Seeds are often found on which part of a gymnosperm? branch leaf cone stem

Consider a stone and a football both at rest and having the same mass. why is it painful to kick the stone than to kick the foot

Kryger [21]

Answer:

stone with lesser mass can not go far because of firm structure but football that has greatet mass can go farther because of the structure that makes it move further when a force was applied to it

6 0

2 years ago

A scientist hypothesizes that the temperature at which an turtle's egg is incubated

just olya [345]

Answer:

thw temperature of the male will be higher than that of the female.

6 0

3 years ago

When a pair of identical resistors are con- nected in series, the

Monica [59]

All three choices are correct.

7 0

3 years ago

Un engrane que gira con una velocidad de 20 rad/s, es acelerado durante 5 segundos hasta alcanzar una velocidad de 35 rad/s

larisa [96]

Answer:

a) La aceleración angular es: $\alpha=2\: rad/s^{2}$

b) El engranaje gira 125 radianes.

c) El engranaje hara aproximadamente 20 revoluciones.

Explanation:

a)

La aceleración angular se define como:

$\alpha=\frac{\Delta \omega}{\Delta t}$

Donde:

Δω es la diferencia de velocidad angular (en otras palabras ω(final)-ω(inicial))
Δt es el tiempo en el que occure el cambio de velocidad angular

$\alpha=\frac{35-25}{5}$

$\alpha=2\: rad/s^{2}$

b)

El desplazamiento angular puede ser calculado usando la siguiente ecuación:

$\theta=\theta_{i}+\omega_{i}t+\frac{1}{2}\alpha t^{2}$

Aqui el angulo inicial es 0, por lo tanto.

$\theta=20(5)+\frac{1}{2}(2)(5)^{2}$

$\theta=125\: rad$

El engranaje gira 125 radianes.

c)

Lo que debemos hacer aquí es convertir radianes a revoluciones.

Recordemos que 2π rad = 1 rev

Entonces:

$\theta=125\: rad \times \frac{1\: rev}{2\pi\: rad}=19.89\: rev$

Por lo tanto el engranaje hara aproximadamente 20 revoluciones.

Espero te haya sido de ayuda!

6 0

2 years ago

The Hubble Space Telescope has a mass of 1.16*10^ 4 kg and orbits the Earth at an altitude of 5.68 * 10 ^ 5 above Earth's surfac

andrezito [222]

Answer:

$E=8.13\times 10^{12}\ J$

Explanation:

Given that,

The mass of a Hubble Space Telescope, $m_1=1.16\times 10^4\ kg$

It orbits the Earth at an altitude of $5.68\times 10^5\ m$

We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

$E=\dfrac{Gm_1m_e}{r}$

Where

$m_e$ is the mass of Earth

Put all the values,

$E=\dfrac{6.67\times 10^{-11}\times 1.16\times 10^4\times 5.97\times 10^{24}}{5.68\times 10^5}\\\\E=8.13\times 10^{12}\ J$

So, the potential energy of the telescope is $8.13\times 10^{12}\ J$ .

5 0

2 years ago