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ohaa [14]
2 years ago
7

The idea that the Earth sits motionless in the Universe at the center of a revolving globe of stars, with the Moon and planets i

n orbit around the Earth, is the ___________ model of the Universe.
Physics
1 answer:
andrew-mc [135]2 years ago
6 0
Earth sits motionless in the universe at the center of a revolving globe of starts , with the moon and planets in orbit around the earth, is the surrounding model of the uninverse
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Sebutkan aplikasi gerak osilasi fisis
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6 0
3 years ago
Read 2 more answers
2. A rock is dropped from rest at the top of a 10 meter tall building. How much time does the rock
ivann1987 [24]

Answer:

1.4 seconds

Explanation:

here u= 0

v= ?

h= 10m

g= 9.8 m/ s*2

using 2as = v^2 - u ^2

2x9.8x10 = v^2 - 0

v^2 = 196

v = 14 m

time = v- u/ a

14 / 98

1.4 seconds

7 0
3 years ago
Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
Goryan [66]

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

4 0
3 years ago
A 4watt night light is left on for 8 hours each night. how much work does it do per night
nlexa [21]

Power = (work) / (time)

Work = (power) x (time)

          = (4 watts) x (8 hours)

1 "watt" = 1 joule/sec .

Work = (4 joule/sec) x (8 hours) x (3,600 sec/hr)

         = (4 x 8 x 3,600) joules  =  1,024 joules .

==================================

Another way:

4 watts = 0.004 kilowatt

Work = energy = (power) x (time)

                        = (0.004 kw) x (8 hrs)  =  0.032 kilowatt-hour .

7 0
3 years ago
Mass m = 0.1 kg moves to the right with speed v = 0.31 m/s and collides with an equal mass initially at rest. After this inelast
nataly862011 [7]

Answer:

VR = 0.26 m/s

Explanation:

m = 0.1 kg                                M = 0.1 kg

v = 0.31 m/s                              Vi = 0 m/s    

the kinetic energy of the system initially is:

Ki = 1/2×m×(v^2) + 1/2×M×(Vi)^2

   = 1/2×(0.1)×(0.31)^2

   = 4.805×10^-3 J

then, we told that the system after collision only retains a fraction 0.69 of its initial kinetic energy. that is the final kinetic energy of the system is:

Kf = 0.69×4.805×10^-3 J

    = 3.31545×10^-3 J

but due equal masses of the bodies, we know that after the collision the only body that would be in motion would be the body that at res and the body that was initially moving will now be at rest.so the kinetic energy is only made by the second body and given by:

                   Kf = 1/2×M×(VR)^2

3.31545×10^-3 = 1/2×(0.1)×(VR)^2

               VR^2 =  0.066309

                   VR = 0.26 m/s

according to the coservation of linear momentum:

6 0
4 years ago
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