cm$

In this way converting the units,

$k = 3.136N/cm(\frac{100cm}{1m})$

$k = 313.6N/m$

Therefore the spring constant is 313.6N/m

3 0

3 years ago

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:

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

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

4 0

3 years ago

9.Water flows'along a horizontal pipe of cross sectional area 48cm2 which has a constriction of

Kamila [148]

Answer:

v₁ = 9 m/s

Explanation:

Here, we can use the continuity equation, as follows:

$A_1v_1 = A_2v_2$

where,

A₁ = Cross-sectional area of the wider section = 48 cm²

A₂ = Cross-sectional area of the constriction = 12 cm²

v₁ = speed of flow in wider section = ?

v₂ = speed flow in constriction = 36 m/s

Therefore,

$(48\ cm^2)v_1 = (12\ cm^2)(36\ m/s)\\\\v_1 = \frac{(12\ cm^2)(36\ m/s)}{(48\ cm^2)} \\\\$

v₁ = 9 m/s

3 0

3 years ago

Explain what happens to an atom when protons are added or removed from its nucleus

dsp73

Neutrons do not carry an electrical charge so adding or removing them from the nucleus does not change the electrical charge of the nucleus. ... So, adding or removing protons from the nucleus changes what element that atom is! For example, adding a proton to the nucleus of an atom of hydrogen creates an atom of helium.

4 0

4 years ago