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Rama09 [41]
2 years ago
7

You slide across a icy sidewalk and you eventually come to a stop according to newtons first law why are you no longer in motion


a. you are subject to unbalanced forces
b. there’s too much air resistance
c. you are subject to balanced forces
d. inertia forced you to stop

SOMEONE PLS HELP!!:(
Physics
2 answers:
katrin2010 [14]2 years ago
7 0

Answer:

the awnser is B

Explanation:

the awnser is B

Natali5045456 [20]2 years ago
6 0

Answer:

the awnser is B

Explanation:

the awnser is B

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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
3 years ago
A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total accelera
AlekseyPX

Answer:

Explanation:

Given

Radius of bicycle wheel r=0.3\ m

Initial angular velocity \omega _0=0

It rotates 3 revolution in 5 s therefore

\omega =2\pi 3=\6\pi =18.85\ rad/s

using \omega =\omega _0+\alpha t

where \alpha =angular\ acceleration

\omega =Final\ angular\ velocity

t=time

\alpha =\frac{18.85}{5}=3.77 rad/s^2

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

\omega at t=1

\omega =0+3.77\times 1=3.77 rad/s

a_c=\omega ^2\cdot r

a_c=(3.77)^2\cdot 0.3=4.26 m/s^2

Tangential acceleration a_t=\alpha \times r

a_t=3.77\times 0.3=1.13 m/s^2

a_{net}=\sqrt{a_t^2+a_c^2}

a_{net}=\sqrt{(1.13)^2+(4.26)^2}

a_{net}=4.41 m/s^2

                       

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3 years ago
How is the medium of a wave and transmit of a wave similar and different?
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Answer:

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I'm assuming we're applying the standard Integral form of the calculation of work. The solution is provided in the image.  

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What is the difference between regional metamorphism and contact metamorphism?
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