For purposes of completing our calculations, we're going to assume that
the experiment takes place on or near the surface of the Earth.
The acceleration of gravity on Earth is about 9.8 m/s², directed toward the
center of the planet. That means that the downward speed of a falling object
increases by 9.8 m/s for every second that it falls.
3 seconds after being dropped, a stone is falling at (3 x 9.8) = 29.4 m/s.
That's the vertical component of its velocity. The horizontal component is
the same as it was at the instant of the drop, provided there is no horizontal
force on the stone during its fall.
Answer:
D
Explanation:
it could be the yellow cloth but we see are shadow on the sidewalk most of the time and shadows are made from reflecting light so your answer should be D
Answer:
448 J/kg/°C
Explanation:
m₁ C₁ (T₁ − T) + m₂ C₂ (T₂ − T) = 0
(0.0414 kg) C (243°C − 20.4°C) + (0.411 kg) (4186 J/kg/°C) (18°C − 20.4°C) = 0
(9.22 kg°C) C − 4129 J = 0
C = 448 J/kg/°C
Answer:
a) v2=4147.72 m/s
b) stotal=5.53x10^6 m
Explanation:
a) the length from the center of the earth is equal to:
L1=1x10^6+((6.37/2)x10^6)=4.18x10^6 m
the velocity is 5.14 km/s=5.14x10^3 m/s
the farthest distance is equal to:
L2=2x10^6+((6.37/2)x10^6)=5.18x10^6 m
As the angular momentum is conserved, we have to:
I1=I2
m*L1*v1=m*L2*V2, where m is the mass of satelite
clearing v2:
v2=(L1*V1)/L2=(4.18x10^6*5.14x10^3)/5.18x10^6=4147.72 m/s
b) Using the Newton 3rd law:
vf^2=vi^2+2as
where:
a=g=9.8 m/s^2
vf=0
vi=5.14 km/s
s=?
Clearing s:
s=(vf^2-vi^2)/(2g)=((0-(5.14x10^3)^2)/(2*9.8)=1.35x10^6 m
the total distance is equal to:
stotal=s+L1=1.35x10^6+4.18x10^6=5.53x10^6 m
