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2 years ago

HELP ASAP

Physics

1 answer:

labwork [276]2 years ago

6 0

Answer:

Explanation:

A is summer, C is winter, B should be fall in-between those 2

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I REALLY NEED HELP PLEASE!!! CAN SOMEONE ANSWER THIS? At 0 ºC, some amount of energy is required to change 1 kg of water from a

Nezavi [6.7K]

Answer:

C)It would require more energy to change solid water into liquid water because there are more molecules in this larger piece of ice.

8 0

2 years ago

Read 2 more answers

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

Anton [14]

Answer:

91.64 km

91.64 km high material would go on earth if it were ejected with the same speed as on Io.

Explanation:

According to Newton Law of gravitation:

$g=\frac{Gm}{r^2}$

Where:

G is gravitational constant= $6.67*10^{-11} m^3/kg.s^2$

For Moon lo g is:

$g_M=\frac{6.67*10^{-11}*8.93*10^{22}}{(1821*10^3)^2m^2} \\g_M=1.7962 m/s^2$

According to law of conservation of energy

Initial Energy=Final Energy

$K.E_i+mgh_i=K.E_f+mgh_f$

$\frac{1}{2}m(v_0)^2+mgh_o= \frac{1}{2}m(v_f)^2+mgh_f\\At\ maximum\ height\ v_f=0\\\frac{1}{2}m(v_0)^2+0=mgh_f\\v_0=\sqrt{2gh_f}$

For Jupiter's moon Io:

Velocity is given by:

$v_0_M=\sqrt{2g_Mh_f_M}$

For Earth Velocity is given by:

$v_0_E=\sqrt{2g_Eh_f_E}$

Now:

$v_o_M=v_o_E$

$\sqrt{2g_Mh_f_M}=\sqrt{2g_Eh_f_E}\\h_f_E=\frac{g_Mh_f_M}{g_E}$

$g_E=9.8 m/s^2$

$g_m=1.7962 m/s^2, As\ Calculated\ above$

$h_f_E=\frac{1.7962*500*10^3m}{9.8} \\h_f_E=91642.85 m\\h_f_E=91.64Km$

91.64 km high material would go on earth if it were ejected with the same speed as on Io.

8 0

2 years ago

A net force of 100 N is moving a mass with an acceleration of 5 m/s2. What is the mass of the object?

olya-2409 [2.1K]

Fnet=ma so m = fnet/a 100/5=20 the mass is 20

7 0

2 years ago

In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs

bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

$V=\frac{I}{nAq}$

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

$\frac{I}{q} = 1.2*10^{18}$

$A= 1.3*10^{-8}m^2$

$n=6.3*10^{28} e/m^3$

$\omicron{O} = 1.2*10^{-4}(m/s)(N/c)$ Mobility

We can find the drift velocity replacing,

$V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}$

$V= 1.465*10^-3m/s$

The electric field is given by,

$E= \frac{V}{\omicron{O}}$

$E=\frac{1.465*10^-3}{1.2*10^{-4}}$

$E=12.2V/m$

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2 years ago

1. A man travels 45 m in 35 seconds. How fast is that?<br> What’s the answer

Natalija [7]

Answer:

1.28

Explanation:

If you want to find the m/s you would divide distance by time, so

45 divided by 35 would equal 1.28571429 and so on.

you can just write the three first numbers.

8 0

3 years ago