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3 years ago

How might the temperatures on Mercury be different if it had the same mass as Earth?

Physics

1 answer:

Reil [10]3 years ago

4 0

Answer:

well, mercury is the closest planet to the sun, making it MUCH hotter than earth. that being said, just because mercury is closest to the sun, it doesnt mean its the hottest planet.

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A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.5 m>s. Two seconds later

ivanzaharov [21]

Answer:

4.28 s

Explanation:

after two seconds (2 s) His friends is

d = 3.5 m/s x 2 s = 7 meter ahead.

in this state, a bicylist start from initial velocity vo = 0 m/s and accelerat 2.4 m/s²

then, when bicylist reach his friend

t friend = t bicyclist = t

d bicylist = d friend + d

-------

d friend = 3.5 . t

d bicylist = vo . t + ½ a t²

d friend + d = vo . t + ½ a t²

3.5 t + 7 = 0 . t + ½ . 2.4 . t²

3.5 t + 7 = 1.2 t²

0 = 1.2 t² - 3.5 t - 7

t = -1.363 and t = 4.28

take the positive one

6 0

3 years ago

What is the relationship between elevation and energy

jasenka [17]

The Energy related to object motion.Potential energy most commonly refers to gravitational potential energy, which is energy associated with an object's elevation. For an object of mass m and elevation z, its potential energy is mgz where g is the gravitational constant .............. HOPE IT HELPS YOU :)

7 0

3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant